UNIVERSITY OF THE PHILIPPINES MANILAComplex-Formation Titrations COMPLEX-FORMATION REACTIONS Copper(II) ion forms a neutral complex with glycine, Cu(NH3CH2COO)2. .. NH2 O Cu2+ O O Cu NH2 NH2 O O CH2 + H H .. .. OH H2 C + 2H+ Most metal ions form stable complexes with EDTA. O O O Mn+ + HO HO O H2 .. .. C N C C N H2 H2 C H2 H2 O C OH C H2 OH O O O O O M N N The Health Sciences Center Billones Lecture Notes UNIVERSITY OF THE PHILIPPINES MANILA Complexometric titrimetry are titrimetric methods based on complex formation. It is based upon a particular class of coordination compounds called chelates. A chelate is produced when a metal ion coordinates with two (or more) donor groups of a single ligand to form a five- or six-membered heterocyclic ring. Multidentate ligands, particularly those having four or six donor groups, have two advantages over the unidentate ones. 1) react more completely with cations and thus provide sharper end points 2) react with metal ions in a single-step process The Health Sciences Center Billones Lecture Notes UNIVERSITY OF THE PHILIPPINES MANILA Curves for Complex Formation Titrations 20 1:1 2:1 4:1 pM 10 Much sharper end point is obtained with a reaction that takes place in a single step EP 0 Volume of Titrant 0 Multidentate ligands are ordinarily preferred for complexometric titrations. The Health Sciences Center Billones Lecture Notes UNIVERSITY OF THE PHILIPPINES MANILA TITRATIONS WITH AMINOCARBOXYLIC ACIDS Ethylenediaminetetraacetic acid (EDTA) EDTA is the most widely used complexometric titrant. The EDTA molecule has six potential sites (i.e. hexadentate) for bonding a metal ion; the four -COOH groups and the two -NH2 groups. 1.0 α0 H 4Y EDTA species: H4Y, H3 H2Y2-, HY3-, and Y4-. Y-, α1 α2 H2Y2- α3 HY3- α4 Y4- α 0.5 0 0 The Health Sciences Center H3Y- 2 4 6 pH 8 10 12 14 Billones Lecture Notes Only at a pH values greater than 10 does Y4. Ag+ + Y4Al3+ + Y41:1 The Health Sciences Center AgY3AlYM : EDTA Billones Lecture Notes .become a major component of solutions.UNIVERSITY OF THE PHILIPPINES MANILA It is apparent that the H2Y2.predominates in moderately acidic media (pH 3 to 6). COMPLEXES OF EDTA AND METAL IONS EDTA reagent combines with metal ions in a 1:1 ratio regardless of the charge on the cation. .. N C C N H2 H2 . ....Structure of H4Y and its dissociation products HOOC H N C C N H H2 H2 -OOC COOH + + UNIVERSITY OF THE PHILIPPINES MANILA COO-OOC COO- H N C C N H H2 H2 -OOC COO- + + H 4Y H2Y-2 -OOC H N C C N H H2 H2 -OOC COOH + + COO- -OOC H N C C N H2 H2 -OOC + COOCOO- H3Y- -OOC -OOC . The Health Sciences Center .. Billones Lecture Notes Y-4 . . HY-3 COOCOO. 92 x 10-7 K4 = 5.50 x 10-11 Formation constants KMY for common EDTA complexes The constant refers to the equilibrium involving the deprotonated species Y4.with the metal ion: Mn+ + Y4The Health Sciences Center MY(n-4) KMY = [MY(n-4)+] [Mn+][Y4-] Billones Lecture Notes .14 x 10-3 K3 = 6.UNIVERSITY OF THE PHILIPPINES MANILA The dissociation constants for the acidic constants for the acidic groups in EDTA are K1 = 1.02 x 10-2 K2 = 2. 