Solubility Equilibria Solubility Product Constant Ksp for saturated solutions at equilibrium

Comparing values.

Solubility Product (Ksp) = [products]x/[reactants]y but..... reactants are in solid form, so Ksp=[products]x i.e. A2B3(s)  2A3+ + 3B2– Ksp=[A3+]2 [B2–]3 Given: AgBr(s)  Ag+ + Br– In a saturated solution @25oC, the [Ag+] = [Br– ]= 5.7 x 10–7 M. Determine the Ksp value.

concentrations of each Problem: A saturated solution of silver chromate was to found contain 0.022 g/L of Ag2CrO4. Find Ksp Eq. Expression: Ag2CrO4 (s)  2Ag+ + CrO42– Ksp = [Ag+]2[CrO42–] So we must find the concentrations of each ion and then solve for Ksp.

Problem: A saturated solution of silver chromate was to contain 0 Problem: A saturated solution of silver chromate was to contain 0.022 g/L of Ag2CrO4. Find Ksp Eq. Expression: Ag2CrO4 (s)  2Ag+ + CrO42– Ksp = [Ag+]2[CrO42–] 0.022 g Ag2CrO4 L Ag+: 1.33 x 10–4 0.022g Ag2CrO4 L CrO4–2: 6.63 x 10–5

Problems working from Ksp values. Given: Ksp for MgF2 is 6.4 x 10–9 @ 25 oC Find: solubility in mol/L and in g/L MgF2(s)  Mg2+ + 2F– Ksp = [Mg2+][F–]2 N/A 0 0 I. C. E. N/A +x +2x N/A +x +2x Ksp= [x][2x]2 = 4x3 6.4 x 10–9 = 4x3 now for g/L: 7.3 x 10–2

The common ion effect “Le Chatelier” overhead fig 17.16 What is the effect of adding NaF? CaF2(s)  Ca2+ + 2F-

Solubility and pH CaF2(s)  Ca2+ + 2F– Add H+ (i.e. HCl) 2F– + H+  HF

Solubility and pH Mg(OH)2(s)  Mg2+ + 2OH– Adding NaOH? Adding HCl?

The common ion effect “Le Chatelier” Why is AgCl less soluble in sea water than in fresh water? AgCl(s)  Ag+ + Cl– Seawater contains NaCl

Problem: The solubility of AgCl in pure water is 1.3 x 10–5 M. What is its solubility in seawater where the [Cl–] = 0.55 M? (Ksp of AgCl = 1.8 x 10–10) AgCl(s)  Ag+ + Cl– Ksp= [Ag+][Cl–] I. C. E. N/A 0 0.55 N/A +x +x N/A +x 0.55 + x Ksp= [x][0.55 + x] try dropping this x Ksp = 0.55x 1.8 x 10–10 = 0.55x x = 3.3 x 10–10 = [Ag+]=[AgCl] “AgCl is much less soluble in seawater”

more Common ion effect: a. What is the solubility of CaF2 in 0.010 M Ca(NO3)2? Ksp(CaF2) = 3.9 x 10–11 CaF2(s)  Ca2+ + 2F– Ksp=[Ca2+][F-]2 [Ca2+] [F–] I. C. E. 0.010 0 +x +2x 0.010 + x 2x Ksp= [0.010 + x][2x]2  [0.010][2x]2 = 0.010(4x2) 3.9 x 10–11 = 0.010(4x2) x = 3.1 x 10–5 M Ca2+ from CaF2 so = M of CaF2 Now YOU determine the solubility of CaF2 in 0.010 M NaF.

Answer: 3.9 x 10–7 M Ca2+ CaF2(s)  Ca2+ + 2F– 0 0.010 +x 2x x 0.010 + 2x Ksp = [x][0.010 + 2x]2  x(0.010)2 3.9 x 10-11 =x(0.010)2 x = 3.9 x 10-7 What does x tell us

Reaction Quotient (Q): will a ppt. occur? Use Q (also called ion product) and compare to Ksp Q < Ksp reaction goes Equilibrium Q = Ksp reaction goes Q > Ksp

