Solutions Manual to Basics of Analytical and Chemical Equilibria - Brian M. Tissue

Chapter 1: End-of-Chapter Solutions1. In order from increasing to decreasing precision: pipet (2 mm i.d.) - most precise buret (1 cm i.d.) graduated cylinder (2.5 cm i.d.) - least precise The precision of the volume measurement increases as the diameter of the glassware gets smaller. 2. a) %-RSD = b) %-RSD = c) %-RSD = 0.02 mL = 0.2 % ⇒ 0.00500 ± 0.00001 M 10.0 mL 0.1 mL = 1 % ⇒ 0.00500 ± 0.00005 M 10.0 mL 1 mL = 10 % ⇒ 0.0050 ± 0.0005 M 10.0 mL 3. The calculations were very precise, they have to be since space travel is difficult. Unfortunately they were inaccurate due to a gross error in calculations that failed to convert values in metric units (newtons) with values in Imperial units (pound-force). 4. The one-point calibration with pH=7 buffer is rapid and will be accurate for pH measurements near pH 7. The disadvantage of the one-point calibration is that measurements at low pH or high pH could be erroneous due to an incorrect slope in the calibration function of the pH meter. The two-point calibration is more time consuming, but will provide more accurate data over the full range of the pH meter. Figure 1.5 shows an example of the error introduced by extrapolating a one-point calibration. Brian M. Tissue, Basics of Analytical Chemistry and Chemical Equilibria, (J. Wiley, New York, 2013). 5. Plots of Table 1.11 data forcing trendline through zero. y = 0.0344x 0.03 Absorbance diamonds: 266 nm squares: 440 nm 0.04 0.02 Note that the blank for the 266 nm data is obscured by the square marker in both charts. 0.01 Plots of Table 1.11 data as given. 0.04 diamonds: 266 nm squares: 440 nm 0.03 y = 0.0117x 0 0 0.2 0.4 0.6 0.8 1 1.2 Concentration (μM) Absorbance y = 0.0321x + 0.002 0.02 y = 0.0118x - 4E-05 0.01 0 0 0.2 0.4 0.6 0.8 1 1.2 Concentration (μM) Sensitivities: zero intercept non-zero intercept 266 nm 0.034 μM−1 0.032 μM−1 440 nm 0.012 μM−1 0.012 μM−1 There is a 6-% difference in the slopes at 266 nm depending on whether or not the trendline is forced through zero. Looking at the data, the 1.2 μM point appears low compared to the trendline. The scatter in the 440 nm data points appears random. The take-home message is that calibration curves should be constructed from at least 5 or 6 measurements. Brian M. Tissue, Basics of Analytical Chemistry and Chemical Equilibria, (J. Wiley, New York, 2013). 6. The individual volume measurements give a mean and standard deviation of: 1.022 ± 0.007 mL. The density is: 2.3291 g = 2.278 g/mL 1.022 mL The RSD of the volume measurements is: 0.007 mL = 0.0071 1.022 mL The final result should have the same RSD (2.278 g/mL × 0.0071 = 0.016 g/mL) to give: density = 2.278 ± 0.016 g/mL (2.28 ± 0.02 g/mL is also correct) 7. a) For either primary standard we need (0.1000 L)(0.150 M) = 0.0150 mol of Na. • For NaCl (f.w. = 58.443 g/mol): (58.443 g/mol)(0.0150 mol Na) = 0.8766 g NaCl • For Na2C2O4 (f.w. = 133.998 g/mol): 1 mol Na2C2O4 133.998 g/mol 0.0150 mol Na = 1.0050 g Na2C2O4 2 mol Na (note an additional significant figure for the higher formula weight standard) b) We need (0.05000 L)(0.0100 mol/L)(58.443 g/mol) = 0.0292 g Na. Since we can weigh to 0.0001 g, the uncertainty is 0.0001 g × 100 % = 0.3 % 0.0292 g c) We need (1.000 L)(0.500 M)(58.443 g/mol) = 29.2215 g of Na. Weighing will not be a limiting factor, so the volume measurement, 0.1 %, is the largest source of uncertainty. d) A weight measurement equivalent to 0.1 % give that the balance weighs to 0.0001 g is: 0.0001 g 0.1 % = × 100 % xg x = 0.1000 g For NaCl (f.w. = 58.443 g/mol): 0.0001 g = 1.711 mmol 58.443 g/mol The volume needed to make 0.1 mM is: 1.711 mmol 0.100 mM = xL x = 17.11 L This result is a large volume, which shows why common lab practice is to make a concentrated stock solutions that can be diluted to the desired concentrations. Brian M. Tissue, Basics of Analytical Chemistry and Chemical Equilibria, (J. Wiley, New York, 2013). 77 79.ration steps that will be necessary for such a complicated sample matrix. (J. using an internal standard will be the preferred method of calibration. Any matrix effects should affect the standard addition equally to the matrix effects of the unknown amount of Pb in the test solutions.050 mL.021 g)/(0. 78. 2013).05 mL × 100 % = 0. The actual volume delivered by the “10-mL” pipet is (10. c) In addition to the usual blanks.99707 g/mL) = 10. (a) Your spreadsheet will look something like: Data: mean = std dev = %-RSD = std err = 95% C.1 M nitric acid should be sufficient.182*C9/SQRT(4) Notes: Brian M.386753 =3. New York. 10. The error is 0.121544 0.308379 =(C9/C8)*100 0. 11. b) “Field” blanks and field spikes are necessary due to the large number of sample prepa. so a calibration curve using standards of lead in 0. Basics of Analytical Chemistry and Chemical Equilibria.93 78.5 %. . 10.8.050 mL 9. Using an internal standard requires some knowledge of the sample composition. a) There should be few if any interferences in drinking water. Wiley. An element similar to Pb that is not present in the sample should be chosen for the internal standard.243088 0.52 78. so blanks to check for contamination and a spike to check for detection limit should be sufficient. a) There should be few if any interferences in drinking water.05 mL or 0.I. Tissue. b) The complexity of the sample matrix warrants using the standard addition method for calibration. a “field” spike will be very important to determine if there is any loss of analyte.09 78.8275 0. c) Given the possible loss of analyte during sample processing. New York. Results 5 394. .5 0 1 2 3 1 2 3 Mean and standard deviation: 98. but it can also be reported as 98. (b) In the following example.• • • To understand the formulas copied to the right of the results. Basics of Analytical Chemistry and Chemical Equilibria.5 90 80 99 70 98.8 N= sum = mean = std dev = %-RSD = std err = 95% C.2.24.I.27 0. the formulas are written to accommodate up to 10 data points. the standard deviation of 0.26 Formulas =COUNT(G$5:G$14) =SUM(G$5:G$14) =I6/I5 =STDEV(G$5:G$14) =(I8/I7)*100 =I8/SQRT(I5) =2. It is not necessary to revise formulas if data is added to the set: Data: 78. cell C10 contains the standard deviation result and cell C9 has the mean. 2013).5 30 20 97 10 96.96 (an additional significant figure is gained due to the size of the sum of the data.11 78.4 ± 1.776*I8/SQRT(I5) 12.09 0. The mean can gain an additional significant figure due to the size of the sum of the data: 78.52 78.93 78. or 78. Brian M. (a) The left figure is Excel’s default scaling on the y axis.243 shows that extended significant figures in the result have little meaning.21 0. 100 100 99.83 ± 0.09 78. Tissue.5 60 50 98 40 97. The result can be reported as 78. Wiley. However.0).82 0.8 ± 0. Note how it exaggerates the differences in the values compared to the right plot that is scaled from 0 to 100.77 79.37 ± 0.828. (J. if you must divide by a sum by N. d) As an example. 2013).b) 100 98 96 94 92 90 88 86 84 82 80 1 2 3 4 5 6 7 8 Mean and standard deviation: 94.95 ± 0. but it can also be reported as 95. Basics of Analytical Chemistry and Chemical Equilibria. Wiley.0 ± 0. .8) c) Q (0. In the first case. New York. (J. so the values must be retained. In the second case.75 (an additional significant figure is gained due to the size of the sum of the data. the outlier is not very different from other data points. Brian M. it is more robust to use a spreadsheet’s COUNT function to get N rather than entering it numerically since it might change on adding or removing data points. Tissue. N is small and the criteria for rejection is high.11) is less than Qc for both cases.63 and 0. 495-0.0054 0.457 0.0034 -0.0006 -0.0084 -0.447 deviation -0.844E-03 0. Basics of Analytical Chemistry and Chemical Equilibria.50 Voltage 0. .5 on the yaxis.457) / (0.488E-03 8.30 0.00 1 2 3 4 5 6 7 Run # (c) Q= (outlier-closest) / (outlier-farthest) Q= (0.104E-05 2.456 ± 0.5: Data Set 1: 0.545E-04 3.444 0.890E-05 N (#): sum/N: average: 7 0.0124 0. a) Your spreadsheet might look something like the following screen capture: Data Set 1 Run # 1 2 3 4 5 6 7 Voltage 0. dev. 2013). Wiley.20 0.495-0.018 V 0.451 0. Tissue.10 0.947E-05 1.265E-07 7.4564 0.4564 sum d^2: std.13.40 0.4 to 0.448 0.453 0.745098 Qc=0. (J.444) Q= 0.0175 0. stdev: 1.495 0.57 Q > Qc so data point 6 may be discarded Brian M. New York.0094 deviation^2 1. exaggerating the difference in the values compared to plotting the y-scale from 0 to 0.0175 b) Note that the automatic scaling in spreadsheets will show approximately 0.176E-05 1.0386 -0. Q is 0.745, larger than Qc of 0.57 for seven data points. The value may be rejected, and the new average is: mean std. dev. 0.4500 V 0.0046 V 0.450 ± 0.005 V is also correct. 14. (a) 5.0 Data Set 2 Signal (V) 4.0 3.0 2.0 y = 0.0488x + 0.0118 R² = 0.9997 1.0 0.0 0 20 40 60 80 100 Conc. (ppb) 0.06 Data Set 2 Deviations Deviation (V) 0.04 0.02 0.00 -0.02 -0.04 -0.06 0 20 40 60 80 100 Conc. (ppb) Note that the randomness in the scatter of the residuals indicates a linear model is appropriate for this data. Brian M. Tissue, Basics of Analytical Chemistry and Chemical Equilibria, (J. Wiley, New York, 2013). (b) Below are two different ways to obtain this data: Conc. (ppb) 0.00 5.00 10.00 25.00 50.00 100.00 unknown signal: Conc. (ppb) 0.00 5.00 10.00 25.00 50.00 100.00 Signal (V) 0.002 0.259 0.489 1.284 2.407 4.903 deviation (d) -0.010 0.003 -0.011 0.052 -0.045 0.010 0.999 Signal (V) 0.002 0.259 0.489 1.284 2.407 4.903 sum(x)= sum(y)= 190.00 9.34 N= 6 avg(x)= avg(y)= 31.67 1.56 zero line 0 0 0 0 0 0 unknown conc: estimated uncertainty: LINEST results slope intercept 0.04881 0.01176 0.00042 0.01969 0.99971 0.03564 13568.93 4 17.23112 0.00508 20.2 ± 0.3 ppb ppb deviation (d) -0.010 0.003 -0.011 0.052 -0.045 0.010 x^2 0.00 25.00 100.00 625.00 2500.00 10000.00 x-d^2 1002.78 711.11 469.44 44.44 336.11 4669.44 y-d^2 2.4191 1.6857 1.1413 0.0747 0.7219 11.1935 (x-d)*(y-d) 49.25 34.62 23.15 1.82 15.58 228.62 sum(x^2)= 13250.00 Sxx= 7233.33 Syy= 17.2362 Sxy= 353.04 sy= 0.0356 0.0488 0.0118 std.dev. 0.0004 0.0197 R.S.D.(%) 0.86 167.44 ± 0.559 ppb ± 2.76 R.S.D.(%) m= b= (c) unknown: 0.999 V 20.227 Note that the units are not easy to show in the spreadsheets, they are: slope = 0.0488 ± 0.0004 V ppb−1 intercept = 0.0118 ± 0.0197 V Brian M. Tissue, Basics of Analytical Chemistry and Chemical Equilibria, (J. Wiley, New York, 2013). Chapter 2: End-of-Chapter Solutions 1. a) methane: gas, nonpolar b) octane: liquid, nonpolar c) water: liquid, polar d) beeswax: solid, nonpolar e) acetic acid: liquid, polar f) sodium acetate: solid, ionic, soluble in water g) K2CrO4: solid, ionic, soluble in water h) methyl tert-butyl ether (MTBE): liquid, moderate polararity i) ethanol: liquid, moderate polararity j) perchloric acid: liquid, polar (ionic in water) k) NaOH: solid, ionic, soluble in water l) KCl: solid, ionic, soluble in water m) CaCO3: solid, ionic, insoluble in water 2. a) dissolve aspirin tablets in water; filter through coffee filter; dry and weigh insoluble starch b) filter gravel from mixture with window screen, weigh either gravel or sand c) weigh water sample; allow water to evaporate; weigh remaining salt d) extract caffeine with organic solvent; allow solvent to evaporate and weigh remaining caffeine (repeat with SPE cleanup after extraction; drying down and weigh each fraction to determine if any other extractable components are present) 3. a) acetic acid - predominantly in water (αS(aq) = 0.6) b) hexane - predominantly in octanol c) 1-hexanol - predominantly in octanol d) methanol - predominantly in octanol e) oleic acid - predominantly in octanol 4. For KD of 1200 and equal volumes of water and octanol, αS(aq) = 8.3×10−4. For 100 mL of each solution this becomes 0.083 mL octanol in 100 mL of water. Using the density of octanol (0.8 g/mL) the solubility is approximately 0.06 g/100 mL solution. (Note that this approach is an approximate prediction since the αS(aq) calculation is dependent on relative volumes.) Brian M. Tissue, Basics of Analytical Chemistry and Chemical Equilibria, (J. Wiley, New York, 2013). 