LARUTAN DAN KOLOIDMacam-Macam Campuran • Larutan • Dispersi Koloid • Suspensi Perbandingan larutan dispersi Koloid & Suspensi LARUTAN DISPERSI KOLOID SUSPENSI Semua bentuk partikel dari atom, ion atau molekul (0,1 – 1 nm) Parikel paling sedikit satu komponen atom, ion atau molekul kecil (1 – 1000 nm) Partikel paling sedikit satu komponen yang dapat dilihat di bawah mikroskop Tidak stabil Tidak Homogen Tidak tembus Tidak transparan Partikel terpisah Stabil terhadap gravitasi Homogen Tembus Cahaya Tidak ada efek Tyndall Tidak ada gerak Brown Tidak dapat dipisahkan dengan penyaringan Kurang Stabil Perbatasan homogen Buram Efek Tyndall Gerak Brown Tidak dapat dipisahkan dengan Dapat dipisahkan penyaringan dengan penyaringan 1 • SUSPENSI Dapat dipisahkan dengan penyaringan atau dengan sentrifugasi • DISPERSI KOLOID JENIS Buas FASA TERDISPERSI Gas MEDIA PENDISPERSI Cair CONTOH Busa sabun Busa Padat Aerosol Cair Emulsi Emulsi Padat Asap Sol Sol Padat Gas Cair Cair Cair Padat Padat padat Padat Gas Cair Padat Gas Cair padat Batu apung Kabut, halimun, awan Krim, susu, saos Mentega, keju Debu, partikulat dalam asap Pati dalam air, jeli, cat Aloy, mutiara 2 Muatan Elektron partikel Koloid Colloidal particle with organic ionic groups Colloidal particle with adsorbed chloride ions FIGURE Colloidal particles often be an electrical charges that stabilize the dispersion. On the left is a particle whose extremely large molecules carry negatively charged groups. On the right the colloidal particles has attracted chloride ions to itself. In either case, these colloidal particles repel each other and cannot join together. 3 4 . 5 . Efek Tyndall The Tyndall effect. and in the ab sence of the ryndall effect we might think they are all solutions. and the third a colloidal dispersion of Fe2O3. Not true solutions LARUTAN CAMPURAN HOMOGEN 6 . A pencil line thin red laser beam passes through the liquid in three Rest tubes. The first contain a colloidal dispersion of starch. However. All three appear transparent. the Tyndilll effect reveals that the fist and third are coilds. the second a solution of sodium chrornate. Gasolin Gula dalam cair Udara Sistem koloid Sistem koloid Gas dalampadat Cair dalam padat Padat dalam padat Aloy hidrogen dim paladium Benzen dalam karet Karbon dalam besi 7 .Larutan Gas Gas dalam gas Cair dalam gas Padat dalam gas Larutan Cairan Gas dalam cair Cair dalam cair Padat dalam cair Larutan Padat Coca-cola Cuka. Mengapa Terbentuk Larutan • Larutan Dalam Cairan + - Ion – ion force of attraction asin sodium chploride + + - Polar molecule Dipole-dipole force of attraction as in sugar of water 8 . experience hydrogen bonding ( …) between themselves in pure alcohol. C2H5 – O – H. When these two liquids for a solution. hydrogen bonds can easily form between molecules of water and those of alcohol thus. and the solution easily forms 9 . Hydrogen bonds also occur in pure water.• Larutan cair Dalam Cair Figure Ethyl alcohol molecules. attractive forces between molecules in the pure liquids are replaced by similar forces in the solution. 10 . and water molecules find ion more attactive than even other water molecules. In the solution. and Na+ and CIions have only each other in crystal to be attracted to. water molecules are attracted only to each other.