1 / 20

Solubility Equilibria

Solubility Equilibria. Solubility Product Constant K sp for saturated solutions at equilibrium. Comparing values. Solubility Product (K sp ) = [products] x /[reactants] y but. reactants are in solid form, so K sp =[products] x. K sp =[A 3+ ] 2 [B 2– ] 3.

abeni
Download Presentation

Solubility Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solubility Equilibria Solubility Product Constant Ksp for saturated solutions at equilibrium

  2. Comparing values.

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

  4. 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.

  5. 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

  6. 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 Ksp = [Mg2+][F–]2 MgF2(s)  Mg2+ + 2F– 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

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

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

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

  10. 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

  11. 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) Ksp= [Ag+][Cl–] AgCl(s)  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”

  12. 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  [0.010][2x]2 = 0.010(4x2) Ksp= [0.010 + x][2x]2 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.

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

  14. 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

  15. 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.

  16. 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

  17. 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.

  18. 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

  19. 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 = but ..... How much NH3 must we add? [NH3]eq = 0.93 [NH3]total= 0.93 + (2 x 0.050) = 1.03 M 2 ammonia’s for each Ag+

  20. 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. Ksp= 3.9 x 10–29 = [Cd2+][1.5 x 10–19] CdS(s)  Cd2+ + S2– So: [Cd2+] = 2.6 x 10–10 M when NiS starts to ppt.

More Related