50 16.9 23.32 8.3 x 108 5.3 x 1021 1.2 x 1013 2.3 x 1025 7.46 21.8 x107 6.6 x 1023 logKMY Ag+ Mg2+ Ca2+ Sr2+ Ba2+ Mn2+ Fe2+ Co2+ Ni2+ 7.1 x 1014 2.2 x 1016 2.UNIVERSITY OF THE PHILIPPINES MANILA Cation KMY 2.2 x 1018 logKMY Cation KMY 6.13 25.79 14.9 x1025 1.62 Cu2+ Zn2+ Cd2+ Hg2+ Pb2+ Al3+ Fe3+ V3+ Th4+ 18.04 16.80 16.33 16.3 x 1018 3.2 Billones Lecture Notes The Health Sciences Center .70 8.9 x 108 5.1 x 107 4.3 x 1016 1.1 25.1 x 1018 1.76 13.0 x 1010 4.0 x 1016 4.63 7.69 10.9 x 1016 6.31 18.80 18. UNIVERSITY OF THE PHILIPPINES MANILA Equilibrium Calculations Involving EDTA A titration curve for the reaction of a cation Mn+ with EDTA consists of a plot of pM versus reagent volume. The α4 for H4Y would be defined as α4 = [Y4-] cT [Y4-] = α4cT where cT is the total molar concentration of uncomplexed EDTA: cT = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] The Health Sciences Center Billones Lecture Notes . Calculating [Mn+] in a buffered solution containing EDTA is a relatively straightforward procedure provided the pH is known. Values for pM are readily computed in the early stage of a titration by assuming that the [Mn+] is equal to cM. we substitute α4cT for [Y4-] in the formation-constant expression: Mn+ + Y4KMY = [MY(n-4)+] [Mn+][Y4-] MY(n-4) but [Y4-] = α4cT KMY = [MY(n-4)+] [Mn+]α4cT The Health Sciences Center Billones Lecture Notes .UNIVERSITY OF THE PHILIPPINES MANILA Conditional Formation Constants. To obtain K’ for the equilibrium shown in the equation. K’ (pH dependent K) Conditional or effective formation constants are pH dependent equilibrium constants that apply at a single pH only. Computation of α4 Values for EDTA Solutions α4 = ___K1K2K3K4_____________ __ [H+]4 + K1[H+]3 + K1K2[H+]2 + K1K2K3[H+] + K1K2K3K4 or αHY3- α4 = K1K2K3K4 D = K1K2K3[H+] αH2Y2.UNIVERSITY OF THE PHILIPPINES MANILA Combining the two constants yields a new constant K’MY K’MY = [MY(n-4)+] = α4KMY [Mn+]cT K’MY = α4KMY K’MY describes equilibrium relationships only at the pH for which α4 is applicable.= K1K2[H+]2 D D Billones Lecture Notes The Health Sciences Center . 00.0 4.2 x 10-2 3.5 x 10-1 9.0 10.0 8.5 x 10-7 2. α4 at Selected pH Values pH 2.UNIVERSITY 3OF THE PHILIPPINES MANILA αH3Y.at pH 2.0 7.6 x 10-9 3. K2.0 3.0 6.7 x 10-14 2.8 x `10-1 Only about 4 x 10-12 percent of the EDTA exists as Y4.5 x 10-1 8.0 11.= K1[H+] αH4Y = [H+]4 D D where K1. Billones Lecture Notes .5 x 10-11 3.0 12.0 The Health Sciences Center α4 3. K3 and K4 are the four dissociation constants for H4Y and D is the denominator.4 x 10-3 5.0 5.8 x 10-4 5.0 9.2 x 10-5 4. we obtain K1 = 1.02 x 10-2 K1K2 = 2.02 x 10-2)(6.56 x 10-33 K1K2[H+]2 = (2.Example UNIVERSITY OF THE PHILIPPINES MANILA Calculate α4 and the mole percent of Y4.20) = 6.31 x 10-11)4 = 1.31 x 10-11)3 = 2.31 x 10-11 From the values for the dissociation constants for H4Y.31 x 10-11)2 = 8.31 x 10-22 Numerical values for the several terms in the denominator: [H+]4 = (6. [H+] = antilog (-10.58 x 10-41 K1[H+]3 = (1.68 x 10-26 The Health Sciences Center Billones Lecture Notes .18 x 10-5)(6.18 x 10-5 K1K2K3 = 1.in a solution of EDTA that is buffered to pH 10.51 x 10-11 K1K2K3K4 = 8.20. 20.47 x 100% = 47% Only the last two terms in the denominator contribute significantly to the sum D at pH 10.51 x 10-11)(6.K1K2K3[H+] = (1.78 x 10-21 mol % Y4.