1 % of the solute in the aqueous phase and meets the state criterion.001 = │ Vorg 1 + 66 └ 100 mL ┘ Vorg = 46.5.5 mL 1 + 30 └ 100 mL αS(aq) = 0.0039 5 0.8 mL. For KD' = 66 and two sequential extractions (n = 2): 1 ┐2 ┌ │ 0.10 = │ 30 mL 1 + KD' └ 100 mL KD' = 30 (b) 1 ┌ αS(aq) = │ 7. The general expression is: 1 ┐n ┌ αS(aq) = │ Vorg │ 1 + KD' └ Vaq ┘ a) 1 ┌ 0.00098 Doing 5 extractions leaves slightly less than 0.016 4 0. New York.4 mL The total volume of organic phase will be twice this amount or 92. 6.25 2 0.0090 (or 0. .9 %) (c) 1 ┌ αS(aq) = │ 10 mL 1 + 30 └ 100 mL ┐1 │ ┘ ┐4 │ ┘ ┐n │ ┘ The easiest way to solve this question is to calculate αS(aq) as a function of n: n αS(aq) 1 0. Brian M. Tissue. Basics of Analytical Chemistry and Chemical Equilibria. (J.062 3 0. 2013). Wiley. Adding acid to lower the pH below the pKa of phenol will protonate the phenol so that it is a neutral molecule and will be soluble in organic solvents. 9. Anions attracted to the stationary phase can be eluted by lowering the pH so that the anions are protonated and become neutral. Brian M.. A weak anion-exchange stationary phase consist of an ammonium species that can lose it’s positive charge at high pH. . water:acetonitrile mixture (mixtures are common because they allow fine tuning of the mobile phase to optimize separations) d) a buffer solution at a basic pH to prevent the anions from becoming protonated. (a) Normal-phase partition chromatography usually uses a non-polar mobile phase such as hexane. propanol. 8.g. Addition of an internal standard before sample preparation for calibration. methanol will have low solubility in hexane and an alternate mobile phase is needed. For these alcohol solutes.. (J. Lowering the pH with time will elute the more strongly retained components faster. which can displace monovalent anions from the stationary phase. A strong anion-exchange stationary phase consist of a quaternary ammonium species that cannot lose it’s positive charge.76.. A possible choice is acetonitrile. 2013). methanol. Adjusting the pH of the buffer can increase the mobile phase concentration of CO32−. Basics of Analytical Chemistry and Chemical Equilibria. The solutes will elute in the order from least polar to most polar: butanol. hexaneisopropanol (99:1) c) reverse-phase chromatography uses a polar solvent. e. Acidifying to a pH near 8 should work well. Anions attracted to the stationary phase can be eluted by raising the pH so that the stationary phase is neutralized. a) A buffer solution at an acidic pH will keep the amino acids present as cations. To avoid extracting acetic acid. hexane or a mixture. b) normal-phase chromatography uses a nonpolar solvent. 11. e.7.g. A field spike to check for analyte loss.g. The carbonate buffer system is a convient eluant for anion-exchange SPE. e. ethanol. New York. Raising the pH with time will elute the more strongly retained components faster. Wiley. the pH should not be lowered to within two pH units of 4. Tissue. • • • • A field blank to check for contamination during sample collection A lab blank to check for cross-contamination or instrument carryover. 10. (J. propanol. Tissue. it’s a joke. detectable but cannot be quantitated) Zn: measurable with no dilution 14.(b) Short chain alcohols are soluble in water.8 0. (c) The concentration of psilocin in the test solution is: psilocin 500 μg = 4130 82110 psilocin = 25. The elution orde will be from most polar to least polar: methanol. A polar bear of course – smile.3 0.3 0. Find the answer to this question by searching manufacturer literature. 12. Basics of Analytical Chemistry and Chemical Equilibria. so a suitable mobile phase will be a water:acetonitrile mixture. we find: Cd: measurable with no dilution Cu: dilute by 5 or more Fe: measurable with no dilution Pb: detectable with no dilution (below LOQ.6 205.000456 17.3 0.48 71. The solution concentration of each analyte can be predicted from: (SRM concentration)×(SRM amount)/(volume) which gives μg/mL or ppm in the test solution. so it is a cation at pH = 6. ethanol. butanol. (a) The online literature gives a psilocin pKa near 8. Wiley.02976 0.8592 2.038 1424 SRM amount (g) 0.088 Comparing these calculations to the method measurement ranges. 15.1 μg Brian M. . New York.4696 0. The results are: Cd Cu Fe Pb Zn SRM (μg/g) 2. The method must be validated by running through the procedure with certified reference materials or spiked samples of known psilocin content.5.3 0. (b) The measurement will be erroneously low if the sample is heated too high. 2013).3 volume (mL) 25 25 25 25 25 solution (ppm) 0. 13. New York. Given obvious evidence of a soda spill. samples might be taken from the front and rear of a car of from the top and bottom of the ore load. Depending how the cars are loaded. What they have in common is that they are each a measurement from a given instrument or procedure. The internal standard will not correct for systematic errors due to incomplete conversion of psilocybin to psilocin or loss of psilocin due to degradation. The sampling sites within each car should be far apart to catch heterogeneity. Basics of Analytical Chemistry and Chemical Equilibria. The systematic approach is useful to try to determine the degree of heterogeneity in a sample batch.For a sample weight of 200 mg. The LOD is the concentration that gives a measurement that is just detectable.200 g sample (d) The internal standard will correct for any inefficiency in the SPE cleanup step. You sample around the soda bottle to determine the extent of the spill. A large number of sampling sites should be selected at random to provide the highest confidence in the results. The standard will usually be at a concentration that is similar to samples being measured and usually much higher than the LOD. a systematic sampling plan is reasonable. Two samples should be taken from each rail car. usually three times the noise level. . The limits on a control chart are usually three times the standard deviation of measurements of a standard. 18. Wiley. the concentration in ppm is: 25. Brian M. 19.1 μg psilocin = 126 ppm psilocin 0. Tissue. 2013). Since the ore has been mixed to some extent. (J. 16. The limit of detection (LOD) and the lower limit on a control chart are two completely different things. (a) hexane (non-polar organic) – may be retained to some extent (b) sodium acetate (mostly an anion at pH = 6) – not retained (c) sodium benzoate (mostly an anion at pH = 6) – not retained (d) phosphoric acid (mostly an anion at pH = 6) – not retained (e) acetone (moderate polarity organic) – should be retained 17. judgmental sampling is appropriate. Wiley. which would otherwise form a precipitate (soap scum) with surfactants. CH3COOH. 2013). Pb(CH3COO)2. The equilibrium constant expressions are: [A]2[B]2 [C]2 K1 = [A][B] [C] K2 = [A]2[B]2 [A][B] [A][B] = = (K1)(K1) = K12 2 [C] [C] [C] K2 = 2. the sodium salt of hypochlorous acid.Chapter 3: End-of-Chapter Solutions 1. HClO: ClO−(aq) + H2O ⇌ HClO(aq) + OH−(aq) Kb' = • [HClO][OH−] [ClO−] Disodium EDTA. Basics of Analytical Chemistry and Chemical Equilibria. New York. The Pb2+ ions react with sulfur in hair protein to form lead sulfide (PbS) precipitates that adhere to hair. Tissue. is very common in soaps and shampoos. (J. EDTA4−(aq) + Ca2+(aq) ⇌ Ca(edta)2−(aq) Kf' = • [Ca(edta)2−] [EDTA4−][Ca2+] Soluble lead acetate. The equilibrium on reacting with water is: CH3COOH(aq) + H2O ⇌ CH3COO−(aq) + H3O+(aq) Ka' = • [CH3COO−][ H3O+] [CH3COOH] Clorox and many cleaners contain sodium hypochlorite. . Some possibilities are: • Vinegar is 8 % acetic acid. The subsequent equilibrium is: PbS(s) ⇌ Pb2+(aq) + S2−(aq) Ksp' = [Pb2+][S2−] Brian M. It complexes the Ca2+ and Mg2+ found in hard water. the disodium salt of ethylenediamminetetraacetic acid. is contained in hair darkening products. a weak acid.01201 mol MgNH4PO4·6H2O 1 mol P 0.05 M HCl + 0.05 M (NH4)2C2O4 a precipitation reaction: CaC2O4(s) ⇌ Ca2+(aq) + C2O42−(aq) Ksp' = [Ca2+][C2O42−] 4.05 M NaOH These species are a strong acid and a strong base. Tissue. 2013). . and the equilibrium is: NH4+(aq) + H2O ⇌ NH3(aq) + H3O+(aq) [NH3][ H3O+] Ka' = [NH4+] d) 0. the H+ and OH− undergo a neutralization reaction. First convert the number of grams of product to number of moles: 2. New York. (a) 0. Cl− and Na+ are spectator ions.01201 mol MgNH4PO4·6H2O 245. respectively.05 M HCl + 0.947 g MgNH4PO4·6H2O = 0. K+ and Cl− are spectator ions: MnO4−(aq) + 5Fe2+(aq) + 8H+(aq) ⇌ Mn2+(aq) + 5Fe3+(aq) + 4H2O K' = [Mn2+][Fe3+]5 [MnO4−][Fe2+]5[H+]8 (b) 0. Wiley.4 g/mol Convert to mol of P: 0.01201 mol P 1 mol MgNH4PO4·6H2O Brian M. (J. The reaction product is NH4+.05 M KMnO4 + 0.05 M AgNO3 a precipitation reaction: AgCl(s) ⇌ Ag+(aq) + Cl−(aq) Ksp' = [Ag+][Cl−] e) 0.05 M NH3 Another neutralization reaction. Basics of Analytical Chemistry and Chemical Equilibria. and the resulting equilibrium is: 2H2O ⇌ H3O+(aq) + OH−(aq) Kw' = [H3O+][OH−] (c) 0.05 M HCl + 0.05 M FeCl2 MnO4− can Fe2+ can undergo a redox reaction. but in this case involving a weak base.05 M CaCl2 + 0.3. 3719 g P Find weight percent: 0.32 g/mol) = 0.485 C/mol (0. Tissue.7671 mmol Fe2O3 wt-% = 2 mol Fe 1 mol Fe2O3 55. .6 s)(0.8045 % 98.88 % 0.01201 mol P) (30.75 g test portion b) parts per thousand = 0.79441 g NaCl × 100 % = 0.1390 M 1.9482 g)/(143.69 g/mol 0.04825 C/s) = 19.75 g)/(1.974 g/mol) = 0. Sample calculation for the first measurement: 0.013593 mol and (0.75 % 3.013593 mol)(58.01 g/mL) 1000 mL = 0.08568 g Fe mol Fe 0.Convert mol of P to grams of P: (0.96 C 19. (J. (1. 2013).96 C 0.10 % or 17.8045 % × 10 = 8. (413.77 ± 0.8 ± 0.045 ‰ c) [NaCl] = 0.08568 g Fe × 100 % = 17.459 g test portion 5.0002068 mol e−)(1 mol I2/2 mol e−) 1 mol vitamin C = 0.4793 g test portion Repeating for the other samples and finding the average gives: 17.845 g Fe = 0.79441 g NaCl a) wt-% = 0.013593 mol NaCl (98.1 % 6. Basics of Analytical Chemistry and Chemical Equilibria.0001034 mol vitamin 1 mol I2 Brian M.3719 g P wt-% = × 100 % = 10. New York.0 L 7.0002068 mol e− 96.1225 g Fe2O3 = 0.7671 mmol Fe2O3 159.443 g/mol NaCl) = 0. Wiley. .845 g/mol) = 0.05000 L (b) phenol red. Wiley. (a) The analyte is Fe2+ and the titrant is MnO4−: 5Fe2+(aq) + MnO4−(aq) + 8H+(aq) ⇌ 5Fe3+(aq) + Mn2+(aq) + 4H2O (b) (0.0700 M 0. Basics of Analytical Chemistry and Chemical Equilibria.0200 g sample 8.02 g sample 9.119×10−4 g Al.0001034 mol vitamin C)(176.002045 M MnO4−) 5 mol Fe2+ = 3.03500 L) = 3.(0.50 mmol OH− [OH−] = 3. The amount of AlPO4 is then (0.02 g sample 6.50 mmol H+ = 3.5695×10−4 mol FePO4)(150.05383 g) = 1.01821 g vitamin C wt-% = 0.5695×10−4 mol Fe)(55.01993 g Fe d) The precipitate contained (3.816 g/mol) = 0. e) weight percent for each element is: wt-% = 0.01993 g Fe ×100 % = 0. Neutralization of a strong base with a strong acid produces a titration end point pH near 7.0663 g − 0.01821 g vitamin C 100 % = 91.119×10−4 g Al wt-% = ×100 % = 0.05383 g FePO4. New York. (a) (0.953 g/mol which equals 6.1000 M)(0.00122 % Al 50. Tissue.03491 L)(0.06 g=mol) = 0.022×104 mol 121. (J.03984 % Fe 50. 