Chapter 11: Solving Equilibrium Problems for Complex Systems 1/30 For simultaneous equilibria in aqueous solutions, BaSO4(s) in water for example, there are three equilibria: BaSO4(s) Ba2+ + SO42(1) SO42- + H3O+ HSO4- +H2O (2) 2H2O H3O+ + OH(3) The addition of H3O+ causes: (2) shift right and (1) shift right. since Ba2+ + OAc- BaOAc+ (4) The addition of OAc- causes: (1) shift right *The introduction of a new equilibrium system into a solution does not change the equilibrium constants for any existing equilibria. 2/30 11 A Solving multiple-equilibrium problems using A systematic method Three types of algebraic equations are used to solve multipleequilibrium problems: (1) equilibrium-constant expressions (2) mass-balance equations (3) a single charge-balance equation 3/30 11 A-1 Mass-Balance Equations Mass-balance equations: The expression that relate the equilibrium concentrations of various species in a solution to one another and to the analytical concentrations of the various solutes. These equations are a direct result of the conservation of mass and moles. A weak acid HA dissolved in water for example: HA+ H2O H3O+ + A(1) 2H2O H3O+ + OH(2) mass equation 1: cHA = [HA] + [A-] cHA is analytical concentration, [HA] and [A-] are equilibrium concentration. mass equation 2: [H3O+] = [A-] + [OH-] since [H3O+] = [H3O+]from HA + [H3O+]from H2O , where [H3O+]from HA = [A-] , [H3O+]from H2O = [OH-] 4/30 * * : conc. of H3O+ at equilibrium 5/30 * * * 6/30 11 A-2 Charge-Balance Equation Charge-Balance Equation: An expression relating the concentrations of anions and cations based on charge neutrality in any solution. Charge balance equation: n1[C1+n1] + n2[C2+n2] + ..... = m1[A1-m1] + m2[A2-m2] + ..... Example: A solution contains H+, OH–, K+, H2PO4–, HPO42–，, and PO43–，what is the charge balance equation? Solution: [H+] + [K+] = [H2PO4–] + 2[HPO42–] + 3[PO43–] + [OH–] 7/30 + [Cl-] 8/30 11A-3 Steps for solving problems with several equilibria Figure 11-1 A systematic method for solving multiple-equilibrium problems. 9/30 11A-4 Using Approximations to Solve Equilibrium Calculations Approximations can be made only in charge-balance and mass-balance equations, never in equilibrium-constant expressions. If the assumption leads to an intolerable error, recalculate without the faulty approximation to arrive at a tentative answer. 11A-5 Use of Computer Programs to Solve MultipleEquilibrium Problems Several software packages are available for solving multiple nonlinear simultaneous equations include Mathcad, Mathematica, Solver, MATLAB, TK, and Excel. 10/30 A simple example of systematic calculations Q Calculate [H3O+] and [OH-] in pure water A Step 1: Step 2: Step 3: Step 4: Step 5: Step 6: Step 7: Step 8: 2H2O H3O+ + OH[H3O+]=? and [OH-]=? 2 unknowns [H3O+][OH-] = 1x10-14 (1) mass-balance equation: [H3O+]=[OH-] (2) charge-balance equation: [H3O+]=[OH-] (3) equations (2) and (3) are identical, omit equation (3) two unknowns two different equations (1) and (2), OK Approximation, omit equation (2) substitute into equation (1) [H3O+] [OH-] = [H3O+]2 = 1x10-14 ∴ [H3O+] = 1x10-7 and [OH-] = 1x10-7 11/30 11B Calculating solubilities by the systematic method 11B-1 Solubility of metal hydroxides for High Ksp value, pH controlled by the solubility Example 11-5 Calculate the molar solubility of Mg(OH)2 in water. Solution Step 1. Balanced equations 2+ Mg (aq) +2OH -(aq) Mg(OH)2(s) 2H 2 O(l) + H 3O(aq) +OH (aq) Step 2. Unknown(s)? s=[Mg 2+ ]=? [H 3O + ]=? [OH - ]=? 12/30 Step 3. Equilibrium-constant expressions K sp =[Mg 2+ ][OH - ]2 =7.1x10-12 (1) K w =[H 3O + ][OH - ]=1x10 14 (2) Step 4. Mass-balance equation(s) [OH - ]=2[Mg 2+ ] [H 3O + ] (3) Step 5. Charge-balance equation [OH - ]=2[Mg 2+ ] [H 3O + ] (4) identical to equation (3) Step 6. Independent equations and unknowns 3 unknowns and 3 independent equations, OK 13/30 Step 7. Make approximations assume [H3O+ ] <<2[Mg 2+ ], equation (3) simplifies to [OH - ] 2[Mg 2+ ] Step 8. Solve the equations substitute (5) into (1) (5) K sp =[Mg 2+ ][OH - ]2 =[Mg 2+ ](2[Mg 2+ ])2 =7.1x10-12 [Mg 2+ ] s=1.21x10-4 M substitute into (5) [OH - ]=2.42x10-4 M substitute into (2) [H 3O + ]=4.1x10-11 M Step 9. Check [H3O+ ] << 2[Mg 2+ ] OK 14/30 for Low Ksp value, pH ≈7, controlled by autoprotolysis of water Example 11-6 Calculate the molar solubility of Fe(OH)3 in water. Solution Step 1. Balanced equations +3OH Fe3+ (aq) (aq) Fe(OH)3(s) 2H 2O (l) + +OH (aq) H 3O (aq) Step 2. Unknown(s)? s=[Fe3+ ]=? [H 3O + ]=? [OH - ]=? Step 3. Equilibrium-constant expressions K sp =[Fe3+ ][OH - ]3 =2x10-39 (1) K w =[H 3O + ][OH - ]=1x10 14 (2) 15/30 Step 4. Mass-balance equation(s) [OH - ]=3[Fe3+ ] [H 3O + ] (3) Step 5. Charge-balance equation [OH - ]=3[Fe3+ ] [H 3O + ] (4) identical to equation (3) Step 6. Independent equations and unknowns 3 unknowns and 3 independent equations, OK Step 7. Make approximations assume 3[Fe3+ ]<<[H 3O + ], equation (3) simplifies to [OH - ] [H 3O + ] (5) 16/30 Step 8. Solve the equations substitute (5) into (2) [OH - ]=[H 3O + ]=1.0x10-7 M substitute into (1) K sp =[Fe3+ ][OH - ]3 =[Fe3+ ](1.0x10-7 )3 =2x10-39 [Fe3+ ] s=2x10-18 M Step 9. Check 3[Fe3+ ] [H 3O + ] OK 17/30 11B-2 The Effect of pH on Solubility *The solubility of precipitates containing an anion with basic properties, a cation with acidic properties, will depend on pH. (simultaneous equilibria) Solubility Calculations When the pH Is Constant ([OH-] and [H3O+] are known) Example 11-7 Calculate the molar solubility of calcium oxalate in a solution that has been buffered so that its pH is constant and equal to 4.00. Solution Step 1. Balanced equations CaC2O4(s) 2+ 2Ca (aq) +C2O4(aq) H 2C2O 4(aq) +H 2O HC2O-4(aq) +H 2O + H 3O(aq) +HC 2O-4(aq) + H 3O(aq) +C2O 2-4(aq) 18/30 Step 2. Unknown(s) s=[Ca 2+ ]=? [C 2O 2-4 ]=? [HC 2O -4 ]=? [H 2 C 2O 4 ]=? Step 3. Equilibrium-constant expressions [Ca 2 ][C 2O 24 ] K sp 1.7 10 19 (1) [H 3O ][HC2 O 4 ] K1 5.60 102 [H 2 C2 O 4 ] (2) [H 3O ][C2 O 42 ] 5 K 5.42 10 2 [HC2 O 4 ] (3) Step 4. Mass-balance equation(s) [Ca 2+ ] = s = [C 2O 42- ]+[HC 2O -4 ]+[H 2C 2 O 4 ] (4) Step 5. Charge-balance equation buffer is not specified, omit 19/30 Step 6. Independent equations and unknowns 4 unknowns and 4 independent equations, OK Step 7. Make approximations omit Step 8. Solve the equations from equation (3) [ H 3O ][C2O42- ] 1.00 10-4 [C2O42- ] 2[ HC2O ] 1.85[ C O 2 4 ] (5) -5 K2 5.42 10 4 substitute (5) into (2) [ H 3O ][ HC2O4- ] 1.00 10-4 1.85[C2O42- ] [ H 2C2O4 ] K1 5.60 10-2 3.30 10 -3[C2O42- ] (6) 20/30 substitute (5) and (6) into (4) [Ca 2 ] [C2O42- ] 1.85[C2O42- ] 3.30 10-3[C2O42- ] 2.85[C2O42- ] [Ca 2 ] [Ca ][C2O ] [Ca ] K sp =1.7 109 2.85 [Ca 2 ] s 7.0 105 M 2 24 2 Also [C2O 42- ]=2.46x10 5 M [HC 2O -4 ]=4.54x105 M [H 2C2O 4 ]=8.11x10 8 M 21/30 Solubility Calculations When the pH Is Variable ([OH-] and [H3O+] are unknown) for High Ksp value (pH controlled by the solubility) omit for Low Ksp value (pH≈7, controlled by autoprotolysis of water) omit 22/30 11B-3 The Effect of Undissociated Solutes on Precipitation Calculations For example, a saturated solution of AgCl(s) contains significant amounts of undissociated silver chloride molecules, AgCl(aq) complexs: AgCl(s) AgCl(aq) + Ag(aq) + Cl -(aq) AgCl(aq) AgCl(s) + Ag(aq) + Cl (aq) K s =3.6x10-7 K d =5.0x10-4 K sp =K s K d 1.8x10-10 Example 11-8 Calculate the solubility of AgCl in distilled water. Solution + solubility = s = [AgCl(aq) ] + [Ag(aq) ] = 3.6x10-7 K sp = = 3.6x10-7 1.34x105 = 1.38x105 M neglecting [AgCl(aq) ] leads to -3% error. 23/30 11B-4 The Solubility of Precipitates in the Presence of Complexing Agents The solubility increase in the presence of reagents that form complexes with the anion or the cation of the precipitate. Ex. F- prevent the precipitation of Al(OH)3 Al(OH)3(s) Al3+ + 6F (aq) (aq) 3+ Al(aq) + 3OH (aq) 3AlF6(aq) for High stability constant omit for Low High stability constant omit 24/30 • Complex formation with a common ion to the precipitate may increase in solubility by large excesses of a common ion. Example Given: AgCl(s) AgCl(aq) AgCl(s) AgCl(aq) + Ag(aq) + Cl(aq) + Ag(aq) + Cl(aq) AgCl(s) + Cl(aq) AgCl 2(aq) + Cl -(aq) AgCl 2(aq) 2 AgCl3(aq) What is the the concentration of KCl at which the solubility of AgCl is a minimum? Solution omit 25/30 Figure 11-2 The effect of chloride ion concentration on the solubility of AgCl. The solid curve sows the total concentration of dissolved AgCl. The broken lines show the concentrations of the various silver-containing species. CKCl = 0.003 M 26/30 11C Separation of ions by control of the concentration of the precipitating agent 11C-1 Calculation of the feasibility of separations Generally, complete precipitation is considered as 99.9% of the target ion is precipitated, i.e., 0.1% left. 27/30 [Fe3+][OH-]3 = 2x10-39 [Mg2+][OH-]2 = 7.1x10-12 Fe3+會先沈澱 設剩餘 0.1% 之Fe3+為 complete precipitation for Fe(OH)3， 則 [Fe3+] = 0.1x0.1% = 1x 10-4 M 1 x 10-4 x [OH-]3 = 2x10-39 [OH-] = 3 x 10-12 M 完全沈澱Fe3+所需之[OH-] Mg(OH)2 開始沉澱之 [OH-]： 0.1 x [OH-]2 = 7.1x10-12 [OH-] = 8.4 x 10-6 M Mg2+開始沈澱之[OH-] 控制水溶液之 [OH-] = 3 x 10-12 ~ 8.4 x 10-6 M，可將 0.1 M 的 Fe3+ 與 0.1 M 的 Mg2+ 分離。 28/30 11C-2 Sulfide Separations Saturated H2S, [H2S](aq) = 0.1 M H 2S H 2O HS H 2O (1) (2) [H 3O ][HS ] H 3O + HS K1 9.6 108 (1) [H 2S] H 3O + S [H 3O ][S2 ] 14 K2 1.3 10 [HS- ] 2 (2) [H 3O ]2 [S2 ] =K1K 2 [H 2S] [H 3O ]2 [S2 ] [H 2S]K1K 2 (0.10)(9.6 10 8 )(1.3 10 14 ) 1.2 x1022 1.2 x1022 [S ] [H 3O ]2 2 [S2-] in H2S saturated solution depend on the pH Metal sulfide solubility for (M2+S2-) in saturated H2S MS(s) 2 K sp =[M 2 ][S2 ] 2+ M (aq) + S(2-aq) [M ] = solubility = K sp 2- [S ] = K sp [H 3O ]2 1.2 1022 * The solubility of MS in H2S saturated solution depend on the pH 29/30 Homework (Due 2014/11/27) Skoog 9th edition, Chapter 11, Questions and Problems 11-5 (e) (g) 11-6 (e) (g) 11-7 (a) 11-14 End of Chapter 11 30/30

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