= 0. At low pH values.31 x 10-22 = 0.31 x 10-11) = 9.466 D 1.53 x 10-22 K1K2K3K4 UNIVERSITY OF THE PHILIPPINES MANILA = 8.78 x 10-21 signiﬁcant conﬁguration The equation α4 = K1K2K3K4 becomes D α4 = K1K2K3K4 = 8. in contrast.31 x 10-22 D = 1. only the first two or three terms are important The Health Sciences Center Billones Lecture Notes . NiY2- Ni2+ + Y4- Kinstab = The Health Sciences Center 1 KMY very small Billones Lecture Notes .0.0 and (b) 8.2 x 1018 [Ni2+][Y4-] The equilibrium concentration of NiY2.UNIVERSITY OF THE PHILIPPINES MANILA Calculation of the Cation Concentration in EDTA Solutions Calculate the equilibrium concentration of Ni2+ in a solution with an analytical NiY2.0150 M at a pH (a) 3.is equal to the analytical concentration of the complex minus the concentration lost by dissociation. Ni2+ + Y4- NiY2- KMY = [NiY2-] = 4.concentration of 0. 5 x 10-11 at pH 3.0.05 x 108 [Ni2+]2 The Health Sciences Center Billones Lecture Notes .UNIVERSITY OF THE PHILIPPINES MANILA [NiY2-] = 0.0150 because Kinstab is small Since the complex is the only source of both Ni2+ and the EDTA species. [Ni2+] = cT = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] Substitution of this equality gives K’MY = α4KMY = [NiY2-] = [NiY2-] [Ni2+]cT [Ni2+]2 (a) α4 is 2.5 x 10-11 x 4. Substitution of this value and the concentration of NiY2.2 x 1018 = 1.0150 = α4KMY = 2.into the equation for K’MY gives K’MY = 0.0150 – [Ni2+] ≈ 0. as assumed b) At pH 8. K’MY = 5. Thus.0150.2 x 10-5 M [Ni2+] << 0.05 x 10 ) = 1.0.0150/1.0 mL of 0.500 M EDTA.27 x 1016) = 8.0150 / 2.0300 M Ni2+ with 50. the conditional constant is much larger.UNIVERSITY OF THE8PHILIPPINES MANILA [Ni2+] = sqrt(0.00.0 mL of 0. small 1/K’MY . The mixture was buffered to a pH of 3.4 x 10-3 x 4. low [ion] Example Calculate the concentration of Ni2+ in a solution that is prepared by mixing 50. The Health Sciences Center Billones Lecture Notes .2 x 1018 = 2.1 x 10-10 M large K’MY.27 x 1016 Substitution into the equation for K’MY followed by rearrangement gives [Ni2+] = sqrt(0. 0 x 0.0500 – 50 x 0.0 x 0.0300 = 0.= 50. Thus.0150 M 100 cEDTA = 50. UNIVERSITY OF THE PHILIPPINES MANILA cNiY2.03000 = 0. the solution has an excess of EDTA.0150 M At this point.0150 – [Ni2+] ≈ 0.0100 M The Health Sciences Center Billones Lecture Notes . the total concentration of uncomplexed EDTA is given by its analytical molarity: cT = 0.Here.0100M 100 Again assume that [Ni2+] << [NiY2-] so that [NiY2-] = 0. and the analytical concentration of the complex is determined by the amount of Ni2+ (limiting reagent) originally present. 0100 = 2.0150 0.4 x 10-8 EDTA Titration Curves Example Derive a curve (pCa as a function of volume of EDTA) for the titration of 50. The Health Sciences Center Billones Lecture Notes .05 x 108 [Ni2+] = 0.0150 = α4KMY [Ni2+] cT [Ni2+] 0.05 x 108 = 1.0100 x 1.2 x 1018 = 1.00500 M Ca2+ with 0.0.5 x 10-11 x 4.Substitution into the equation gives UNIVERSITY OF THE PHILIPPINES MANILA K’MY = [NiY2-] = 0.0100 M EDTA in a solution buffered to a constant pH of 10.0 mL of 0. 00 mL reagent: [Ca2+] = 50. α4 = 0.0100 + cT ≈ 2.00500 – 10.0 The Health Sciences Center Billones Lecture Notes . which is equal to cT.75 x 1010 Preequivalence Point Values for pCa Before the EP is reached.35 and KCaY = 5. and assumed to be small. after addition of 10.35 x (5.Calculation of Conditional Formation Constant UNIVERSITY OF THE PHILIPPINES MANILA K’CaY = [CaY2-] = α4KCaY [Ca2+] cT From Table. For example. the [Ca2+] is equal to the sum of the contributions from the untitrated excess of the cation and from the dissociation of the complex.0 x 0.0 x 0.0 x 1010 Substitution gives K’CaY = 0.0 x 1010) = 1.50 x 10-3 60. 