2013).5695×10−4 mol Fe 1 mol MnO4− (c) (3. Brian M.50 mmol OH− = 0.0 % 0. Back titration result: 1 mol Ag+ = 1.01104 L)(0. the endpoint has been passed. Cl− (OH− is present at < 10−7 M) (e) Add a known amount of strong acid and back titrate the excess acid with a standard base.6437×10−3 mol Ca2+ 1 mol EDTA Brian M.6437×10−3 mol EDTA Amount of Ca2+ in test portion: 1 mol Ca2+ 1.2994×10−3 mol Ag+ = 2.(c) At 30.2994×10−3 mol Ag+ 1 mol SCN− (0. Ca2+.3089×10−3 mol Ag+ = 2. Wiley. Ca2+. .05000 L 11.0 mL of titrant. H+. New York.7238×10−3 mol EDTA = 1.3089×10−3 mol Cl− 1 mol Ag+ Concentration of Cl− in test portion: 2. OH−.6083×10−3 mol Ag+ − 1.01805 L)(0.6083×10−3 mol Ag+ Amount of Ag+ reacted with Cl−: 3. (J.04618 M Cl− 0.3675×10−3 mol EDTA Amount of EDTA reacted with Ca2+: 3. Cl− (H+ is present at < 10−7 M) (d) At 40.3675×10−3 mol EDTA − 1.1347 M EDTA) = 3. Back titration result: (0.6437×10−3 mol EDTA = 1.0955 M Mg2+) 1 mol EDTA = 1.03555 L)(0.3089×10−3 mol Ag+ Amount of Cl− in test portion: 1 mol Cl− 2. the endpoint has not been reached.7238×10−3 mol EDTA 1 mol Mg2+ Total EDTA added to test portion: (0.3089×10−3 mol Cl− = 0.02500 L)(0.1177 M SCN−) Total Ag+ added to test portion: (0. 2013). Basics of Analytical Chemistry and Chemical Equilibria.1015 M Ag+) = 3. Tissue.0 mL of titrant. 10. 0 mL that the brass was dissolved in.3 ± 1.124×10−3 mol EDTA 1 mol Pb2+ 1 mol Cu2+ = 1.015 Titration result: (0.546 g Cu = 0. (67. The average of the three titrations is 0.45 % CaCO3 1.05620 L)(0.05620 L.00082 0.000 g solid 12.26 %)(0. (J. Tissue.0 % Cu 13.0 %.26 % 1.062 g brass Propagating the RSD from the three titrations. trial 1 2 3 average stdev RSD titrant vol (L) 0.015) = 1.0200 M Pb2+) 1.124×10−3 mol EDTA 1 mol EDTA = 1.16452 g CaCO3 mol CaCO3 0. .124×10−3 mol Cu2+)(10) wt-% = 63.087 g CaCO3 = 0.16452 g CaCO3 × 100 % = 16.0 mL of solution was determined from the 250.1. Basics of Analytical Chemistry and Chemical Equilibria. gives a final result of: 67. 2013). Titration of the H2SO4 with OH− is another neutralization reaction. New York. there is a 10-fold factor to multiply: (1.71426 g Cu mol 0.0571 0.05620 0.71426 g Cu × 100 % = 67.124×10−3 mol Cu2+ 1 mol EDTA Since 25. Conversion of SO2 to H2SO4 is a redox reaction. (b) 1 mol SO32− 1 mol H2SO3 1 mol H2SO3 1 mol SO2 1 mol SO2 1 mol H2SO4 1 mol H2SO4 2 mol OH− Brian M.6437×10−3 mol Ca2+ wt-% = 1 mol CaCO3 1 mol Ca2+ 100. Wiley.056 0.0555 0. (a) • • • The initial reaction is a neutralization to convert the analyte to a new form. 2013). Not all analytes can be determined by electrochemical methods.Overall: 1 mol SO32− 2 mol OH− (c) Since the titration is a strong acid and a strong base. 16. Wiley. phenol red is the best choice. gravimetry Requires only a balance for measurement. If the analyst overshoots the endpoint. titration The endpoint is often difficult to observe. the purity of a primary standard will always be a limit in a chemical analysis. and technique. (J. titration Requires standard solutions and volumetric glassware. a pure. phenolphthalein. Even with the best glassware. gravimetry Some analyte is left in solution. Brian M. Can be automated. 14. weighable form. New York. (d) Ba2+ and SO42− form a BaSO4 precipitate. 15. which goes from colorless to pink at pH 8 is also suitable. the result will be erroneously high. The acid-base titration is not specific for sulfuric acid. . it can be automated for rapid and unattended analysis. balance. disadvantages Like other classical methods. it is not sensitive enough for trace analysis. In practice. Tissue. Requires a reaction that goes to completion and Requires a reaction that converts the analyte to a means of detecting the end point. Basics of Analytical Chemistry and Chemical Equilibria. so the presence of sulfate should be confirmed. advantages Coulometry is an absolute measure and does not depend on other standards for calibration. Since it is an electrochemical method. Adding the soluble BaCl2 should show a precipitate to confirm the presence of SO42−. 17. The result will be erroneously low due to the lost analyte. New York. 2013). Tissue. Wiley. Basics of Analytical Chemistry and Chemical Equilibria.Brian M. . (J. 650 nm: E = (6. 2013).998×108 m/s ν= = 7. is a common unit in spectroscopy because of the convenient scale in the infrared through visible regions.40×1014 Hz or 740 GHz 405×10−9 m (b) Using E = hν. λ= 1 3100 cm−1 1m = 3.Chapter 4. 6. Basics of Analytical Chemistry and Chemical Equilibria.61×1014 Hz or 461 GHz 650×10−9 m 405 nm: 2. End-of-Chapter Solutions 1. You will want to develop a sense of scale of the different spectral regions. where h is Planck’s constant.626×10−34 J s)(4. cm−1. infrared.06×10−19 J E= 1 650×10−9 m 1m = 15. γ-ray (highest energy) 3. The highest charge that it can have is +1. Wiley.626×10−34 J s)(7. ultraviolet. X-ray. (J.2×10−6 m = 3. The H atom has one proton and one electron.626×10−34 J s. . The wavenumbers or inverse centimeter unit." 4.40×1014 s−1) = 4.998×108 m/s ν= = 4.700 cm−1 100 cm Both wavelengths are in the visible spectral region and are of the same order of magnitude in energy.400 cm−1 100 cm 405 nm: E = (6. visible. (a) Rearranging c = λν 650 nm: 2. Maybe the writer meant "highly excited. New York.61×1014 s−1) = 3. 2. microwave. (lowest energy) radio.2 μm 100 cm Brian M.90×10−19 J E= 1 405×10−9 m 1m = 24. Tissue. 700 cm−1 = 11. In the scattering example. (a) a filled antibonding orbital – No. and most scattered photons return to their original energy level. 5. once an orbital is full it is full and no further electrons may be added. . will be the strongest and require the highest IR energy to be excited.) 7. the ground state in this example. 6. Brian M. such as absorbing collisional. The shortest bond in the series. thermal.The energy of a 280 nm photon is 1 1m E= = 35. C−F. the photon is shown at an angle to indicate a change in direction after interacting with the electronic system. Bond length is inversely related to bond energy.700 cm−1 280×10−9 m 100 cm Dividing this energy by the IR photon energy gives: 35. The electron symbols show the occupied levels after the transition occurs. In the emission example. Schematic of emission (left) and Raman scattering (right). New York.5 IR photons 3100 cm−1 This example shows that the scale of energies between the infrared and the visible differs by roughly ten-fold. (J. (Raman scattering is a weak process. 2013). or light energy. Tissue. Basics of Analytical Chemistry and Chemical Equilibria. an empty orbital can accept an electron. one electron was placed in the excited state by some means. (b) an empty antibonding orbital – Yes. Wiley. Brian M.111 mg/mL. 2013).8. (c) In a 2-cm cuvette A will be 2(0. C6H14 has no non-bonding or π electrons and therefore can only have the short wavelength σ → σ* transitions. Absorbance is proportional to path length.916. New York. and therefore a transmittance of 0.333) = 0. The stain on the cuvette will appear as an absorbance in addition to the analyte absorbance.6 %transmittance. (J. Tissue.67 = 0.666 = 0. 12.8432 mL/mg)[glucose] + 0.521 the expression becomes 0.0-cm absorbance or 0. (a) A = abc.0 cm. The equation for the best fit line to the calibration data is y = (−0. so in a 5-cm cuvette A will be 5(0.666 AU and T = 10−0. .6148 where x is glucose concentration in mg/mL and y is the unitless absorbance measurement. Basics of Analytical Chemistry and Chemical Equilibria.521 = (−0.0 cm.916 = −log 1000 leads to 121 photons at 2. so the measurements will be erroneously higher than the true value. n-hexane.67 AU. 9. 10. so first determine the absorbance at 1.458 A = −log 1000 The absorbance at 2.1 %. not for transmittance. Wiley.16 %-transmittance.0 cm will be twice the 1.8432 mL/mg)x + 0. (b) A = −log(T) so T = 10−1.6148 and the result is [glucose] = 0. which is equivalent to 2. 11. Now reversing our calculation: photons 0. P = 398 photons and Po = 1000 so: 348 = 0.0216.333) = 1.216 or o 21. Note that the linear relationship for A versus b or c is true only for A. For a measurement of 0.121 or percent-transmittance of 12. 009 must be subtracted from the measurement. Their disadvantage is that the spectral resolution is fixed for the pixel width of the detector.418 mg/mL 17.0 = −log(T).271 = −0. Scanning instruments are more flexible and can provide higher resolution and greater sensitivity by varying the slit width and changing the type of detector.408 mg/mL (b) The background of 0. (J. A = 0. New York.6148 [riboflavin] = 0. Brian M.6148 [riboflavin] = 0. T = 0.009) = −0. then calculating as before: (0.79 and A = 1. .1 = −log(T). The noise fluctuations of the lamp source will limit the ratio of P and Po that can be distinguished.271 − 0. respectively.10 These transmittance values provide a range where lamp noise in subtracting two large values (for the high T region) and stray light (for the low T region) are unlikely to affect the measurement.458 1000 Using the Beer-Lambert law: 0.0 cm: 348 A = −log = 0. Tissue. 15. The light measurements of P and Po will be nearly identical.8432[glucose] + 0.0×10−5 M) ε = 11.458 = ε(1. 14. 0.8432[glucose] + 0.13. Array spectrometers acquire the whole spectrum simultaneously and are therefore very rapid. Using the data point for a distance of 1. Basics of Analytical Chemistry and Chemical Equilibria.00 cm)(4. A very weak absorbance will attenuate the light power passing through the test portion a very small amount.500 M−1 cm−1 The result should be the same for any data point. (a) Inserting the unknown measurement into the equation for the calibration curve. 16. T = 0. but doing a number of calculations and taking the mean will average random scatter. Wiley. 2013). 00 cm)(4.455 = (13720 M−1 cm−1)(1. Note that the trendline equation does not show the units of the slope and intercept.00 cm)[ Fe(phen)32+] 0. 0.455 [Fe(phen)32+] = (13720 M−1 cm−1)(1.466 ppm−1)[riboflavin] + 0.) (b) Again using the Beer-Lambert law and the molar absorptivity from the calibration above. Tissue. 