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Equilibrium and Solubility

Equilibrium and Solubility. Solubility rules for common ions Using the solubility product, K sp , to calculate solubility - molar solubility, gram solubility 2. The common ion effect 3. Precipitation and the solubility product - the criterion for solubility

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Equilibrium and Solubility

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  1. Equilibrium and Solubility Solubility rules for common ions Using the solubility product, Ksp, to calculate solubility - molar solubility, gram solubility 2. The common ion effect 3. Precipitation and the solubility product - the criterion for solubility 4. Effects of pH on solubility - solubility of hydroxides - influence of hydrolysis on solubility

  2. Solubility Rules Soluble: all Group I (Li+, Na+, K+, Rb+, Cs+) and NH4+ salts all nitrates, most acetates, and most sulfates most halides, except as noted below Slightly Soluble: fluorides: SrF2, BaF2, PbF2, other halides: PbCl2, PbBr2, HgBr2 sulfates: CaSO4, Ag2SO4, Hg2SO4 hydroxides: Ca(OH)2, Sr(OH)2 Insoluble (except as noted above): sulfides, carbonates, sulfites, phosphates, hydroxides MgF2, CaF2, AgCl, AgBr, AgI, HgI2 SrSO4, BaSO4, PbSO4

  3. Solubility and Ksp The equilibrium constant for the solubility reaction has a special name, the solubility product, Ksp. BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = 1.1 x10-10 Ag2CrO4(s) 2 Ag+(aq) + CrO42-(aq) Ksp = 1.2 x10-12 Example 1: What is the solubility of Ag2CrO4(s) in water? Ksp = 1.2 x10-12 = [Ag+]2 [CrO42-] (x = moles per Liter of CrO42-) 2x x = [2x]2 [x] x = (0.3 x 10-12)1/3 = 6.7 x 10-5 M gram solubility = 2.2 x 10-2 g-L-1 molar solubility = 6.7 x 10-5 M

  4. The Common Ion Effect Example 2: What is the solubility of Ag2CrO4(s) is 0.100 M AgNO3? Ag2CrO4(s) 2 Ag+(aq) + CrO42-(aq) Ksp = 1.2 x10-12 Ksp = 1.2 x10-12 = [Ag+]2 [CrO42-] 0.100+2x x (x = moles per Liter of CrO42-) = (0.100 + 2x)2 (x) ≈ (0.100)2 (x) x = 1.2 x 10-10 M molar solubility = 1.2 x 10-10 M gram solubility = 4.0 x 10-8 g-L-1 An example of the common ion effect: the solubility of Ag2CrO4(s) is lowered in the presence of added Ag+(aq)

  5. Precipitation and Ksp Example 3: Will Ag2CrO4(s) precipitate if 300 mL of 1.2 x10-5 M AgNO3 and 200 mL of 3.5 x10-4 M Na2CrO4 are mixed? Ans: yes, if Q0 = [Ag+]02 [CrO42-]0 exceeds Ksp in the mixture. Ag2CrO4(s) 2 Ag+(aq) + CrO42-(aq) Ksp = 1.2 x10-12 (0.200 x3.5 x10-4)/(.500) (0.300 x1.2 x10-5)/(.500) Q0 = (0.72 x10-5)2 (1.4 x10-4) = 7.3 x10-16 Precipitation does not occur, since the ion product is less than Ksp.

  6. Mg(OH)2(s) Mg2+(aq) + 2 OH-(aq) Ksp = 2 x10-13 Effects of pH on Solubility of Hydroxides Example 4: What is the solubility of Mg(OH)2 at pH 10? x 1 x10-4 M Ksp = 2 x10-13 = [Mg2+][OH-]2 = x (1 x10-4)2 molar solubility = 2 x 10-5 M at pH 10 = 2 x 10-9 M at pH 12 = 20 M at pH 7 The solubility of Mg(OH)2 increases as pH decreases. Solubility varies as (pH)-2 in solutions more basic than pH 8.

  7. [CO32-] pH = pKa2 + log10 [HCO3-] Effects of pH on Solubilities of Salts (1) Example 5: What is the solubility of CaCO3 at pH 8.32? at pH 10.32? CaCO3(s) Ca2+(aq) + CO3-2(aq) Ksp = 8.7 x10-9 x depends on pH Ksp = 2 x10-13 = [Ca2+][CO32-] x depends on pH 10.32 When pH is more acidic than pKa2, the solubility of CaCO3 increases because most of the carbonate present is in the form of bicarbonate. When pH<10, solubility varies linearly with (1/pH).

  8. Effects of pH on Solubilities of Salts (2) Example 5: What is the solubility of CaCO3 at pH 8.32? at pH 10.32? (1) at pH 8.32, [CO32-] = 0.01* [HCO3-] CaCO3(s) Ca2+(aq) + CO32-(aq) Ksp = 8.7 x10-9 x 0.01 x solubility = 9 x10-4 M (2) at pH 10.32, [CO32-] = [HCO3-] CaCO3(s) Ca2+(aq) + CO32-(aq) Ksp = 8.7 x10-9 x 0.5 x solubility = 9 x10-5 M

  9. Effects of pH on Solubilities of Salts (3) Example 6: Describe how the solubility of CaF2 depends on pH. CaF2(s) Ca2+(aq) + 2 F-(aq) Ksp = 3.9 x10-11 x depends on pH Ksp = 3.9 x10-11 = [Ca2+][F-]2 x depends on pH [F-] pH = pKa + log10 [HF] 3.18 When pH is more acidic than pKa, the solubility of CaF2 increases because most of the fluorine present is in the form of HF rather than F-.

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