60 Other pre-EP data are derived in this way.= 50.33 x 10-3 50.0 The only source of Ca2+ ions is the dissociation of the CaY2. cCaY2.33 x 10-3 M Substituting into the conditional formation constant expression gives K’MY = [CaY2-] = 0.complex.0 x 0.50 x 10-3) = 2.00500 = 3. At EP. [Ca2+] = cT [CaY2-] = 3.UNIVERSITY OF THE PHILIPPINES MANILA pCa = -log (2.75 x 1010 [Ca2+]cT [Ca2+]2 The Health Sciences Center Billones Lecture Notes .00333 = 1. The Equivalence-Point pCa Compute first the analytical concentration of CaY2-.0 + 25.33 x 10-3 – [Ca2+] ≈ 3. 0 x 0.94 x 10-3 from ionization The Health Sciences Center Billones Lecture Notes .0 + 35.0 mL of reagent.18 x 10-3 85.94 x 10-3 M 50. we can write [CaY2-] = 2.36 x 10-7) = 6. For example.0 x 0.36 x 10-7 M pCa = -log (4.36 Postequivalence-Point pCa Beyond the equivalence point.00500 = 1. analytical concentrations of CaY2and EDTA are obtained directly from the stoichiometric data. after the addition of 35.0 x 0.0100 – 50.UNIVERSITY OF THE PHILIPPINES MANILA [Ca2+] = sqrt(0.0 As an approximation.00500 = 2. cCaY2.= 50.00333 / 1.94 x 10-3 – [Ca2+] ≈ 2.0 the complex is diluted cEDTA = 35.75 x 1010) = 4. 18 x 10-3 2.42 x 10-10) = 9.94 x 10-3 = 1.94 x 10-3 = 1. The Health Sciences Center Billones Lecture Notes .85 The approximation that [Ca2+] is small is clearly valid. Curve A (red) in the following figure is a plot of the data for the Ca titration in the example.42 x 10-10 1.UNIVERSITY OF THE PHILIPPINES MANILA cT = 1. Curve B is the titration curve for a solution of magnesium ion under identical conditions.18 x 10-3 x 1.18 x 10-3 M from ionization Substitution into the K’CaY expression gives K’CaY = [Ca2+] = 2.18 x 10-3 + [Ca2+] ≈ 1.75 x 1010 [Ca2+] x 1.75 x 1010 pCa = -log (1. 10 pCa pM 8 6 pMg The shaded areas show the transition range for Eriochrome Black T (EBT).00500 M Ca2+ (K’ for CaY2.72 x 108) at pH 10.UNIVERSITY OF THE PHILIPPINES MANILA EDTA titration curves for 5.75 x 1010) and Mg2+ (K’ for MgY2.+ MgY2blue 4 2 blue 25.+ HY3red HIn2.+ CaY2- MgIn.0.= 1.= 1.+ HY3red EP HIn2.0100 M EDTA The Health Sciences Center Billones Lecture Notes .0 mL 0 Volume of 0.0 mL of 0.0 mL EP 35. CaIn. and hence K’CaY.0100 M EDTA 10 8 pH 12 pH 10 pH 8 pCa 6 pH 6 4 2 0 50. smaller α4 • Smaller α4.00 mL Volume of 0. Influence of pH on the titration 50.Titration curves for Ca UNIVERSITY OF THE PHILIPPINES MANILA +2 solutions buffered to various pH levels Recall that α4.0 mL of 0. • Lower pH. becomes smaller as the pH decreases.p • End points for EDTA titrations become less sharp as pH decreases because the complex formation reaction is less complete under these circumstances. smaller K’MY The Health Sciences Center Billones Lecture Notes .0100 M EDTA • Smaller ΔpM at e.0100 M Ca2+ with 0. 0100 M EDTA • smaller KMY.1 x 1014 KCaY2.0 20 16 12 KFeY. Billones Lecture Notes .0 mL of 0.= 6.2 x 1014 KFeY2.3 x 1021 pM 10 8 4 0 50.= 1.3 x 1025 KHgY2.00 mL Volume of 0.0100 M cation solutions at pH 6.= 5.