• Larutan padat dalam cair Portion of surface and edge of NaCl cristal in contact with water FIGURE Hydration of ions Hydration involves a complex redirection of force of attraction and repulsion. the ions have water molecules to takes the places of their oppositely charged counterparts. Before this solution forms. Panas Larutan Terjadi pertukaran energi sistem dan sekelilingnya apabila 1 mol zat terlarut dilarutkan dalam bentuk ( pada tekanan konstan) untuk membuat larutan encer. H : Fungsi keadaan yang tidak bergantung pada jalannya perubahan FIGURE Enthalpy diagram for a solid dissolving in liquid. The energy change along the direct path is the algebraic sum of step 1 and step 2. We can analyze the energy change by imagining the two separate steps. because entrhalpy changes are fuctions of state and are independent of path. the solution is formed directly as indicated by the red arrow. 11 . In the real word. FIGURE The Formation of aqueous potassium iodide Step 1 : Kl(s) Step 2 : K+(g) + l-(g) Net : Kl(s) K+(g) + l-(g) K+(g) + l-(g) K+(g) + l-(g) H = +632 kJ H = -619 kJ Hlarutan = +13 kJ 12 . • Larutan cairan dalam cairan • Larutan gas dalam cairan Energi solvasi eksoterm 13 . Pengaruh Suhu pada Kelarutan Kelarutan : Massa zat terlarut yang membentuk larutan jenuh dengan massa pelarut pada suhu tertentu Satuan : gram zat terlarut / 100 gram pelarut Solut (tidak larut) Solut (larut) Kelarutan naik jika mengabsorpsi panas Solut (tidak larut) + Panas Solut (larut) 14 . FIGURE Solubilty in water versus temperature for several substances Gas larut secara eksoterm dalam cairan pada semua konsentrasi Gas (tidak larut) Gas (larut) + Panas 15 . equilibrium exists between the vapor phase and the solution. (b) An Increase in pressure puts stress on the equilibrium. (c) More gas dissolves and equilibrium is restored 16 .Pengaruh Tekanan Pada Kelarutan dalam Gas Kelarutan gas dalam cairan naik dengan niaknya tekanan Gas + pelarut Larutan FIGURE Solubility in water versus pressure for two gases FIGURE How pressure inueases the solubility of a gas in a liquid. (a) At some specific pressure. SO2 & CO2 bereaksi dengan air kesetimbangan CO2(aq) + H2O H2CO3(aq) H+(aq) + HCO3-(aq) SO2(aq) + H2O H+(aq) + HSO3-(aq) NH3(aq) + H2O NH4+(aq) + OH-(aq) 17 . Pg Kelarutan gas yang terhidrasi kuat SO2.Hukum Henry : Konsentrasi gas dalam cairan pada suhu yang diberikan secara langsung sebanding dengan tekanan gas pada larutan. NH3 & CO2 lebih mudah larut dibanding S2 & N2 NH3 Ikatan H. Cg = Kg . % berat (% b/b) : jumlah gram zat terlarut / 100 g larutan % volume (% v/v) : jumlah mL zat terlarut / 100 mL larutan 18 . V Hukum gas ideal : nA = R. nA XA = nA + nB nC + … dst PA . Molal m kg pelarut PA XA Ptot * % konst. T mol zat terlarut * Konst.Konsentrasi * Fraksi mol dan % mol Mol fraksi : Perbandingan jumlah mol suatu komponen terhadap jumlah mol total komponen yang ada. • • • • • • Perubahan di antara satuan dan konsentrasi Merubah dari % berat ke molal Merubah dari % berat ke fraksi mol Menghitung % berat dari fraksi mol Merubah molal ke fraksi mol Merubah % berat ke molar Merubah dari molar ke % berat Sifat Koligatof Larutan Penurunan Tekanan Uap • • • Tek uap campuran turun dengan adanya komponen lain Tek uap larutan (zat terlarut : non volatil) < tek uap pel murni Hukum Raoult Plarutan = Xpelarut . Popelarut 19 . 9 mole percent of N2 20 .789. X N2 PN2 Ptotal 600 torr Ptotal But the total pressure is the sum of the partial pressures. so X N2 600 torr 600 torr 160 torr = 0.CALCULATING THE MOLE FRACTION OF A GAS FROM PARTIAL PRESSURE Problem : What are the mole percents of nitrogen and oxygen in air when the partial pressure are 160 torr for oxygen and 600 torr for nitrogen ? (Asumme no other gases are present) Solution : Let us use Equation to find the mole fraction of N2 fist. or 78. 