2013).00×10−5 M) ε = 13. The calculated concentration will be less than the true iron concentration in the water sample. The test portion should be diluted until the absorbance is less than or near 1.00 cm) [Fe(phen)32+] = 3. Inserting the unknown measurement into the equation for the calibration curve: 14. the slope is 50. so [Fe2+] = (3.720 M−1 cm−1 (I carry an extra significant digit in intermediate calculations that I will drop in the final result. Since the fluorescence signal is given in arbitrary units. (a) Using the Beer-Lambert law: 0.32×10−4 M (d) This absorbance value is above the linear range for most spectrophotometers.466 ppm−1. Wiley. (e) This absorbance measurement will be lower than a test portion that was all Fe2+. New York. Brian M. (J.18.272 ppm 19. .5 = (50.0.32×10−5 M)(10) = 3. Basics of Analytical Chemistry and Chemical Equilibria.7505 [riboflavin] = 0.32×10−5 M (c) The water sample was diluted by 10 to add the other reagents for the absorbance measurement.549 = ε(1. but OH− is present. (a) 0. NH4+. calcium perchlorate is a strong electrolyte (c) KI: neutral. so we call it a weak acid. Tissue. Wiley. OH− Note that NO3− is a strong electrolyte so we assume that HNO3 dissociates completely. H3O+. in this case the solution will be acidic (g) ammonium acetate: amphiprotic. CH3COOH.3 and 9. H3O+. is a weak acid (f) NH4F: amphiprotic. Basics of Analytical Chemistry and Chemical Equilibria. F−. for pH greater than 9. [CH3COOH] is significant and much larger than [H3O+] or [CH3COO−].01 M HClO4 (perchloric acid is a strong acid and produces a lower pH than an equal amount of a weak acid) (c) 1×10−4 M HClO4 (pH = 4. End-of-Chapter Solutions 1.Chapter 5. (a) bucket of deionized water: H2O. 4.8) Brian M. is a weak base (e) NH4Br: acidic. in this case Ka ≈ Kb and the solution will be close to neutral 3. barium nitrate is a strong electrolyte (b) Ca(ClO4)2: neutral. NO3−. Again.9 the charge is −1 (b) extracting into an organic solvent will not occur if the amino acid is charged. ammonium ion. 2. H3O+. CH3COOH. .01 M HCl (hydrochloric acid is a strong acid and produces a lower pH than an equal amount of a weak acid) (b) 0. fluoride ion. (c) small amount of acetic acid. so adjust the pH to between 3-9. but OH− is present. (a) Ba(NO3)2: neutral. (J. OH− (b) small amount of HNO3 added: H2O. (a) below pH of 2. added: H2O. [OH−] will be very small relative to the other concentrations in solution. compare Ka of NH4+ to Kb of F−. OH− CH3COOH dissociates to a small extent. pH between 2. New York. [OH−] will be very small relative to the concentrations of the other ions in solution. One way to define a weak acid is simply as an acid that does not dissociate completely in aqueous solution. 2013). 0. the larger dominates. potassium iodide is a strong electrolyte (d) NaF: basic.3 the charge is +1.01 M HClO has pH 4. CH3COO−.9 the overall charge is 0 (zwitterionic form). so compare compare Ka of NH4+ to Kb of CH3COO−. 250 M)} = 0. inserting the di for each ion: i..001 M KOH (a lower concentration of strong base will have lower pH. (a) 0.602 Brian M. 2013).250 M) +(−1)2(0.01 M CH3COONa (weak base.500 M) +(−2)2(0. Basics of Analytical Chemistry and Chemical Equilibria.900 (b) Ic = 0.509(−1)2(0.5 log γOH = 1 + (3. use the Debye-Hűckel equation. Sample calculation for OH−: −0. use the Debye-Hűckel equation.010 M.250 M) +(−1)2(0. OH−: γOH = 0. i.223 ii.250 M) +(−1)2(0.35)(0. but if you write the equilibria you'll see that there is no change in the number of ions in solution. H3O+: γH3O = 0.752 iv.750 M)} = 1.750 M (c) 0.5. H3O+: γH3O = 0.4.010 M (b) 0.29(0. (a) Ic = 0.50 M (we neglect the reaction of Al3+ with water) (d) 0.250 M CH3COONa Ic = 0.899 ii.5{(+1)2(0. K+: γK = 0. 7. Ca2+: γCa = 0.899 iii.250 M Ca(NO3)2 Ic = 0.010)0. Wiley. more acidic.250 M AlCl3 Ic = 0. New York. (J.578 iii. lower pH than strong base) (c) 0.250 M)} = 0.914 iv. than higher concentration of strong base) (b) 0. OH−: γOH = 0. .5{(+3)2(0. 1×10−4 M NaOH has pOH = 4 and pH = 10) 6.5 i.01 M CH3COONa (pH ≈ 8.5{(+1)2(0. (a) 0. NO3−: γNO3 = 0.010)0. I−: γI = 0.250 M For d) and e) there is some reaction of NH4+ and CH3COO− with water.750 M (e) 0.010 M KI Ic = 0.500 M)} = 0.5{(+1)2(0.e. Tissue. inserting the di for each ion.750 M.250 M (NH4)2SO4 Ic = 0.010 M) +(−1)2(0.5{(+2)2(0.010 M)} = 0. 4 M 1. acid dissociation is 0.23×10−14 (0.01×10−14 Kw' = = 1.2 M CH3COONa in 0. 0. 0.0. Kw = (aH3O+)(aOH−) = (γH3O+)[ H3O+](γOH−)[OH−] = (γH3O+)(γOH−) Kw' Kw' = Kw (γH3O+)(γOH−) 1.2 M CH3COOH. . 1.8.2 M CH3COONa in 0.0 M CH3COOH. 0. since the OH− has the same charge as CH3COO−.2 M NaCl 0.2 M NaCl (b) Activity coefficients decrease with increasing Ic. Brian M. 0. set up the expression for Kw and substitute activity coefficients and concentrations for the activities. 0. New York.2 M NaCl. will produce activity coefficients farthest from the ideal case of 1. (J.2 M CH3COONa.2 M NaCl.2 M CH3COOH in 0.0 M CH3COOH. so the solution with the highest Ic.602) 9. Ic = 0. Tissue.2 M CH3COOH. Ic slightly higher than 0.5%. there is no effect on ionic strength) 0.914)(0.2 M NaCl.002 M 0.900) b) 1. Ic 0. acid dissociation is approximately 1%. 2013).23×10−14 Kw' = (0.005M (a) Ranking the solutions from lowest to highest ionic strength. Basics of Analytical Chemistry and Chemical Equilibria.2 M CH3COONa in 0.2 M CH3COOH in 0.752)(0.2 M (a small amount of acetate reacts with water to form CH3COOH and OH−.2 M due to acid dissociation in addition to the 0. 0.2 M NaCl.01×10−14 = 2. Wiley. Ic = 0. (a) Using the activity coefficients from the previous question. Ic = 0.2 M CH3COONa. 01 M − [H3O+] [H3O+] = 0. (a) 4. Activity coefficients are 0. Basics of Analytical Chemistry and Chemical Equilibria.1 % (b) 27 % (c) 62 % 11.01 = [H3O+]2 = 0. (J. The ionic strength is approximately 0.0062 M × 100 % = 62 % 0. 12. Correct Ka to obtain Ka'.75 for H3O+ and 0. Percent-dissociation is equal to: [H3O+] %-dissoc = × 100 % cHA Sample calculation: [H3O+]2 Ka' = cHA − [H3O+] Enter Ka' and cHA. For Ka' = 1×10−2. 2013).01 M (a rather strong “weak acid”) %-dissoc = Other results: (a) 3. .0062 M 0. New York.6 % (b) 37 % (c) 75 % You can see there can be a significant difference even for monoprotic acids at high ionic strength. then do the calculation in the same way as above.75×10 = cHA − [H3O+] −5 Brian M. Wiley. Tissue. CH3COOH(aq) + H2O(aq) ⇌ CH3COO−(aq) + H3O+(aq) Ka' = [CH3COO−][H3O+] [CH3COOH] [H3O+]2 1.62 for A−.10. Ka' = 0.75 M (slightly higher due to acid dissociation). then rearrange and solve with the quadratic equation. 00 − 5. and solve with the quadratic equation.1×10−4 M)} = 4.75×10−5 5.10×10−4 M p[H3O+] = 3.39. [H3O+] = 5. Brian M.19×10−4 M p[H3O+] = 3. so Ka' = 3. New York.765 for H3O+ and 0. [H3O+] = 4. Basics of Analytical Chemistry and Chemical Equilibria. .977 for CH3COO−. 14.0100 M − [OH−] Solve for [OH−] using the quadratic equation. rearrange. Ic = 0. 13. Using this value in the calculation results in an insignificant change in the result.5{(+1)2(4. then convert to [H3O+].0100 M for the acetic acid formal concentration. The equilibrium is: CH3COO−(aq) + H2O ⇌ CH3COOH(aq) + OH−(aq) Kb' = Kw' [CH3COOH][OH−] = Ka' [CH3COO−] Kb' = 1.38.69×10−4 M p[H3O+] = 3.Enter 0.43×10−5. Using this value in the calculation results in a change of 0. [H3O+] = 4.83×10−5.1×10−4 M Activity coefficients are 0.978 for H3O+ and 0. (J.62 p[H3O+] = 14.15 pH units. 2013). so Ka' = 1.62 = 8. Activity coefficients are 0.667 for CH3COO−.77×10−10 = [OH−]2 0. Wiley. 15. [OH−] = 2.01×10−14 = 5.40×10−6 M p[OH−] = 5.1×10−4 M) + (−1)2(4. Tissue.77×10−10 1.25.38. New York.001 M − [H3O+] −2 [H3O+] = 0.35×10−2 M p[H3O+] = 1. Wiley.38 quadratic equation: 1. (J. Basics of Analytical Chemistry and Chemical Equilibria. Brian M. etc.1 M. Activity coefficients are 0.001 M [H3O+] = 4.90 for both CH3COO− and OH−.06 For acetic acid the approximate calculation is very close to the quadratic result. 2013).001 M [H3O+] = 2.864×10−2 M p[H3O+] = 2.10×10−4 M p[H3O+] = 3. Find the result by tabulating p[H3O+] for each weak acid for cHA = 0.16. If cHA is high and Ka' is not large.5×10 = 0.75×10−5 = [H3O+]2 0.63 [H3O+]2 5. Using the spreadsheet allows you to do multiple calculations quickly. 0.001 M. Here I set up the calculation using acetic acid and dichloroacetic acid at cHA = 0. For the strong dichloroacetic acid. Given the form of the Kb' expression Kw' [CH3COOH][OH−] Kb' = = Ka' [CH3COO−] The effect of the activity coefficients cancel and Kb' = Kb.39 dichloroacetic acid 5.18×10−4 M p[H3O+] = 3. 17. The result is the same as in question 15. the approximation introduces a significant error.01 M.01 M.5×10−2 = [H3O+]2 0. Tissue.001 M − [H3O+] [H3O+] = 4.75×10−5 = 0. . we expect the approximate solution to give the same result as using the quadratic equation. 0. acetic acid approximate solution: [H3O+]2 1. 79 [H3O+] = 10−3.85 [H3O+] = 10−3. so the quick approximation mostly fails for weak acids with relatively large Ka values except at fairly high formal concentrations. Setting pH = 3.85×10−5) results in γ2 = 0.75×10−5 = γ2(2.91 using Ka = 1.79.85 = 1. 2013). Tissue.62×10−4)2 0.xls leads to an ionic strength of 0. Basics of Analytical Chemistry and Chemical Equilibria.001 M acetic acid has a pH of 3. Inserting values for [Na+] and [Cl−] in ionic-strengthactivitycoefficients.01 − 1. 19.62×10−4 Ka' = 2.79 = 1. (J.1 M.02×10−6 Repeating for the measured pH: p[H3O+] = 3.41×10−4 Ka = 2. This problem is a Ka' calculation worked backwards. Wiley.85×10−5.01 − 1.67×10−6 Brian M.625 and γ = 0.41×10−4)2 0.41×10−4 M Ka = [H3O+]2 cHA − [H3O+] Ka = (1.11 M. 0.75×10−5.81 and working backwards requires a Ka' of 2. Using the usual means of correcting Ka: Ka = γ[A−]γ[H3O+] = γ2Ka' [HA] 1. 18.62×10−4 M Ka' = [H3O+]2 cHA − [H3O+] Ka' = (1. New York. p[H3O+] = 3.(a) ≈ 1×10−4 M (b) ≈ 0. . and use the activity coefficients to determine Ka'.29(0. (J.509(1)2(Ic)0.Usually we look up Ka from a reference table. 2013).5 0. Wiley.0605 = −0.0605 = 0.0605 + 0.509(Ic)0. .5 1 + (3.5 (Ic)0.5 = 0.32(Ic)0.67×10−6)0.0796(Ic)0. calculate Ic to find activity coefficients.870 Now use the Debye-Huckel expression to find Ic −0.5 1 + 1.5 log(0.5 γ = (2. Here we know Ka and Ka' and we can calculate the ionic strength of the solution.85 = 1.429(Ic)0.4)(Ic)0.509(Ic)0.0605 = 0.02×10−6/2. Tissue.5 0. [H3O+] = 10−3.62×10−4 M Ka = γ[A−]γ[H3O+] = γ2Ka' [HA] γ = (Ka / Ka' )0.5 0.020 M Brian M. which provides a more realistic prediction of a weak acid equilibrium.870) = −0.5 1 + 1.32(Ic)0.509(Ic)0.5 = 0.5 γ = 0.141 (Ic) = 0. Basics of Analytical Chemistry and Chemical Equilibria. New York. Basics of Analytical Chemistry and Chemical Equilibria.5 mol of NaOH 5.0.60 mol OH− 0. This condition is always true and it does not depend on ionic strength. Wiley.0: acetic acid.76.99. End-of-Chapter Solutions 1. pKa1 = 6. pKa = 4. (a) pH = 5.76.) 3. H3PO4 + OH− → H2PO4− H2PO4− + OH− → HPO42− HPO42− + OH− → PO43− 0.33. is an alternate inorganic choice. Individual fractions might change slightly as Ic changes.8 to buffer at this pH.25. pKa2 = 4. Note that the question asked for monoprotic acids only.050 mol H+ ----------------total = 0. pKa1 = 6. Tissue.0: there is not a good monoprotic organic acid in Table 5. (J. (b) pH = 7.0.050 mol H+ 0. .30 mol OH− ----------------total = 1. but the best choice in Table 5. 4. pKa = 9.0: carbonic acid.60 mol OH− 0.35. PO43− + H+ → HPO42− HPO42− + H+ → H2PO4− 0. 2013). pKa = 9.0: carbonic acid.0. but the total must still be 1.35.Chapter 6. pKa = 7. (b) pH = 7.8.0: phenol. Alternate choices are the inorganic hypochlorous acid. (c) pH = 9. pKa2 = 10.54. the total of all alpha’s must be one. Σ αi = 1. New York. 2. (A bit far from 9.0: citric acid. ammonium ion. (a) pH = 5. Phenol is not used as a pH buffer in practice due to toxicity. or the diprotic carbonic acid.10 mol of HCl Brian M. (c) pH = 9. Since alpha is a fraction. 5(6. so basic but not a buffer solution. 0.78 b) p[H3O+] = 0.30 M Contents of solution after adding HCl: (0.15 M) + (−1) 2 (0.10 M) + (−1)2(0.15 M Na+.6.040 moles of a monoprotic weak acid: excess OH−.050 mol of H3PO4.070 moles of a potassium hydrogen phthalate: Buffer solution at pH = 2.500 mol of HCl.9. Protonating all of the HPO42− to obtain H2PO4− requires another 0. From the phosphate alpha plots (Figure 6.5(7. we will need 0.20 + 12.0 or just slightly higher.450 moles of H2PO4− to achieve this ratio. 9. Finally we add 0.1 and H2PO4− = 0. New York.35) = 9. 0. (a) Adding 0.0×10−8 M is less than the intrinsic [H3O+] of pure water.050 mol of H2PO4− to 0. (d) Adding 0. 