= 3. smaller ΔpM The Health Sciences Center KZnY2.UNIVERSITY OF THE PHILIPPINES MANILA Titration curves for 50.0 x 1012 Cations with larger formation constants provide good end points even in acidic media.= 2. Y3+ Cd2+ Al3+ La3+ Fe3+ Th4+ Hg2+ Ga3+ VO2+ Cu2+ Sm3+ Zn2+ Co2+ Fe2+ Mn2+ Ca2+ moderately acidic environment is satisfactory for many divalent heavy-metal cations UNIVERSITY OF THE PHILIPPINES MANILA Consider a mixture of Ca2+ and Zn2+. log KMY Titration w/ EDTA at pH 10 (both Ca2+ and Zn2+) Titration w/ EDTA at pH 6 (only Zn2+ will titrate) mmol Zn2+ = mL EDTA x M EDTA @ pH 6 mmol Ca2+ = (mL x M EDTA @ pH 10) – mmol Zn pH The Health Sciences Center Billones Lecture Notes .Minimum permissible pH for a satisfactory end point in the titration 26 24 In3+ Sc3+ 22 20 18 16 14 12 10 8 0 2 4 Mg2+. 6 Sr2+ 8 10 12 La3+ Ni2+ Pb2+. zinc(II) is ordinarily titrated in a medium that has fairly high concentrations of ammonia and ammonium chloride. For example. Zn(NH3)22+. Zn(NH3)42+ + HY3- ZnY2. ammonia forms ammine complexes with zinc(II) and prevents formation of the sparingly soluble zinc hydroxide.UNIVERSITY OF THE PHILIPPINES MANILA THE EFFECT OF OTHER COMPLEXING AGENTS ON EDTA TITRATION CURVES Many cations form hydroxide precipitates when the pH is raised to the level required for their successful titration with EDTA. These species buffer the solution to a pH that ensures complete reaction between cation and titrant In addition. The Health Sciences Center Billones Lecture Notes .+ 3NH3 + NH4+ The solution also contains such other zinc/ammonia species as Zn(NH3)32+. an auxiliary complexing agent is needed to keep the cation in solution. and Zn(NH3)2+. When this problem is encountered. 0 mL Volume of 0.00500 M Zn2+. pZn 10 8 6 4 0 cNH3 = 0.0100 M EDTA The Health Sciences Center Billones Lecture Notes . UNIVERSITY OF THE PHILIPPINES MANILA 14 12 Influence of ammonia concentration on the end point for the titration of 50.0 mL of 0.Complexation of a cation by an auxiliary-complexing reagent causes preequivalence pM values to be larger than in a comparable solution with no such reagent.100 cNH3 = 0.010 25. Titration Curves When a Complexing Agent is Present UNIVERSITY OF THE PHILIPPINES MANILA A quantitative description of the effects of an auxiliary complexing agent (ACA) can be derived by a procedure similar to that used to determine the influence of pH on EDTA titration curves. [Mn+] = αMcM where cM is the sum of the concentrations of species containing the metal ion exclusive of that combined with EDTA. cM = [Zn2+] + [Zn(NH3)2+] + [Zn(NH3)22+] + [Zn(NH3)32+] + [Zn(NH3)42+] The value of αM can be expressed readily in terms of the ammonia concentration and the formation constants for the various ammine complexes. The Health Sciences Center Billones Lecture Notes . A quantity αM is defined that is analogous to α4: αM = [Mn+] cM . it is readily shown that [Zn(NH3)22+] = K2[Zn2+][NH3]2 [Zn(NH3)32+] = K3[Zn2+][NH3]3 [Zn(NH3)42+] = K4[Zn2+][NH3]4 Substitution of these expressions into the cM equation gives cM = [Zn2+](1 + K1[NH3] + K2[NH3]2 + K3[NH3]3 + K4[NH3]4) Substituting this expression for cM (here. leads to αM = 1 1 + K1 [NH3] + K2[NH3]2 + K3[NH3]3 + K4[NH3]4 The Health Sciences Center Billones Lecture Notes .UNIVERSITY OF THE] PHILIPPINES MANILA K = [Zn(NH )2+ 1 3 [Zn2+][NH3] [Zn(NH3)2+] = K1[Zn2+][NH3] Similarly. [Mn+] = [Zn2+]). The Health Sciences Center Billones Lecture Notes . and 2. Example Calculate the pZn for solutions prepared by adding 20.00500 M Zn2+. 