150 m NaCl” gives us the following. two rations.Now we do the same for oxygen 160 torr X O2 600 torr 160 torr 0. or 21.150 m solution of sodium chloride.1 mole percents add up to O 2 You can easlly see that the two mole percents add up to 100% CALCULATING MOLAL CONCENTRATION Problem : An experiment calls for an aqueous 0. our first step is to prepare a conversion factor. how many grams of NaCl would have to be dissolved in 500 g of water ? Solution : As with almost all problems involving concentrations. “0. Thus. where we substitute 1000 g for 1 kg.211. To prepare a solution with this concentration. 21 . of course.5 g NaCl/mol NaCl) 58.150 mol NaCl 1000 g H2O and 1000 g H2O 0.0750 mol of NaCl to grams of NaCl. we use the first ratio.150 mol NaCl To calculate the moles of NaCl we need for 500 g of H2O. (The formula weight of NaCl is 58.0. We next convert 0.39 g NaCl 1 mol NaCl 22 .0750 mol NaCl x 4.150 mol NaCl 500g H2O x 0. because then the units will cancel property 0.0750 mol NaCl 1000 g H2O This gives us the moles of NaCl needed.5 which means.5 g NaCl 0. 58. 0.39 g NaCl 1000 g H2O 1 mol NaCl USING WEIGHT/WEIGHT PERCENT Problem : How many grams of a 4.00% (w/w) solution of NaCl needed to obtain 0. With a little practice. when 4.500 g of NaCl ? Solution : The given concentration gives us the following conversion factors 23 .39 g of NaCl is dissolved in 500 g of H2O. For example. the concentration is ). you will be able to set up a string of conversion factors and do the calculation at the end. 150 m NaCl.150 mol NaCl 58.5 g NaCl 500 g H2O x 4.Thus. 500 g of NaCl from this solution.00 g NaCl We want 0.00 g NaCl 100 g solution and 100 g solution 4. so we use the second conversion factor 100 g solution 0.500 g of 24 . we will also be taking 0.00% (w/w) solution 4.5 g of 4.5 g to the 4.500 g NaCl x 12.00 g NaCl Thus.4. if we take 12.00% (w/w) NaCl solution. FIGURE The vapor pressure of an ideal. two-component solution of volatile compounds 25 . 26 .• Larutan ideal dan penyimpangan hukum Raoult FIGURE Typical deviations from ideal behavior. of the total vapor pressure of real. two-componen solutions of volatile substances. m FIGURE Boiling point elevation: Shown here are plots of vapor pressures versus temperatures for a solvent (upper curve) and for a solution of a non volatile solute in the same solvent (lower curve).KenaikanTitik Didih th = kb . 27 . 07 2.6 5.7 Cyclohexane 80.4 3.53 3.2 - 3.63 - 16.86 Acetic acid Benzen Chloroform Camphor 118.57 5.69 6.5 20.Tabel. molal boiling point elevation and freezing point depression constants Solvent Water Bp (oC) 100 Kb 0.7 2.0 • Menghitung kenaikan titik didih dari molar harga konstanta kenaikan titik dan molal • Menghitung BM dari kenaikan titik didih 28 .3 802 61.45 178.07 37.15 Mp (oC) 0 Kf 1. konst akan berubah hingga setimbang. m • Menghitung penurunan titik beku dari molar harga konstanta penurunan titik dan molal • Menghitung BM dari kenaikan titik beku Dialisis dan Osmosis Dialisis : Jika 2 larutan dengan konstrasi berbeda dipisahkan oleh suatu membran.Penurunan Titik Beku th = kb . Membran bersifat “semipermiabel” (hanya ion dan molekul kecil yang dapat lewat) Osmosis : Jika hanya molekul pelarut yang dapat lewat pada membran Tekanan Osmosis : Tekanan untuk menjaga aliran osmosis = MRT 29 . 30 . 31 . Sifat – sifat Koligatif pada Larutan Elektrolit • Memperkirakan sifat koligatif pada larutan elektolit • Ineraksi ion-ion dalam larutan cairan Δt f pengukuran faktor Van' t Hoff : i Δt f penghitungannonelektrolit % ionisasi elektrolit – elektrolit lemah Δt mol asam terionisas i m %ionisasi Kf mol asam yang ada 32 .