8.05 M PO43−): Ic = 0.040 moles of NaOH to a solution containing 0.10 M Cl−.500 mol + 0. Tissue. Wiley. Contents of initial solution: (0. pH depends on pKa of HA. we see that we need H3PO4 = 0.35 + 10.500 moles of phosphate.050 mol of HCl to convert 0.15 M Na+.34 c) 1.5(pKa1 + pKa2): a) p[H3O+] = 0. . Protonating all of the PO43− to obtain HPO42− requires 0.33) = 8. (c) Adding 0. Using p[H3O+] = 0. (J.070 moles of a monoprotic weak acid: significant amounts of a weak acid and conjugate base.500 mol of HCl.4.6) at pH = 3.070 moles of NaOH to a solution containing 0.040 moles of NaOH to a solution containing 0.0500 moles of H3PO4 and 0.15 M) + (−3)2(0.5{(+1)2(0.5{(+1)2(0.05 M)} = 0.05 mol HCl. The total amount of strong acid added is 0.500 mol + 0. (b) Adding 0.040 moles of HCl to a solution containing 0. so p[H3O+] is expected to be 7. 0.070 moles of a potassium hydrogen phthalate: Buffer solution at pH = 5.9. Brian M. Since we are starting with 0. so a buffer solution.05 M H2PO4−): Ic = 0.05 M)} = 0. 2013).050 mol = 1.15 M 7. Basics of Analytical Chemistry and Chemical Equilibria. The predominant form of phosphate can be viewed from the alpha plots directly.05 M C8H5O4−. Wiley.57×10−3 M and p[H3O+] = 2. and the ionic strength is 0.91×10−6 [C8H4O42−] 2.854) Now using this Ka' .95 + log 1. 2013). Find the desired pH on the x-axis and then determine which curve has the largest alpha value at that pH.02 = 8. (J.15 pH units.12×10−3 = 1.05 M Na+ and 0. 12.18 (c) the second equivalence point: the solution contains C8H4O42−. .97 11.95. [H3O+] = 1. Correcting Ka: Ka' = [H3O+]2 cHA − [H3O+] p[H3O+] = 2. Basics of Analytical Chemistry and Chemical Equilibria.834)(0. Activity coefficients are 0. The difference in results is 0.05 M.03 p[H3O+] = 14. respectively.10.0333 M − [OH−] [OH−] = 9.5(pKa1 + pKa2) = 4.0 − 5. (a) halfway point (for the first acidic proton): the solution contains equal amounts of C6H4(COOH)2 and C8H5O4−. so p[H3O+] = pKa1 = 2. so solve as a weak base problem correcting for dilution during titration: Kb = Kw [C8H5O4−][OH−] = 3. New York. (b) the first equivalence point: the solution contains predominantly C8H5O4−.834 and 0. Determining the error in the previous answer requires predicting p[H3O+] after correcting pKa for ionic strength.57×10−3 (0.28×10−6 M p[OH−] = 5. The solution at the halfway point contains 0.58×10−9 = [OH−]2 0. a) pH = 4: H2PO4− b) pH = 6: H2PO4− c) pH = 8: HPO42− d) pH = 10: HPO42− Brian M. so p[H3O+] = 0. Tissue.854 for C8H5O4− and H3O+.80. 05 9. (J. . Tissue.17.015 M = 3.90 11.13 M 0. Since the pH of a saturated solution is approximately halfway between pKa1 and pKa2.10 9. Interpolating in the raw data values in alpha-plot-3protic. 2013).5×10−6)(0. the solution contains [C6H4(COOH)2] = 0. New York.6 18. Wiley.015 M. 14.15 9. The concentration of any given species is the fraction of that species times the total concentration: [HPO42−] = (4.3 0. Basics of Analytical Chemistry and Chemical Equilibria.95 11. the form of NTA in the solution must be H2NTA−.5×10−6 M.5×10−6.13.50 10. Some pH values are as follows: HCO3−:CO32− p[H3O+] 0. The value can also be found using Equation 6.3 0. Brian M. Using alpha-plot-3protic or a direct calculation shows that αCO3 is 1.3 0.010 M and [C8H5O4−] = 0.85 11.0 0. After the neutralization reaction.xls gives αHPO42− = 4.5×10−5.5 0.95 + log 0. which is much smaller than [HCO3−].1 0. Using the Henderson-Hasselbalch expression gives: p[H3O+] = 2.010 M 15. The [CO32−] concentration is 1. The values can be obtained by Henderson-Hasselbalch calculations or from alpha plot data.2×10−7 M HPO42− 17.0500 M) = 2. 16. New York. 2013). which is done by adding a small amount of strong acid or strong base to this buffer solution.5{(+1)2(1. Tissue. Note that the recipe specifies adjusting the pH to 7.101 mol −2 ions in a solution volume of 1 L. (J. Using formula weights and combining common ions. the solution contains: 1. Basics of Analytical Chemistry and Chemical Equilibria.413 mol −1 ions 0.19.616 mol +1 ions 1. . it is easier to measure and adjust the pH than to try to calculate the pH.616 M) + (−1)2(1.413 M) + (−2)2(0. Due to the high ionic strength and day-to-day temperature fluctuations.4. Ic = 0.72 M Brian M. Wiley.101 M)} = 1. 0 (no effect) 5. The base has unbonded pairs of electrons and the proton has the 1s orbital empty. The bonding between a base and a proton is analogous. (b) MnCl63−(aq): Mn(III).001 M − [H3O+] and the results are tabulated in the last column of the table.8 12. so these two ions have no effect on p[H3O+]. The measurement of the excess EDTA will be erroneously high. the charge of a Mn(VII) ion is so high that it will react with water to form an oxyanion. Basics of Analytical Chemistry and Chemical Equilibria. When using the indicator.0. The pKa and the conversion to Ka for each cation is listed in the table. I assume pKa' = pKa. Ligands have unbonded electrons (usually pairs) and metal ions have empty orbitals. and there is no charge to this neutral solid material. and the overall complex charge is −3. O has the usual −2 oxidation state. Cl has a −1 oxidation state.2×10−8 p[H3O+] 7.5 Ka 1. the indicator remains uncomplexed until reaching the endpoint. 3. (d) Mn(H2O)67+(aq): Mn(VII). 2.5×10−13 3. (e) MnO2(s): Mn(IV). New York. water is neutral. Calculation of the Ca2+ concentration will then be erroneously low. . Coordinate covalent bonds form between the electron pairs on the ligands and the empty orbitals of the metal ions. For 1.6×10−14 2. 2013). When using a Pb2+-sensitive electrode. (c) Mn(H2O)62+(aq): Mn(II). The Cl− is a strong electrolyte and has no direct effect on p[H3O+]. The calculation is the same as for any weak acid as done in Chapter 5: [H3O+]2 Ka' = 0. metal ion will bind to the indicator. If we assume that the indicator binds either the Ca2+ analyte or the Pb2+ titrant equally. and the overall charge of this ion is +2. Pb2+ titrant that displaces Ca2+ from Ca(edta)2− will cause an overshoot of the endpoint. (J. 4.0 (no effect) 7.2 Brian M.Chapter 7: End−of−Chapter Solutions 1. MnO4−. Tissue.6 7. Wiley. (a) KMnO4(s): Mn(VII) in a covalent oxyanion. Li+ Ca2+ Cu2+ pKa 13. At the endpoint where all EDTA is complexed. then the indicator will still change color at the endpoint. The calculation for Li+ and Ca2+ gives a result higher than 7.0 mM ionic strength. this complex as written does not exist. 2×10−7)2 3. Fe(OH)2+.5 we see that at pOH = 9 we have significant fractions of Fe(OH)3.92) = 9. 2013). (J.5.92.2×10−7 M Setting up the Ka' expression and solving for c: [H3O+]2 Ka' = 2+ [Ca ] − [H3O+] 2.001.6 and it is 0.5×10−13 = (3.2×10−7 [Ca2+] = 0. Common interferences in EDTA titrations are copper or iron. 7. Wiley.5 or: [H3O+] = 10−6. 8. and FeOH2+. These metals have very large formation constants with cyanide.5 = 3. and the concentration is (1. or by interfering with the complexometric indicator.0×10−3 M)(0. In this question we find the concentration of the metal ion that produces p[H3O+] = 6. New York. At pOH = 9 we can get the Fe3+ alpha from Figure 7. They can interfere by appearing as analyte. giving erroneously high results.2×10 = [Cu2+] − 3.2×10−6 M 6.2×10−4 M Brian M. The alpha at pOH = 13 is 0. The concentration is then (1. Tissue. (a) These pH values correspond to pOH values of 9 and 13. (b) Mn(H2O)62+(aq): at high pH the Mn2+ can form hydroxide complexes to reduce the Mn(H2O)62+ concentration.2×10−7 −8 [Cu2+] = 3. At pOH = 13 there is FeOH2+ and Fe3+. Adding cyanide to a test portion will mask these metals so they do not complex with the EDTA titrant. . At low pH there is no effect.4 M (3. (c) Cu(NH3)42+(aq): at high pH the Cu2+ can form hydroxide complexes and at low pH the NH3 can be protonated.0×10−3 M)(0. (b) The concentration is found by multiplying the total concentration by the alpha fraction. either extreme will reduce the Cu(NH3)42+ concentration. Using Figure 7.001) = 1. (a) Mn(edta)2−: at high pH the Mn2+ can form hydroxide complexes and at low pH the EDTA can be protonated. Basics of Analytical Chemistry and Chemical Equilibria. either extreme will reduce the Mn(edta)2− concentration.2×10−7)2 [Ca2+] − 3.0×10−6 M. Taking the case of EDTA titrant and a metal ion as analyte.00−7.33−4.33 = 100.95 = 8.(c) You are starting with 0.4 (b) The 1. New York. Tissue. To reach that form it has reacted with water to form H3O+.38 = 102.38 = 2. 7. 11.115 M Cl−.115 M 9. . This change in electron distribution can shift the electronic energy levels.73 = 5. Basics of Analytical Chemistry and Chemical Equilibria. (J. and 8. there will either be complete depletion of a metal being titrated or the presence of a metal from the titrant. 8.5{(−1)2(0. The alpha plots show that the different complexes appear to be separable for Cu2+ and Fe3+ oxalate but not for Al3+ oxalate. We see the shift in wavelength of the absorbed light as a color change.8).005 M Fe3+ and 0.115 M NH4+ and 0. Complexometric indicators have different colors when a metal is bound or unbound.4×104 107. 10. all of the iron is in the form of Fe(OH)3. (Suitable indicators have Kf' values lower than for the metal ion and EDTA.9×102 108. 2013).73. The NH3 of the buffer system neutralizes this H3O+ to produce 0. Brian M. At the end point of a titration.67 = 4. The log cumulative formation constants for Ag+ and Br− are 4.015 M Cl− from the FeCl3.33.015 mol of NH4+.10 M Cl− from the buffer and 0.10 M NH4+ and 0.115 M) + (+1)2(0.00.10−phenanthroline has larger formation constant values for Ag+ than does Br−.2 (pOH = 4. so for comparable concentrations. which has no charge.38. it will displace Br− to complex with the Ag+. at the end point the EDTA complexes all of the metal ion. After the solution reaches equilibrium.115 M)} Ic = 0. At pH = 9. binding of the metal changes the distribution of electrons in the complex compared to the electron distribution in the absence of the metal. Less than the maximum number of ligands might be stable in solution if the stepwise formation values are well separated. The ionic strength is thus: Ic = 0. it is now 0.73−8. removing metal from a metal−indicator complex. Wiley.00 = 100. (a) The stepwise formation constants are 104.7 108.) Electronically. βeff' = αM αL β1 Ca2+ Pb2+ Zn2+ β1 ' 1×1011 2×1018 3×1016 αM pH = 5 αL βeff' −7 3. Basics of Analytical Chemistry and Chemical Equilibria. .25×10 2. (b) Mixing these two soluble salts will form a Cu(OH)2 precipitate and the copper ammine complex: Cu(OH)2(s) ⇌ Cu2+(aq) + 2OH−(aq) Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+(aq) There will also be the acid−base equilibrium for the NH4+. 15. Predict the precipitation order for the following solutions (a) The solubility products for these precipitates are Cd(OH)2: Ksp = 7. Tissue. Brian M.54×10−7 3. Zn(OH)2 second. (J. Wiley. 2013).54×10 3. equilibria Ksp = (aLa)(aIO3)3 Ksp = (γLa)[La3+](γIO3)3[IO3−]3 Ksp = (γLa)[La3+](γIO3)3Ksp' Ksp Ksp' = (γLa)(γIO3)3 14. and Cd(OH)2 last.2×10−20 Zn(OH)2: Ksp = 3×10−17 Since the stoichiometry is the same for all of these precipitates. (a) The Cu3(PO4)2 is insoluble and the precipitation equilibrium is: Cu3(PO4)2(s) ⇌ 3Cu2+(aq) + 2PO43−(aq) The NH3 can complex with copper so you also have: Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+(aq) There will also be the base hydrolysis equilibria for the ammonia and PO43− to form the protonated form of each.2×10−15 Cu(OH)2: Ksp = 2.12. we can predict the precipitation order to be Cu(OH)2 first.54×10−7 αM pH = 6 αL βeff' −5 2. New York.25×10−5 2.25×10−5 13. AlPO4 will precipitate first.8×10−17 M So. pH effects: (a) BaCO3: Ba2+ is a strong electrolyte so there is no effect at high pH.1 mM each of Cu2+ and Al3+. The Ksp values are Cu3(PO4)2: 1. 