25. The logarithms of the stepwise formation constants for the four zinc complexes with ammonia are 2.29. 2.100 M in NH3 and 0.21.0.175 M in NH4Cl to provide a constant pH of 9. Assume that both the Zn2+ and EDTA solutions are 0. and 30.0. 2.0 mL of 0.03.0100 M EDTA to 50.Finally.36. Thus.0 mL of 0. a conditional constant for the equilibrium between EDTA and zinc(II) in an ammonia/ammonium chloride buffer is UNIVERSITY OF THE PHILIPPINES MANILA K”ZnY = [ZnY2-] = α4αMKZnY cMcT where K”ZnY is a new conditional constant that applies at a single pH as well as a single concentration of ammonia.0. 21 + 2.17 x 10-5 1 + 16 + 316 + 7. thus.17 x 10-5 x 3.76 x 108 Calculation of Conditional Constant.62 x 102 UNIVERSITY OF THE PHILIPPINES MANILA K2 = antilog (2.K1 = antilog 2.36 + 2.76 x 104 K”ZnY = α4 x αM x KZnY = 5. K” A value of the αM can be obtained by assuming that the equilibrium molar and the analytical concentrations of ammonia are essentially the same.29 + 2.2 x 1016 K”ZnY = 1.29 + 2.03) = 7.36) = 7.100.21 = 1.2 x 10-2 x 1.16 x 104 K3 = antilog (2. αM = 1 = 1.24 x 103 + 7.29) = 3.9 x 1010 The Health Sciences Center Billones Lecture Notes .21 + 2. for [NH3] = 0.24 x 106 K4 = antilog (2.21 + 2. = 50.0100 = 7.35 x 10-9 M pZn = 8. the analytical concentration of ZnY2.00500 = 3.14 x 10-4 M 70.0 Substitution of this value gives [Zn2+] = cMαM = (7. cM = 50.14 x 10-4)(1. The remainder is present as Zn2+ and the four ammine complexes.0 x 0.0 mL of EDTA UNIVERSITY OF THE PHILIPPINES MANILA At this point. only part of the zinc has been complexed by EDTA.is cZnY2.0 mL of EDTA At the equivalence point.00500 – 20.0 The Health Sciences Center Billones Lecture Notes .17 x 10-5) = 8.Calculation of pZn after Addition of 20.0 + 25.0 x 0.33 x 10-3 M 50.0 x 0.08 Calculation of pZn after Addition of 25. 33 x 10-3 – cM ≈ 3.33 x 10-3 = 1.The sum of the concentrations of the various zinc species not combined with EDTA equals the sum of the concentrations of the uncomplexed EDTA species: UNIVERSITY OF THE PHILIPPINES MANILA cM = cT and [ZnY2-] = 3. we obtain The Health Sciences Center Billones Lecture Notes . we have [ZnY2-] = K”ZnY cM2 3.9 x 1010 cM2 cM = 4.33 x 10-3 M Substituting.18 x 10-7 M Employing [Zn2+] = cMαM . = [ZnY2-] = 50.18 x 10-7)(1.0 Rearranging K”ZnY = [ZnY2-] = α4αMKZnY cMcT gives.0 x 0. thus. cEDTA = cT = EDTA – sample VT = 6.00500 = 3. Billones Lecture Notes The Health Sciences Center .17 x 10-5) = 4.25 x 10-4 M and since essentially all of the original Zn2+ is now complexed. 90 x 10-12 pZn = 11.12 x 10-3 M 80.31 Calculation of pZn after Addition of 30.0 mL of EDTA The solution now contains an excess of EDTA.UNIVERSITY OF THE PHILIPPINES MANILA Zn2+] = cMαM = (4. cZnY2. 12 x 10-3 = 2.51 INDICATORS FOR EDTA TITRATION Eriochrome Black T (EBT) is a typical metal-ion indicator that is used in the titration of several common cations.17 x 10-5) = 3.UNIVERSITY OF THE PHILIPPINES MANILA cM = [ZnY2-] cTK”ZnY = 3.63 x 10-10 M (6.25 x 10-4)(1. [Zn2+] = cMαM = (2.63 x 10-10)(1. Its behavior as a weak acid is described by the equations: H 2O + H2InRed HIn2blue + H3O+ K1 = 5 x 10-7 H 2O + HIn2blue In3.9 x 1010) From the equation [Mn+] = αMcM.