Chapter 11: Solving Equilibrium Problems for Complex Systems

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.

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

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-] In this system, the charge-balance equation also is [H3O+] = [A-] + [OH-]

* * : conc. of H3O+ at equilibrium

* * *

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–] 每一個溶液系統只有一個charge balance equation 電中性物種不可書寫於charge balance equation Charge balance equation 左右兩邊應等值，可作為平衡運算結果是否正確的判斷依據。

+ [Cl-]

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

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.

A simple example of systematic calculations Q Calculate [H3O+] and [OH-] in pure water A Step 1: 2H2O  H3O+ + OH- Step 2: [H3O+]=? and [OH-]=? 2 unknowns Step 3: [H3O+][OH-] = 1x10-14 (1) Step 4: mass-balance equation: [H3O+]=[OH-] (2) Step 5: charge-balance equation: [H3O+]=[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) [H3O+] [OH-] = [H3O+]2 = 1x10-14 ∴ [H3O+] = 1x10-7 and [OH-] = 1x10-7

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

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

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

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)

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: Example 11-8 Calculate the solubility of AgCl in distilled water. Solution

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 for High stability constant omit for Low High stability constant

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

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

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.

[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。

11C-2 Sulfide Separations Saturated H2S, [H2S](aq) = 0.1 M [S2-] in H2S saturated solution depend on the pH Metal sulfide solubility for (M2+S2-) in saturated H2S * The solubility of MS in H2S saturated solution depend on the pH

Homework (Due 2014/11/27) End of Chapter 11 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