2013).7×10−13 M Ksp = [Al3+][PO43−] 9.5[PO43−] (d) ZnF2: s = [Zn2+] = 0. Brian M. Neglecting competing equilibria: (a) BaCO3: s = [Ba2+] = [CO32−] (b) Ba(OH)2: s = [Ba2+] = 0.84×10−21. . at low pH the OH− is neutralized so more Ba(OH)2 dissolves (c) Pb3(PO4)2: high pH will increase solubility due to formation of lead hydroxide complexes and low pH will increase solubility due to protonation of PO43− (d) ZnF2: high pH will increase solubility due to formation of zinc hydroxide complexes and low pH will increase solubility due to protonation of F− to form HF 18. Tissue. Wiley.333[Pb2+] = 0. 16.(b) adding PO43− to a solution that is 0. at low pH the CO32− is protonated and solubility increases (b) Ba(OH)2: Ba2+ is a strong electrolyte so there is no effect at high pH. Basics of Analytical Chemistry and Chemical Equilibria.40×10−37 = (1×10−4)3[PO43−]2 [PO43−] = 3.5[OH−] (c) Pb3(PO4)2: s = 0. The PO43− concentration at which each precipitate forms is: Ksp = [Cu2+]3[PO43−]2 1.5[F−] (e) Ca(C2O4)·H2O: s = [Ca2+] = [C2O42−] (the waters of hydration in the solid have no effect) 17.40×10−37 and AlPO4: 9. New York.84×10−21 = (1×10−4)[PO43−] [PO43−] = 9. (J. such as lanthanum hydroxide complexes. and activity coefficients from the Debye-Hückel equation are: γLa = 0.90 7.Calculate Q.010 M (neglecting the concentrations of La3+ and IO3− ions from the lanthanum iodate). such as precipitates. the concentrations have exceeded the solubility limit and a precipitate will form. the concentration of this \common” ion will have a direct effect on solubility. There are different ways to think qualitatively about the effect of ionic strength on ionic equilibria.00010 M)(0.0×10−12 M Q < Ksp' So the concentrations are below the level at which a precipitate will form. . 2013). increasing the ionic strength of an aqueous solution will reduce the tendency of ions to recombine.9×10−11 (0. 20. Basics of Analytical Chemistry and Chemical Equilibria. we need only consider: La(IO3)3(s) ⇌ La3+(aq) + 3IO3−(aq) Ksp' = [La3+][IO3−]3 7.44 and γIO3 = 0. but we will correct Ksp for activity effects. This direct effect will decrease solubility and will usually be much larger than the indirect effect caused by ionic strength affecting equilibrium constants.50×10−12.44)(0. Ksp = (aLa)(aIO3)3 Ksp = (γLa)[La3+](γIO3)3[IO3−]3 Ksp = (γLa)[La3+](γIO3)3Ksp' Ksp Ksp' = (γLa)(γIO3)3 The ionic strength is 0. We will neglect competing equilibria. If some amount of an ion involved in a precipitation equilibrium is present or added from an additional source other than the precipitate.00010 M)2 = 1. Ksp' = Ksp = 7. Q = [Ca2+][F−]2 Q = (0. the reaction quotient. One way is to think of the high ion concentration as “screening” the electrostatic attraction of oppositely charged ions that are involved in an equilibrium. Wiley. 19.9×10−4 M 21. New York. in higher ionic strength environments.90)3 Brian M.3×10−9). Tissue. Thus. Neglecting competing equilibria.50×10−12 = (s)(3s)3 = 27s4 s = 6. The result is higher ion concentrations relative to neutral species.50×10−12 Ksp' = = 1. and compare it to Ksp' (since we are neglecting activity effects we will use Ksp' = Ksp = 5. If Q exceeds Ksp'. (J. Neglecting ionic strength effects. BaCO3(s) ⇌ Ba2+(aq) + CO32−(aq) + H2O ↕ HCO3−(aq) + OH−(aq) Brian M. Since CO32− has the largest Kb' of these anions. [IO3−] = 0.010 M+ 3s Thus Ksp' = (s)(0. Basics of Analytical Chemistry and Chemical Equilibria. it will react with water to the greatest extent.1×10−12 Since this metal hydroxide has the highest Ksp.5×10−15 (b) Mg(OH)2 Ksp = 7. (J. .010 M + 3s)3 = 1. Thus for the listed insoluble Ba salts.Setting up the equilibrium problem Ksp' = [La3+][IO3−]3 = 1.010 M The approximation in the calculation is reasonable and our answer is 1. 2013). (a) Cd(OH)2 Ksp = 4. the solubility of BaCO3 will be affected the most by a competing equilibrium.9×10−11 s = 1. it is predicted to be the least soluble and will precipitate first as an acidic solution is made more basic. New York. it is predicted to be the most soluble and will dissolve first as a basic solution is made more acidic.9×10−11 The equilibrium concentration of IO3− is the total that comes from both the sodium iodate and the lanthanum iodate.2×10−15 (d) Zn(OH)2 Ksp = 1×10−17 Since this metal hydroxide has the smallest Ksp.010 M)3 = 1. Tissue.9×10−5 M Checking our assumption: 3(1.9×10−11 Try neglecting 3s compared to 0.9×10−5 M.010 M Ksp' = (s)(0. (c) Pb(OH)2 Ksp = 1. 23. 22.9×10−5) << 0. Wiley. : FeS2(s) + 14Fe3+(aq) + 8H2O ⇌ 15Fe2+(aq) + 2SO42−(aq) + 16H+(aq) where the S is oxidized from −1 in disulfide. aluminum is produced by reducing aluminum oxide: Al2O3 + 6e− → 2Al in molten cryolite (sodium aluminum fluoride) in a giant steel furnace at 960 C. etc. batteries. Wiley. S22−. then a carbon species is oxidized to carbon dioxide. corrosion. Sn2+ will be oxidized to Sn4+ (b) Cl2 will be reduced to Cl−. 5. (b) A significant environmental redox process is the oxidation of minerals. Pb(s). There is no chemical species directly oxidized in the furnace. Sn(s).g. 3. 8 e− transferred. (b) CH4(g) + 2O2(g) ⇌ CO2(g) + 2H2O(g). (a) Technologically important redox processes include smelting. If the electric current is supplied by a fossil fuel burning generating plant. thereby reducing H+(aq) to H2(g).0 V can be oxidized. to the +7 oxidation state in sulfate and Fe3+ is reduced to Fe2+. New York. I− will be oxidized to I2 2. (d) 4Zn(s) + NO3−(aq) + 6H2O + 7OH−(aq) ⇌ 4Zn(OH)42−(aq) + NH3(aq). (c) Zn(s) + 2MnO2(s) ⇌ ZnO(s) + Mn2O3(s). . 2013). 15 e− transferred. (J. All of the metals that have an E value more negative than 0.5Cl2(g) + 5NO3−(aq) + 2H+( aq).Chapter 8: End-of-Chapter Solutions 1. Brian M. fuel cells. (a) 3ClO3−(aq) + 5NO(g) + H2O(l) ⇌ 1. As one example. Basics of Analytical Chemistry and Chemical Equilibria. (a) Cl2 will be reduced to Cl−. 2 e− transferred. (a) no reaction (b) no reaction (c) Ni(s) + Cu2+(aq) → Ni2+(aq) + Cu(s) (d) no reaction 4. etc. e. Tissue. 8 e− transferred. An electric current passed through the cell causes the reduction. 763V (c) The contents of the two half-cells are identical.68×10−2 mol Fe2+ 1 mol MnO4− 8. Wiley.257V (b) All species are at standard concentrations. Using the Nernst equation to calculate E for the overall equation: E = Eo − 0. New York.150 M) = 3.001 M) + Ag(s) where the concentrations in parentheses are the starting concentrations.02240 L)(0. (a) All species are at standard concentrations.0 = 0.257) + 0. Tissue. Brian M.763) + 0. (a) MnO4−(aq) + 8H+(aq) + 5Fe2+(aq) ⇌ Mn2+(aq) + 5Fe3+(aq) + 4H2O (b) From the balanced reaction. so E = −(−0. 9. the stoichiometric factor is 2 mol 2S2O32− to 1 mol ClO−.0592 V 0.16 V 7. the stoichiometric factor is 1 mol MnO4− to 5 mol Fe2+.0V (d) The same species are in each half cell (so E = 0. photosynthesis.36×10 mol MnO4−) 5 mol Fe2+ = 1. 2013).0 V).5 M) → Ag+(aq) (0.(c) Biological redox processes include cellular respiration (citric acid cycle).001 M log 1 0.36×10−3 mol MnO4− −3 (3. (c) (0.16 V) = 0. so there is no driving force for a reaction and E = 0.0 = 0. so E = −(−0. and a whole bunch more. but there is a driving force to equalize Ag+ concentrations. (a) ClO−(aq) + 2I−(aq) + 2H+(aq) ⇌ Cl−(aq) + I2(aq) + H2O (b) I2(aq) + 2S2O32−(aq) ⇌ 2I−(aq) + S4O62−(aq) (c) From the combination of the two balanced reactions. . (J. The reaction is: Ag(s) + Ag+(aq) (0.0V − (−0.5 M E = 0. Basics of Analytical Chemistry and Chemical Equilibria. 6. 10. 2013). . The spontaneous reaction is: Pb(s) + PbO2(s) + 2SO42−(aq) + 4H+(aq) ⇌ 2PbSO4(s) + 2H2O Recharging drives this reaction in reverse: 2PbSO4(s) + 2H2O ⇌ Pb(s) + PbO2(s) + 2SO42−(aq) + 4H+(aq) During charging solid lead sulfate is converted to solid lead metal and solid lead oxide on the plates of the battery and the concentration of sulfuric acid in solution increases. Because vitamin C can undergo a one-electron oxidation to a stable intermediate species. The advantages and disadvantages of potentiometric measurements versus wet chemical methods are similar to those of any instrumental method compared to a classical method: sensitivity: potentiometry can be very sensitive (μM or ppm) wet chemical less sensitive Brian M.Since concentrations are 1. 1. we can work with E values: Ecell = 0. 12.0 M.8×1037 As expected for the large cell potential.7626 = 1. 11. New York. Wiley.340 + 0. (J. the equilibrium constant is large. Vitamin C is easily oxidized.103 V. Tissue. Both are limited to measurement of analytes in solution and both can be specific for a given oxidation state. it reacts faster than other biochemical species that require a two-electron reaction. measurement of: experimental: set up: analytes: general: potentiometry voltage passive measurement indicator and reference electrodes ions only electrode is selective for a given ion voltammetry current variable voltage applied usually three electrodes ions and neutral species can be used for analyte identification 13. It therefore serves as a sacrificial reagent to prevent strong oxidizing agents from damaging other molecules in biological cells. Basics of Analytical Chemistry and Chemical Equilibria.103 V = RT ln(K) nF K = 1. Voltammetry provides more parameters to vary and can be used to study physical processes or for analyte identification. 055 V.055 V = 2. interfering ions might require masking or removal. field kits can measure limited sample amounts 14. (a) Mg2+ in seawater: very high ionic strength. 15.0592 V)log[NO3−] (c) Ecell = Econst + (0.303. can analyze many samples field use: easily portable depends on specific analytes/interferences interferences can require prior removal traceable to primary standards rather slow due to solution manipulation mostly lab based. (J.0592 V)log[NH4+] (d) Ecell = Econst − (0. Inserting the constants and n = −1 (for F−): 8. Brian M. Per our convention in the chapter. equals (RT/nF)×2. . I use concentration symbols rather than activity for clarity in these expressions: (a) Ecell = Econst + (0. Since standards and samples were treated the same and measured on the same day. the deviation from the theoretical slope should not affect the accuracy of the unknown measurement. (b) The measured slope.0296 V)log[Mg2+] (b) Ecell = Econst − (0. Wiley. or there is some deviation from the theoretical slope (quite common). (a) For this data the lowest concentration point is not linear with the other points. so the linear range of the calibration data is from 10−4 to 10−1 M F−. Either the lab was quite chilly. standards must match samples (b) NO3− in a freshwater stream: should be free of interferences and measurable directly with an ISE (c) NH4+ in a sample of solid fertilizer: sample must be dissolved and pH controlled (d) Cl− in stainless steel: sample must be dissolved (not trivial for SS).3145 J mol−1 K−1(T) −0.selectivity: usually very selective interferences: calibration: general use: levels of interferences are documented requires calibration standards rapid. Basics of Analytical Chemistry and Chemical Equilibria. −0. 