+ H3O+ orange K2 = 2.8 x 10-12 Billones Lecture Notes The Health Sciences Center .07 x 10-15 pZn = 14. predominates in the absence of a metal ion.+ MY2blue TITRATION METHODS EMPLOYING EDTA Direct Titration The Health Sciences Center Billones Lecture Notes . for metal ion detection. HIn2-. When the EDTA becomes present in slight excess. the indicator complexes the excess metal ion. Thus. the solution turns blue as a consequence of the reaction MInred + HY3- HIn2. UNIVERSITY OF THE PHILIPPINES MANILA Until the equivalence point in a titration.The metal complexes of Eriochrome Black T are generally red as in H2In-. it is necessary to adjust the pH to 7 or above so that the blue form of the species. so the solution is red. MgY-2 + Ca+2 (less stable) CaY-2 + Mg+2 Mg(EBT) red MgY-2 + EBT blue Billones Lecture Notes Mg+2 + EBT Mg(EBT) + Y4red The Health Sciences Center . Methods Based on Indicators for an Added Metal Ion A small amount of a cation is introduced that forms an EDTA complex that is less stable than the analyte complex and for which a good indicator exists.Methods Based on Indicators for the Analyte Ion UNIVERSITY OF THE PHILIPPINES MANILA Mn+ + EDTA color 1 M(EDTA) color 2 Over 40 elements can be determined by direct titration with EDTA using metal ion indicators. A measured excess of standard EDTA solution is added to the analyte solution. M + Y(xs) Y(unreacted) + Mg ➝ MY + Y(unreacted) MgY (should be less stable than MY) Displacement Methods An unmeasured excess of a solution containing the magnesium or zinc complex of EDTA is introduced into the analyte solution. The method is also useful for cations that react only slowly with EDTA. The Health Sciences Center Billones Lecture Notes . the excess EDTA is back-titrated with a standard magnesium or zinc ion solution to an Eriochrome Black T or Calmagite end point. After the reaction is judged complete.Back-Titration Methods UNIVERSITY OF THE PHILIPPINES MANILA Back-titration is useful for the determination of the cations that form stable EDTA complexes and for which a satisfactory indicator is not available. UNIVERSITY OF THE PHILIPPINES MANILA MgY2. zinc. The Health Sciences Center Billones Lecture Notes . cyanide ion is often employed as a masking agent to permit the titration of magnesium and calcium ions in the presence of ions such as cadmium. copper. and palladium. nickel.If the analyte forms a more stable complex than that of magnesium or zinc.+ M2+ ➝ MY2- + Mg2+ Interference from a particular cation can sometimes be eliminated by adding a suitable masking agent. cobalt. an auxiliary ligand that preferentially forms highly stable complexes with the potential interference. the following displacement reaction occurs: analyte The liberated cation is then titrated with the standard EDTA. All of the latter form sufficiently stable cyanide complexes to prevent reaction with EDTA. For example. which we will formulate R(SH)2. After the equivalence point has been reached. a solution of the complexing agent BAL (2-3-dimercapto-1-propanol. This bidentate ligand reacts selectively to form a complex with Pb2+ that is much more stable than PbY2-: PbY2- + 2R(SH)2 ➝ Pb(RS2)22- + Y4Billones Lecture Notes The Health Sciences Center .