Chapter 17 Solubility and Simultaneous Equilibria

Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO4 in water: BaSO4(s) Ba2+(aq) + SO42(aq) A saturated solution is one in which no more solute can dissolve. Saturated solution – no more solute will dissolve

Solubility Product Constant – equilibrium constant for ionic compounds that are only slightly soluble The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42] Ksp is the equilibrium constant for ionic compounds that are only slightly soluble.

Solubility Products Ksp is not the same as solubility. Solubility generally expressed as (g/L), (g/mL), or in mol/L (M). Ksp - Only one value for a given solid at a given temperature Temperature dependence Solubility and thus Ksp changes with T The solubility of barium sulfate is how much of this material will dissolve in a given amount of solution. The solubility product, in the case of barium sulfate will be the square of that [Ba][SO4].

Solubility Equilibria When ionic salt dissolves in water It dissociates into separate hydrated ions Initially, no ions in solution CaF2(s)  Ca2+(aq) + 2F–(aq) As dissolution occurs, ions build up and collide Ca2+(aq) + 2F–(aq)  CaF2(s) At equilibrium CaF2(s) Ca2+(aq) + 2F–(aq) We now have a saturated solution Ksp = [Ca+][F–]2

Writing Ksp Equilibrium Laws AgCl(s) Ag+(aq) + Cl–(aq) Ksp = [Ag+][Cl–] PbI2(s) Pb2+(aq) + 2I–(aq) Ksp = [Pb2+][I–]2 Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) Ksp = [Ag+]2[CrO42–] AuCl3(s) Au3+(aq) + 3Cl–(aq) Ksp = [Au3+][Cl–]3 Note that the solid does not appear in the Ksp expression because it is has a constant concentration.

Molar Solubility – # moles of ionic solid that dissolves per 1 L DI water to form a saturated solution at 25 C Determine the final equilibrium concentration of Ca2+ (aq) and CO32- (aq) in a saturated solution Molar solubility always equals “X” 6.7 x 10-5 moles of CaCO3 (s) will dissolve per 1 L DI water producing 6.7 x 10-5 M Ca2+ (aq) and 6.7 x 10-5 M CO32- (aq)

Given Solubilities, Calculate Ksp At 25 °C, the solubility of AgCl is 1.34 × 10–5 M. Calculate the solubility product for AgCl. AgCl(s) Ag+(aq) + Cl–(aq) Ksp = [Ag+][Cl–] AgCl(s) Ag+(aq) + Cl–(aq) I 0.00 0. 00 C E 1.34 × 10–5 M 1.34 × 10–5 M 1.34 × 10–5 M 1.34 × 10–5 M Ksp = (1.34 × 10–5)(1.34 × 10–5) Ksp = 1.80 × 10–10

Learning Check The solubility of calcium fluoride, CaF2, in pure water is 2.15 × 10‑4 M. What is the value of Ksp? A. 1.99 × 10–11 B. 3.98 × 10–11 C. 9.94 × 10–12 D. 1.85 × 10-7 CaF2 Ca2+ + 2F– [Ca2+] = (2.15 × 10–4) [F-] = 2(2.15 × 10–4) Ksp = [Ca][F]2 = (2.15 × 10–4) (4.3 x 10-4)2 Ksp = 3.98 × 10–11

Given Ksp, Calculate Solubility What is the molar solubility of CuI in water? Determine the equilibrium concentrations of Cu+ and I– CuI(s) Cu+(aq) + I–(aq) Ksp = [Cu+][I–] Ksp = 1.3  10–12

Molar Solubilities from Ksp Solve Ksp expression Ksp = 1.3 × 10–12 = (x)(x) x2 = 1.3 × 10–12 x = 1.1 × 10–6 M = calculated molar solubility of CuI = [Cu+] = [I– ] Conc (M) CuI(s) Cu+(aq) + I–(aq) Initial Conc. 0.00 Change Equilibrium Conc. No entries No entries No entries +x +x x x

Given Ksp, Calculate Solubilities Calculate the solubility of CaF2 in water at 25 °C, if Ksp = 3.4 × 10–11. CaF2(s) Ca2+(aq) + 2F– (aq) 1. Write equilibrium law Ksp = [Ca2+][F–]2 2. Construct concentration table Conc (M) CaF2(s) Ca2+(aq) 2F–(aq) Initial Conc. (No entries 0.00 Change in this Equilibrium Conc. column) +x +2x x 2x

Molar Solubilities from Ksp 3. Solve the Ksp expression Ksp = [Ca2+][F–]2 = (x) (2x)2 3.4 × 10–11 = 4x3 x = 2.0 × 10–4 M = molar solubility of CaF2 [Ca2+] = x = 2.0 × 10–4 M [F–] = 2x = 2(2.0 × 10–4 M) = 4.0 × 10–4 M

Factors Affecting Solubility The Common-Ion Effect If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease: BaSO4(s) Ba2+(aq) + SO42(aq)

Factors Affecting Solubility pH If a substance has a basic anion, it will be more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions.