Problem: A solution is 1. 5 x 10–6 M in Ni2+ Problem: A solution is 1.5 x 10–6 M in Ni2+. Na2CO3 is added to make the solution 6.0 x 10–4 M in CO32–. Ksp(NiCO3) = 6.6 x 10–9. Will NiCO3 ppt? We must compare Q to Ksp. NiCO3  Ni2+ + CO32– Ksp = [Ni2+][CO32–] Q = [Ni2+][CO32–] Q = [1.5 x 10–6][6.0 x 10–4] = 9.0 x 10–10 Q < Ksp no ppt.

a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–14 Problem: 0.50 L of 1.0 x 10–5 M Pb(OAc)2 is combined with 0.50 L of 1.0 x 10–3 M K2CrO4. a. Will a ppt occur? Ksp(PbCrO4) = 1.8 x 10–14 Pb(OAc)2(aq) + K2CrO4(aq)  PbCrO4(s) + 2KOAc(aq) then: PbCrO4(s)  Pb2+ + CrO42– Ksp= [Pb2+][CrO42–] [Pb2+]: 0.50 L 5.0 x 10–6 1 L [CrO42-]: 0.50 L 5.0 x 10-4 1 L Q = [5.0 x 10-6][5.0 x 10-4] = 2.5 x 10-9 compare to Ksp: Q > Ksp so a ppt. will occur

Try dropping the “+ x” term. b. find the Eq. conc. of Pb2+ remaining in solution after the PbCrO4 precipitates. Ksp(PbCrO4) = 1.8 x 10–14 Since [Pb2+] = 5.0 x 10-6 and [CrO42-] = 5.0 x 10-4 and there is a 1:1 stoichiometry, Pb2+ is the limiting reactant. PbCrO4(s)  Pb2+ + CrO42– I. (after ppt.) C. E. 0 5.0 x 10-4 - 5.0 x 10-6 = 5.0 x 10-4 +x +x x 5.0 x 10–4 + x Ksp = [x][5.0 x 10–4 + x] Try dropping the “+ x” term. 1.8 x 10–14 = x(5.0 x 10-4) x = [Pb2+] = 3.6 x 10–11 This is the concentration of Pb2+ remaining in solution.

Complex ion formation: AgCl(s)  Ag+ + Cl– Ksp= 1.8 x 10–10 Ag+(aq) + NH3(aq)  Ag(NH3)+(aq) Ag(NH3)+(aq) + NH3(aq)  Ag(NH3)2+(aq) Ag+(aq) + 2NH3(aq)  {Ag(NH3)2}+(aq) formation or stability constant: For Cu2+: Cu2+ + 4NH3  [Cu(NH3)4]2+(aq) K1 x K2 x K3 x K4 = Kf = 6.8 x 1012

Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq) Solubility and complex ions: Problem: How many moles of NH3 must be added to dissolve 0.050 mol of AgCl in 1.0 L of H2O? (KspAgCl = 1.8 x 10–10 ; Kf[Ag(NH3)2]+ = 1.6 x 107) AgCl(s)  Ag+ + Cl– Ag+(aq) + 2NH3(aq)  Ag(NH3)2+(aq) AgCl(s) + 2NH3  Ag(NH3)2+(aq) + Cl– sum of RXNS: = 2.9 x 10–3 Now use Koverall to solve the problem: Koverall= 2.9 x 10–3 = [NH3]eq = 0.93 but ..... How much NH3 must we add? [NH3]total= 0.93 + (2 x 0.050) = 1.03 M 2 ammonia’s for each Ag+

[Cd2+] = 2.6 x 10–10 M when NiS starts to ppt. Fractional Precipitation: “ppting” one ion at a time. Compounds must have different Ksp values (i.e. different solubilities) Example: Ksp CdS = 3.9 x 10–29 and KspNiS = 3.0 x 10–21 ? Which will ppt. first? least soluble Problem: A solution is 0.020 M in both Cd2+ and Ni2+. Just before NiS begins to ppt., what conc. of Cd2+ will be left in solution? Approach: Find conc. of S2– ion when Ni2+ just begins to ppt. since Cd2+ will already be ppting. Then use this S2– conc. to find Cd2+. NiS(s)  Ni2+ + S2– Ksp= 3.0 x 10–21 = [0.020][S2–] [S2–] = 1.5 x 10–19 M when Ni2+ just begins to ppt. So: CdS(s)  Cd2+ + S2– Ksp= 3.9 x 10–29 = [Cd2+][1.5 x 10–19] [Cd2+] = 2.6 x 10–10 M when NiS starts to ppt.