2013).0592 V)log[Cl−] 16. Tissue.303 96485 C/mol gives an absolute temperature of 277 K or 4 C. New York. Basics of Analytical Chemistry and Chemical Equilibria. (b) Taking the full range of the x scale. Based on Figure 8.0. Wiley. where a is the activity of the analyte ion. Setting up a ratio for the second analyte being 1. Cyclic voltammetry will detect all analytes present as the voltage is scanned. By doing a two-point calibration.0x−149. An ion-selective electrode will be selective for only one specific analyte ion. Voltammetry will detect neutral species in addition to ions. 17. they have equal diffusion constants. The intercept will change if reference or junction potentials in the circuit change. i.e. 2013).3×10−4 M for the first analyte. If voltages of different components in a mixture overlap. both the slope and intercept are corrected to establish an accurate linear relationship for the pH readout.0 A. Inserting 48 mV for y gives an x of −3. . The slope will change if the temperature of the experiment changes. other components will interfere with measurement of the analyte. New York.13: (a) The width of the transition for each analyte is approximately 60 mV. the scan range of 700 mV (−100 mV to −800 mV) takes 350 s. Brian M. The slope and intercept in this relationship are variable. The scan rate is (700 mV)/(350 s) = 2. which are also sensitive temperature. so potentials > 60 mV are necessary to measure the waves separately. Two analytes with reduction potentials closer than 60 mV would overlap and be difficult to distinguish.58. 18. The first analyte has a current signal of ≈ 9.6 10−4 M. 19.(c) The calibration curve is y = −55. (c) The second analyte has a current signal of ≈ 7. In any potentiometric measurement there is a linear relationship between measured potential and log(a). Taking the inverse log gives [F−] = 2. (J.0 A. Other ions will not interfere unless they are present at higher concentrations than the analyte. (d) The direct ratio used in the previous answer assumes that the analytes are similar..00 mV/s. Tissue.0×10−4 M gives a concentration of 1. 1 ppm) usually very selective interferences for specific ion-selective electrodes are usually known and correctable with selectivity factors and additional measurements requires calibration standards oxidation-state specific atomic absorption spectroscopy lab based destroys sample more expensive very sensitive ( 0. i.e. so intrinsically selective high concentrations of refractory substances can reduce atomization efficiency requires calibration standards total analyte concentration 3. 2013). Tissue. New York.1 ppm typical) narrow line. (J. neither provides information on analyte speciation. Atomic spectroscopy is generally more sensitive and provides total analyte concentration. general sensitivity selectivity interferences calibration speciation potentiometry can be portable and used in-situ preserves sample simpler and less expensive sensitive (10−5 M. . both methods destroy the sample and measure the total concentration of a given element. Due to the atomization requirement. In this case ashing the sample will reduce interferences. Brian M. Obtaining speciation-specific results is possible by separating analytes before measurement. The two methods are often used together to provide complementary information. AES and AAS are both very sensitive for alkali and alkaline earth metals and will be the faster and preferred method. Basics of Analytical Chemistry and Chemical Equilibria. Ca2+ and Mg2+ ISEs can be used after digesting the sample and using the standard addition method for calibration. The key difference between potentiometry and atomic spectroscopy is that an ion-selective electrode can be used in situ and it can be specific for a given speciation of an element. Wiley. but both atomic absorption spectroscopy (AAS) and atomic absorption spectroscopy (AES) will have analyte-specific exceptions. scope sensitivity atomization sources AAS usually single analyte ppm range for flame AAS can be more sensitive than AES for easily ionized elements sufficient signal at lower temperature (simpler and cheaper) AES simultaneous multi-element analysis ppb range for ICP-OES in general AES is more sensitive than AAS excitation efficiency increases with temperature 2.. The following comparison provides general trends. (a) Ca2+ and Mg2+ in leafy vegetables: The matrix is expected to contain multiple substances that can complex the analytes.Chapter 9: End-of-Chapter Solutions 1. (f) Co in steel: Spark source AES is sensitive and rapid (does not require dissolving steel).1251 ppm−1)[Pb2+] + 0. ICPMS is suitable with prior separation of perchlorate from other anions. The inductively coupled plasma (ICP) generates a much higher temperature than a flame. Surface water will be a reasonably clean matrix.1 ppm for flame AES). (c) multiple metals: ICP-AES will provide simultaneous multi-element quantitation at the ppb level.1251 ppm−1 and the calibration equation is y = (0. 2013). New York. (d) NH4+ ISE.033.50 ppm 6. Wiley. 4.1251 ppm−1)[Pb2+] +0. Brian M. 5.90276 u = 0. Basics of Analytical Chemistry and Chemical Equilibria.91460 u − 111. 60008000 K versus 1700-2700 K. The higher temperature results in greater atomization and excitation efficiencies.284 – 0. so detection limits are lower (1 ppb for ICP-AES vs. (J.0 ppm) = 0. For the unknown measurement: 0. These two measurements will distinguish the analytes from each other and from other nitrogen species.(b) NO3− and NH4+ in surface water: nitrate and ammonium ISEs will be the preferred analysis.01184 u The required resolution is approximately 10. same reason as for the NO3− ion (e) ClO4− in sea water: A perchlorate ISE is suitable with correction for Cl− in the seawater. A simple interpolation between the calibration data is sufficient.000. Using this value and the approximate m of 112: m 112 u = = 9460 ∆m 0. The difference in these two masses is 111.033)/(10.033 [Pb2+] = 3. . The slope is (1. but the equation is also easy to determine.471 = (0. Tissue.01184 u. 0. .9969 0.7 0.1 0 0 20 40 60 80 100 Ca concentration (ppm) Brian M.3 0.952.6 0.0140 ppm−1)[Ca2+] [Ca2+] = 32. I chose to force the calibration function through zero.0140x R² = 0.5 y = -1E-04x2 + 0. (J.2 0.7 0.4 0. 0. Tissue.9 0. Wiley.2 0. There is only a small difference in the result if the trendline is allowed to vary the intercept.455 = (0. (c) The unknown measurement is outside of the linear range of the experiment. New York. The analyst should be conservative and quote a result of > 50 ppm Ca. If you discover the situation as you make measurements.9 0.7. a matrix effect appears to create a systematic error compared to the calibration data. You have two options.5 ppm Given that the blank measurement was zero. For a measurement of 0.8 Absorbance 0. If you discover the situation after the instrument is turned off. The response in atomic spectroscopy is known to be non-linear at higher concentrations.4 0. 0.5 0.1 0 0 20 40 60 80 100 Ca concentration (ppm) (b) Using the trendline shown in the plot: 0. 2013). you can fit the calibration data to a 2nd-order polynomial (see plot below).9984 0. you can dilute the unknown by a factor of 2 and measure the AES signal again.0181x R² = 0. Basics of Analytical Chemistry and Chemical Equilibria.6 0.3 y = 0.8 Absorbance 0. (a) A plot shows that the data is linear from 0 to 50 ppm. Tissue. decreases rapidly as a function of DISC. . Basics of Analytical Chemistry and Chemical Equilibria. On the right vertical axis is S/N for the open circles. Wiley. The purpose of this question is to think about data and the resulting signal-to-noise ratio. 2013). Subtracting the high background at lower DISC introduces greater uncertainty in the result. The following plot shows signal counts on the left vertical axis for the data in the table. New York. The details of setting a discriminator level for a detector is more advanced than we need to consider. 10000000 3000 9000000 2500 8000000 Signal (cps) 7000000 2000 6000000 5000000 1500 S/N 4000000 1000 3000000 2000000 500 1000000 0 0 0 5 10 15 20 25 30 35 40 DISC (mV) Brian M. The open circles are found from (test solution – blank) √blank S/N = where the square root of the blank serves as an estimate of the noise. The signal of the test solution is the summation due to the analyte and the blank.8. It also falls off rapidly versus DISC. Looking at the tabulated data. due to background. the analyte signal at DISC = 40 mV is 179000 cps – 5920 cps = 173000 cps. but levels off at a higher level than the blank. the signal of the blank. For example. (J. Multiple GFAAS measurements can be made if sufficient sample is available. Refractory sample constituents can be digested with aggressive acid and redox reagents. The result is a less precise measurement than when using a constant nebulizer. 13.9. Wiley. In atomic absorption spectrometry. the test portion is introduced as a small plug of liquid or solid directly into the furnace tube. The laser ablation sampling method is a third alternative.1 nm or less to resolve closely spaced lines. (J. New York.1 nm and less. or plasma. Molecular spectra consist of broad bands. A background correction technique will also remove this source of error. Tissue. Atomic spectra have very narrow linewidths of 0. requiring a longer time to complete the measurement. which has the advantage of providing spatially resolved elemental analysis on a heterogeneous sample. which keeps the analyte in solution at room temperature before measurement. The matrix effects in GFAAS can also require more extensive method validation for a given analyte and sample matrix. minimizes this loss for more accurate Hg determinations. creates background light emission that will reach the detector. In graphite-furnace atomic spectroscopy (GFAAS). The cold-vapor technique. They are best suited for repetitive analysis of similar samples so that matrix effects can be corrected after thorough method validation. 2013). Other methods include direct introduction of the sample in graphite-furnace AAS or spark source AES. 11. The main disadvantage of GFAAS is that it measures a transient signal and it is not possible to signal average during any one measurement. These methods are rapid. An atomic spectrometer must have a resolution on the order of 0. a flame. These background sources can vary depending on the matrix of the sample. The downside is the potential loss of analyte in this process. The atomization source. Since only microliters of the test portion are needed. Brian M. . This background signal is not related to analyte concentration and is subtracted from the detector signal. Basics of Analytical Chemistry and Chemical Equilibria. the absorption signal might be affected by absorption due to molecular species. measurements can be made when the amount of sample is very limited. 12. often 50 to 100 nm wide. Spectrometers for molecular absorption and fluorescence will have resolution on the order of 1 nm. furnace. The high vapor pressure of Hg leads to loss of analyte when handled in conventional ways. 10. . The atomic emission lines of different isotopes will have slightly different wavelengths. Wiley. on the order of 500. Isotopes will differ by one or more mass units. Tissue. that absorb this light strongly. Basics of Analytical Chemistry and Chemical Equilibria. Brian M. This region is called the vacuum ultraviolet (VUV) region because working in this region requires evacuation of atmospheric gases. and for stable isotopes of elements requires a mass spectrometer with moderate resolution. 2013). The nonmetals have atomic transitions at wavelengths shorter than 190 nm. ICP-MS is usually the method of choice. Measuring isotopic ratios is usually done by mass spectrometry. but the differences are small and requires a very high resolution spectrometer to resolve. mainly oxygen. New York. (J.14. Although ICP-OES is capable of measuring nonmetals with a suitable evacuated spectrometer. 15. g. (J. GC detector TCD (thermal conductivity detector) FID (flame ionization detector) ECD (electron capture detector) mass spectrometer advantage universal non-destructive sensitive high dynamic range selective for electronegative functional groups very sensitive for select analytes non-destructive provides structural information for solute identification disadvantage less sensitive than other detectors destroys analyte limited dynamic range more complex maintenance and operation destroys analyte There are a number of other more specialized GC detectors that are not discussed in the text. ethylbenzene. but many organic analytes have similar spectra. (c) The refractive index and UV-vis absorption detectors (for LC) and the thermal conductivity detector (for GC) are non-destructive. New York. Some examples include:  photoionization (PID): selective for aromatic compounds. Tissue. (a) Chromatographic detector ECD (electron capture detector) FID (flame ionization detector) fluorescence mass spectrometer RI (refractive index) detector TCD (thermal conductivity detector) UV/Vis absorption detector Use GC only GC only LC only GC and LC LC only GC only LC only (b) Of the choices. Basics of Analytical Chemistry and Chemical Equilibria. UV-Vis absorption can provide some identification. 2.Chapter 10: End-of-Chapter Solutions 1. and xylenes (BTEX)  flame photometric detector (FPD): flame chemiluminescence for compounds containing sulfur and phosphorous  nitrogen phosphorous detector (NPD): plasma ionizer for compounds containing nitrogen and phosphorous  dry electrolytic conductivity detector (DELCD): an ECD replacement for chlorine and bromine containing compounds Brian M. only the mass spectrometer provides molecular structure information.. 2013). touluene. Wiley. e. . benzene. (J. Basics of Analytical Chemistry and Chemical Equilibria. benzene. 5. either GC or HPLC should be suitable techniques. which can approach the stationary phase closer. Ions in solution have an electrostatic attraction to immobilized ions on the stationary phase. the disadvantage of GC is that there may not be an optimum stationary phase to achieve narrow symmetric peaks for all analytes.p. (c) The analytes are volatile organic compounds so use capillary GC column.p. Based simply on the structures. (a) The analytes are anions so use the anion-exchange HPLC column. A blood matrix will require an extraction step and the barbiturates are often derivatized. = 157 C) The deciding factor between reversed-phase and normal-phase partition chromatography is usually based on solubility of the solutes. then GC will be advantageous. The expected order is: n-hexane (b. benzene. Tissue. and n-hexanol (less polar to more polar) the more polar analytes will interact more strongly and be retained longer (c) capillary GC with nonpolar stationary phase In this case we expect the analytes to elute in order of boiling point (b. reversed-phase will be the preferred method for these analyte. . (b) The analytes are cations so use the cation-exchange HPLC column. Wiley.p. Brian M. choose the nonpolar polydimethyl siloxane stationary phase. Attempting to inject inorganic ions into a GC results in decomposition or depositing non-volatile salts in the injector. Given that the alcohol is more polar than the other solutes. The main advantage of GC is the superior resolution compared to HPLC. Since the analytes are nonpolar. 4. are attracted more strongly and are retained longer on the column.p.p. If these solutes must be separated from other components. The rationale is the same as for anions.3. (a) reverse-phase partition liquid chromatography: n-hexanol. = 68 C) benzene (b. = 80 C) n-hexanol (b.) from lowest to highest b. The stationary phase will depend on the product of the derivatization reaction. 2013). Ions of higher charge and smaller size. (d) Answering this question requires some research into the structure of barbiturates. Earlier analytical methods used GC analysis. The strength of the interaction depends on the charge and size of the analyte ions. New York. Give the reasonable solubility of the listed solutes in a water-based mobile phase. and n-hexane (more polar to less polar) the nonpolar analytes will interact more strongly and be retained longer (b) normal-phase partition liquid chromatography: n-hexane. 8.85). All analytes are anions at the initial pH. 2013). If they were injected into a gas chromatograph they would decompose rather than vaporize intact.0 2.9 ccaffeine = 0.2 7. a strong anion exchange stationary phase is preferred. 6. .2 ctheobromine = 1. (a) To maintain these acids as anions.4 ctheophylline = 0. As the weakly retained analytes elute from the column.2 7.3 peak height (arb.7 4. New York. To be safe. (J.0 µg/mL = 2.6. The data is summarized here: analyte theobromine theophylline caffeine retention time.7 µg/mL ctheophylline 1.6.0 12. Basics of Analytical Chemistry and Chemical Equilibria. the pH can be lowered to begin protonating the more strongly retained analytes. Tissue. The calculation for each analyte is a simple proportionality: ctheobromine 1.74 µg/mL Brian M. pKa = 4.7. Again SPE cleanup will be needed due to the complex blood matrix. units) 12. (a) The peak height of each analyte is taken from Table 10.Since these molecules are water soluble. The large size of proteins makes them very non-volatile.9 3. the C18 HPLC column is being developed as a standard analytical method. Wiley. (b) Since these analytes are weak acids. Since the weak acids become neutral when protonated. the mobile phase pH must be greater than the analyte with the highest pKa (niacin. 7. The identity of the peaks is made by matching the retention times to the standard data in Table 10. a pH > 6. tR (min) 1.56 µg/mL ccaffeine 1. The stationary phase will not change as a function of mobile phase pH.0 1.9 The concentration of each analyte is determined by converting the test portion peak height to concentration using the calibration data in Table 10.85 will keep > 99 % of the niacin deprotonated.0 µg/mL = 12.0 µg/mL = 7. these analytes will then elute from the column. Solutes that do not have a mass fragment at the selected m/z will not appear in the chromatogram.7 µg/mL)(30 mL) = 50. Table of MS designs mass analyzer design magnetic-sector advantage simultaneous detection of several m/z – best precision for isotope ratio measurements disadvantage slow scan rate quadrupole compact and inexpensive rapid scan rate moderate resolution limited scan range (≈ 5000 Da maximum) Brian M. making it simpler to interpret and reducing the chance of interferences affecting a quantitative measurement. The largest signal provides the best signal-to-noise ratio and thus the greatest sensitivity.8 µg wttheophylline = (0.9 µg 50. Selective ion monitoring (SIM) displays the ion current signal at only one m/z.045 % caffeine 9.8 µg 50. .10 % theobromine 16. so the amount of each analyte was: wttheobromine = (1.74 µg/mL)(30 mL) = 22.(b) Given the results in part (a). Basics of Analytical Chemistry and Chemical Equilibria.000 µg × 100 % = 0. 10. The total ion chromatogram (TIC) is the summation of all peaks and provides the largest analytical signal. This mode provides greater selectivity by picking out the parent ion or an intense fragment ion of the analyte of interest.9 µg wtcaffeine = (0.000 µg × 100 % = 0. (J.000 µg × 100 % = 0.3 µg Dividing each weight by the 50 mg sample portion and multiplying by 100 % gives: 50. Wiley.56 µg/mL)(30 mL) = 16.3 µg 50. The analyte was contained in 30 mL of extracting solvent. the weight percent of the analytes in the cocoa sample is determined by correcting for sample preparation. 2013). New York. Tissue.034 % theophylline 22. 12. Smaller proteins migrate faster and travel farther in gel electrophoresis. it can often be difficult to know what method to choose. The buffer maintains the system at a pH where the proteins are stable. Brian M. The acrylamide concentration and the relative amount of cross linker (usually bisacrylamide) added during preparation of the gel determines the pore size. in this case amino acids. In some applications it is possible to irradiate the gel with ultraviolet light to detect the biomolecules using fluorescence. The biomolecules in a gel are usually visualized by adding a stain. 2013). In many cases the method selection is determined by practical reasons. Basics of Analytical Chemistry and Chemical Equilibria. New York. the pore size must be appropriate to match the proteins being separated.time-of-flight high resolution (15. I provide a general discussion here to help you understand the methods that you might have discovered. In discontinuous gel electrophoresis. Wiley. 14. When multiple analytical methods are suitable for a given class of analytes. Since proteins have a large range of sizes. Since preparation of the gel can vary day-to-day. and reagents to fine tune the polymerization. (J. 13. measurements are made relative to molecular weight standards in one or more separate lanes of the gel. . Tissue. or by constraints such as time or cost. a reducing agent to break disulfide bonds. The most common stains are coomassie brilliant blue for proteins and ethidium bromide for nucleic acids. Using SDS surfactant gives all of the proteins a negative charge.000) rapid scan rate pulsed operation requires interface to some ion sources ion trap can accumulate ions moderate resolution useful as an interface between limited scan range (≈ 5000 a continuous ion source and a Da maximum) time-of-flight mass analyzer orbitrap highest resolution expensive 11. Most labs that use SDSPAGE will have detailed recipes and protocols that are optimized for their applications. Silver stain works for either type of biomolecule and provides greater sensitivity. what instrument do you have. Other additives besides the buffer and SDS surfactant include urea to denature proteins. so they migrate from the negative electrode towards the positive electrode. the change in buffer pH “stacks” the proteins before entering the running gel. The electrodes provide the electrostatic driving force to move the charged proteins (or other type of macromolecules). Large proteins are hindered to a greater extent by the pore size of the gel. A chief scientific reason is the complexity of the sample matrix. 15. methylarsonic acid. There is certainly many. Wiley. 2013). Capillary electrophoresis provides the highest resolution. Brian M. As(V). Tissue. Basics of Analytical Chemistry and Chemical Equilibria. As(III). they can be separated by anion-exchange chromatography. The effluent of the chromatography column can enter an ICP-MS to measure each arsenic species separately. Despite the greater complexity to achieve repeatable results. and dimethylarsinic acid.The high resolution of GC allows rapid chromatography. As(V). I did a Google search on “capilliary electrophoresis application note” to find manufacturer literature. many more papers in the primary literature. arsenate. 16. CE might be the best choice is the amino acids must be separated from similar interferences. but requires derivatization of the amino acids. Since all of these compounds can form anions at high pH (≈11 or higher). but will usually require longer run times than GC. HPLC is readily amenable to amino acids in aqueous solution. As(V). New York. Common forms of arsenic are arsenite. (J. A few examples that I found are: solutes carbonic anhydrase and related proteins antibodies # 5 detection limit Agilent < 10 Horiba therapeutic proteins < 10 Beckman-Coulter speed notes isoelectric point determination detection of glycosylation 30 min A common aspect that I found in my limited search is the focus on biomedical analytes. .