➝ Zn(CN)42- Zn is masked. no reaction with EDTA 1. magnesium. and zinc can be determined on a single sample by two titrations with standard EDTA and one titration with standard Mg2+. which masks Zn2+ and prevents it from reacting with EDTA. The Pb2+ and Mg2+ are then titrated with standard EDTA.Illustration Showing How Masking and Demasking Agents Can Be Used to Improve The Selectivity of EDTA Titrations UNIVERSITY OF THE PHILIPPINES MANILA Lead. Zn2+ + 4CN. is added to the solution. CH2SHCHSHCH2OH). The sample is first treated with an excess of NaCN. 007657 M Mg2+.4085 g sample. Zn2+ is titrated with EDTA.63 mL of the EDTA. The Health Sciences Center Billones Lecture Notes .22 mL of 0. It is determined by titration with Mg2+ The liberated Y4. Y4. Finally.35 mL of 0. Zn(CN)42. Calculate the percentages of the three elements in a 0. Titration of the Y4. the zinc is demasked by adding formaldehyde.liberated by the BAL consumed 19. after addition of formaldehyde. The liberated Zn2+ is then titrated with the standard EDTA solution.is then titrated with a standard solution of Mg2+.UNIVERSITY OF THE PHILIPPINES MANILA 2.02064 M EDTA. Finally. Example Suppose the initial titration of Mg2+ and Pb2+ required 42. the liberated Zn2+ was titrated with 28.+ 4HCHO + H2O ➝ Zn2+ + 4HOCH2CN + 4OH- 3.is related to Pb2+. no.59092 To obtain the percentage.007657 = 0. mmol Zn2+ = 28.14816 no. from the third titration we obtain no.02064 = 0.515% Pb 0.63 x 0.87142 The second titration gives the number of millimoles of Pb2+. we write 0.2072 g Pb / mmol x 100% = 7. Thus.02064 = 0.4085 g sample The Health Sciences Center Billones Lecture Notes . mmol Pb2+ = 19.87142 – 0.72326 UNIVERSITY OF THE PHILIPPINES MANILA Finally.22 x 0. mmol (Pb2+ + Mg2+) = 42. no.14816 = 0.The initial titration reveals the number of millimoles of Pb2+ and Mg2+ present.35 x 0.14816 mmol Pb x 0. mmol Mg2+ = 0. 59092 mmol Zn x 0.4085 g sample 0.06539 g Zn / mmol x 100% = 9.UNIVERSITY OF THE PHILIPPINES MANILA 0.72326 mmol Mg x 0.4085 g sample The Health Sciences Center Billones Lecture Notes .303% Mg 0.024305 g Mg / mmol x 100% = 4.459% Zn 0. Exercise UNIVERSITY OF THE PHILIPPINES MANILA A 1. The solution remaining at the end of this titration was treated with excess formaldehyde.05331 M EDTA.00 mL of 0. mercury(II).0 mL of solution. The magnesium in this sample required 16. which reacts with the free CN. The Health Sciences Center Billones Lecture Notes .+ HCHO +H2O H2C(OH)(CN) + OH- The liberated Zn+2 required 28.00-mL aliquot was treated with 10 mL of an NH3/NH4Cl buffer at pH 10 followed by 25. A 50. complexing both the mercury and the zinc. the excess EDTA was back-titrated with 11. and zinc was dissolved in 250. Calculate the percentage of each metal in the sample.83 mL of 0.and with Zn(CN)42-: CN. A second 50.01816 M MgCl2.43 mL of 0.005583 M EDTA for titration.3174-g sample containing the chloride salts of magnesium.47 mL of the EDTA for titration. After a few minutes of mixing.00-mL aliquot was made basic and treated with excess NaCN. 00 mL x 0.12523 mmol ABANGAN .20757 mmol mmol M+2 = 1...43 mL x 0.Solution UNIVERSITY OF THE PHILIPPINES MANILA mmol xs EDTA = 25.3328 mmol mmol unreacted EDTA = 11. The Health Sciences Center Billones Lecture Notes .3328 mmol .0.20757 mmol = 1.05331 M = 1.01816 M = 0.