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Solubility Equilibria

Solubility Equilibria. CAN YOU? / HAVE YOU? Write a balanced chemical equation to represent equilibrium in a saturated solution. Write a solubility product expression. Answer questions about Ksp and various missing concentrations using I.C.E. tables.

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Solubility Equilibria

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  1. Solubility Equilibria

  2. CAN YOU? / HAVE YOU? • Write a balanced chemical equation to represent • equilibrium in a saturated solution. • Write a solubility product expression. • Answer questions about Ksp and various missing concentrations using I.C.E. tables.

  3. C6H12O6(s) C6H12O6(aq)

  4. Even the most insoluble ionic solids are actually soluble in water to a limited extent There are 3 actions that affect solubility: 1. Nature of the solute and solvent “like dissolves like” Polar / ionic solute dissolve in polar solvent. Non-polar dissolve in non-polar.

  5. 2. Temperature Solids in liquids: ↑ temperature - ↑ solubility. Gases in liquids: ↑ in temperature - ↓ solubility. 3. Pressure Does not affect the solubility of (s)/(l). (g): ↑ pressure ↑ solubility.

  6. AaBb(s) aA+(aq) + bB¯(aq) Kc = [A+]a[B-]b [AaBb] Ksp = [A+]a[B-]b Ksp, called the solubility product constant. Product of ion concentrations in a saturated solution.

  7. Write the dissociation and the product constant equation for the solubility of calcium hydroxide. Pb3(PO4)2(s)3 Pb2+(aq) + 2 PO43-(aq) Ca(OH)2 (s) Ca2+(aq) + OH-(aq) 2 Ksp = [Ca2+][OH-]2 Write a solubility product expression for Pb3(PO4)2. Ksp = [Pb2+]3[PO43-]2

  8. At equilibrium, the [Ag+] = 1.3 x 10-5 M and the [Cl-] = 1.3 x 10-5 M, what is the Ksp of silver chloride? AgCl (s) Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] Ksp =(1.3 x 10-5)(1.3 x 10-5) Ksp = 1.7 x 10-10 *NOTE: Ksp has no units.

  9. Solubility And I.C.E. Tables Solubility - maximum amount of solute that can dissolve in a certain amount of solvent at a certain temperature. (Yeah!)

  10. Calculate Ksp of lead (II) chloride if a 1.0 L saturated solution has of lead ions. PbCl2(s) Pb2+(aq) + 2 Cl-(aq) 1.62 x 10-2 M 1.62 x 10-2 M I--- 0 0 C-x+x +2x E 0 2(1.62 x 10-2) Ksp = [Pb+2][Cl -]2 Ksp = [1.62 x 10 -2][ 3. 24 x 10 -2]2 Ksp = 1.70 x 10 -5

  11. The solubility of PbF2 is . What is the value of the solubility product constant? 0.466 g/L PbF2(s) Pb2+(aq) + 2 F¯(aq) Pb – 207 + 2 (19) = 245g/mol Ksp = [Pb2+][F-]2 0.466 g 1 mol = 1.90 x 10-3 M PbF2 1 L 245.2 g

  12. PbF2(s) Pb2+(aq) + 2 F¯(aq) 1.9 x 10-3 M 0 0 [I] [C] - x + x + 2x [E] 0 3.8 x 10-3 M 1.9 x 10-3 M Saturated – all solid reactant dissociates. Ksp = [Pb2+][F-]2 Ksp = (1.90 x 10-3)(3.80 x 10-3)2 Ksp = 2.74 x 10-8

  13. Calculate Ksp if 50.0 mL of a saturated solution was found to contain 0.2207 g of lead (II) chloride. PbCl2(s) Pb2+(aq) + 2 Cl-(aq) 0.2207 g 1 mol = 0.0159 M PbCl2 0.05 L 278.1g I0.0159M 0 0 C-x+x +2x E00.0159 M 0.0318 M Ksp = [Pb2+][Cl-]2 Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

  14. Mg(OH)2 (s) Mg2+(aq)+ 2 OH-(aq) Ksp of magnesium hydroxide is 8.9 x 10-12. What are the [equilibrium] of ions in saturated solution? I--- 0 0 C-x+x +2x E0 x 2x Ksp = [Mg2+][OH-]2 8.9 x 10-12 = [x][2x]2 8.9 x 10-12 = [x]4x2 8.9 x 10-12 = 4x3

  15. 8.9 x 10-12 = 4x3 4 4 3√ 3√ 2.23 x 10-12 = x3 1.3 x 10-4 = x [Mg2+] = x = 1.3 x 10-4 mol/L [OH-] = 2x = 2.6 x 10-4 mol/L

  16. Estimate the solubility in g/L of Ag2CrO4 if the Ksp is 1.1 x 10-12. Ag2CrO4(s) 2 Ag+(aq) + CrO42-(aq) Ix 0 0 C-x+2x +x E0 Ksp = [Ag+]2[CrO42-] 1.1 x 10 -12 = [2x]2[x] 1.1 x 10 -12 = 4x3 x = 6.50 x 10 -5 M 1.30 x 10 -4 M 6.50 x 10 -5 M

  17. