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Chapter 11: Solving Equilibrium Problems for Complex Systems 1/30.

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1 Chapter 11: Solving Equilibrium Problems for Complex Systems 1/30

2 For simultaneous equilibria in aqueous solutions, BaSO 4(s) in water for example, there are three equilibria: BaSO 4(s)  Ba 2+ + SO 4 2- (1) SO H 3 O +  HSO 4 - +H 2 O(2) 2H 2 O  H 3 O + + OH - (3) The addition of H 3 O + causes: (2) shift right and (1) shift right. since Ba 2+ + 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

3 11 A Solving multiple-equilibrium problems using A systematic method Three types of algebraic equations are used to solve multiple- equilibrium problems: (1) equilibrium-constant expressions (2) mass-balance equations (3) a single charge-balance equation 3/30

4 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+ H 2 O  H 3 O + + A - (1) 2H 2 O  H 3 O + + OH - (2) mass equation 1: c HA = [HA] + [A - ] c HA is analytical concentration, [HA] and [A - ] are equilibrium concentration. mass equation 2: [H 3 O + ] = [A - ] + [OH - ] since [H 3 O + ] = [H 3 O + ] from HA + [H 3 O + ] from H2O, where [H 3 O + ] from HA = [A - ], [H 3 O + ] from H2O = [OH - ] 4/30

5 * * 5/30 : conc. of H 3 O + at equilibrium

6 * * * 6/30

7 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: n 1 [C 1 +n1 ] + n 2 [C 2 +n2 ] = m 1 [A 1 -m1 ] + m 2 [A 2 -m2 ] Example: A solution contains H +, OH –, K +, H 2 PO 4 –, HPO 4 2– ,, and PO 4 3– , what is the charge balance equation? Solution: [H + ] + [K + ] = [H 2 PO 4 – ] + 2[HPO 4 2– ] + 3[PO 4 3– ] + [OH – ] 7/30

8 8/30 + [Cl - ]

9 Figure 11-1 A systematic method for solving multiple-equilibrium problems. 11A-3 Steps for solving problems with several equilibria 9/30

10 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 Multiple- Equilibrium Problems Several software packages are available for solving multiple nonlinear simultaneous equations include Mathcad, Mathematica, Solver, MATLAB, TK, and Excel. 10/30

11  A simple example of systematic calculations Q Calculate [H 3 O + ] and [OH - ] in pure water AStep 1: 2H 2 O  H 3 O + + OH - Step 2:[H 3 O + ]=? and [OH - ]=? 2 unknowns Step 3:[H 3 O + ][OH - ] = 1x (1) Step 4:mass-balance equation: [H 3 O + ]=[OH - ](2) Step 5:charge-balance equation: [H 3 O + ]=[OH - ](3) Step 6:equations (2) and (3) are identical, omit equation (3) two unknowns two different equations (1) and (2), OK Step 7: Approximation, omit Step 8:equation (2) substitute into equation (1) [H 3 O + ] [OH - ] = [H 3 O + ] 2 = 1x ∴ [H 3 O + ] = 1x10 -7 and [OH - ] = 1x /30

12 11BCalculating solubilities by the systematic method 11B-1Solubility of metal hydroxides  for High K sp value, pH controlled by the solubility Example 11-5 Calculate the molar solubility of Mg(OH) 2 in water. Solution 12/30

13 13/30

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15  for Low K sp value, pH ≈7, controlled by autoprotolysis of water Example 11-6 Calculate the molar solubility of Fe(OH) 3 in water. Solution 15/30

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18 11B-2The 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 [H 3 O + ] 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 Solution 18/30

19 19/30

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21 21/30

22  Solubility Calculations When the pH Is Variable ([OH - ] and [H 3 O + ] are unknown) for High K sp value (pH controlled by the solubility) omit for Low K sp value (pH≈7, controlled by autoprotolysis of water) omit 22/30

23 11B-3The 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: Example 11-8 Calculate the solubility of AgCl in distilled water. Solution 23/30

24 11B-4The 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 for High stability constant omit for Low High stability constant omit 24/30

25 Complex formation with a common ion to the precipitate may increase in solubility by large excesses of a common ion. Example What is the the concentration of KCl at which the solubility of AgCl is a minimum? Solution omit 25/30

26 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. C KCl = M 26/30

27 11CSeparation of ions by control of the concentration of the precipitating agent 11C-1Calculation 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

28 [Fe 3+ ][OH - ] 3 = 2x [Mg 2+ ][OH - ] 2 = 7.1x Fe 3+ 會先沈澱 設剩餘 0.1% 之 Fe 3+ 為 complete precipitation for Fe(OH) 3 , 則 [Fe 3+ ] = 0.1x0.1% = 1x M 1 x x [OH - ] 3 = 2x [OH - ] = 3 x M 完全沈澱 Fe 3+ 所需之 [OH - ] Mg(OH) 2 開始沉澱之 [OH - ] : 0.1 x [OH - ] 2 = 7.1x [OH - ] = 8.4 x M Mg 2+ 開始沈澱之 [OH - ] 控制水溶液之 [OH - ] = 3 x ~ 8.4 x M ,可將 0.1 M 的 Fe 3+ 與 0.1 M 的 Mg 2+ 分離。 28/30

29 11C-2Sulfide Separations  Saturated H 2 S, [H 2 S] (aq) = 0.1 M [S 2- ] in H 2 S saturated solution depend on the pH  Metal sulfide solubility for (M 2+ S 2- ) in saturated H 2 S * The solubility of MS in H 2 S saturated solution depend on the pH 29/30

30 Homework (Due 2014/11/27) Skoog 9 th edition, Chapter 11, Questions and Problems 11-5 (e) (g) 11-6 (e) (g) 11-7 (a) End of Chapter 11 30/30


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