Common Ion Effect What happens if another salt, containing one of the ions in our insoluble salt, is added to a solution? Consider PbI2(s) Pb2+(aq) + 2I–(aq) Saturated solution of PbI2 in water Add KI PbI2 (yellow solid) precipitates out Why? Le Chatelier’s Principle Add product I– Equilibrium moves to left and solid PbI2 forms

Learning Check What effect would adding copper(II) nitrate have on the solubility of CuS? A. The solubility would increase B. The solubility would decrease C. The solubility would not change

Common Ion Effect Consider three cases What is the molar solubility of Ag2CrO4 in pure water? What is the molar solubility of Ag2CrO4 in 0.10 M AgNO3? What is the molar solubility of Ag2CrO4 in 0.10 M Na2CrO4? Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) Ksp = [Ag+]2[CrO42–] = 1.1 × 10–12 x = 9.0 10–10 M

Common Ion Effect A. What is the solubility of Ag2CrO4 in pure water? Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries 0.00 M C in this E column) +2x +x 2x x Ksp = [Ag+]2[CrO42–] = (2x)2(x) = 1.1  10–12 = 4x3 x = 9.0 10–10 M x = Solubility of Ag2CrO4 = 6.5 x 10-5 M [CrO42–] = x = 6.5 x 10-5 M [Ag+] = 2x = 1.3 × 10–4 M

Common Ion Effect B. What is the molar solubility of Ag2CrO4 in 0.10 M AgNO3 solution? Ksp = 1.1 × 10–12 Ksp = 1.1 × 10–12 = (0.10 M)2[x] x = Solubility of Ag2CrO4 = 1.1 × 10–10 M [Ag+] = 0.10 M [CrO42–] = 1.1 × 10–10 M Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries 0.10 M 0.00 C in this E column) +2x +x ≈0.10 x

Common Ion Effect Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries C. What is the solubility of Ag2CrO4 in 0.100 M Na2CrO4? Ag2CrO4(s) 2Ag+(aq) + CrO42–(aq) I (No entries 0.00 M 0.10 M C in this E column) +2x +x 2x ≈0.10 Ksp = (2x)2(0.10) = 1.1  10–12 = 4x2(0.10) x = Solubility of Ag2CrO4 = 1.66 × 10–6 M [CrO42–] = 0.10 M [Ag+] = 2x = 3.3 × 10–6 M

Common Ion Effect What have we learned about the solubility of silver chromate? Dissolving it in pure water the solubility was 3.0 × 10–4 M Dissolving it in AgNO3 solution solubility was 1.1 × 10–10 M Dissolving it in Na2CrO4 solution solubility was 3.2 × 10–5 M Common ion appearing the most in the formula of the precipitate decreases the solubility the most

Will a Precipitate Form? In a solution, If Q = Ksp, the system is at equilibrium and the solution is saturated. If Q < Ksp, more solid can dissolve until Q = Ksp. If Q > Ksp, the salt will precipitate until Q = Ksp.

Predicting Precipitation Will a precipitate of PbI2 form if 100.0 mL of 0.0500 M Pb(NO3)2 are mixed with 200.0 mL of 0.100 M NaI? PbI2(s) Pb2+(aq) + 2I–(aq) Ksp = [Pb2+][I–]2 = 9.8 × 10–9 Strategy for solving Calculate concentrations in mixture prepared Calculate Qsp = [Pb2+][I–]2 Compare Qsp to Ksp

Predicting Precipitation Step 1. Calculate concentrations Vtotal = 100.0 mL + 200.0 mL = 300.0 mL [Pb2+] = 1.67 × 10–2 M [I–] = 6.67 × 10–2 M

Predicting Precipitation Step 2. Calculate Qsp Qsp = [Pb2+][I–]2 = (1.67 × 10–2 M)(6.67 × 10–2 M)2 Qsp =7.43 × 10–5 Step 3. Compare Qsp and Ksp Qsp = 7.43 × 10–5 Ksp = 9.8 × 109 Qsp > Ksp so precipitation will occur

pH and Solubility Mg(OH)2(s) Mg2+(aq) + 2OH–(aq) Increase OH– shift equilibrium to left Add H+ shift equilibrium to right Le Châtelier’s Principle Ag3PO4(s) 3Ag+(aq) + PO43–(aq) Add H+ increase solubility H+(aq) + PO43–(aq)  HPO42–(aq) AgCl(s) Ag+(aq) + Cl–(aq) Adding H+ has no effect on solubility Why? Cl– is very, very weak base, so neutral anion So adding H+ doesn’t effect Cl– concentration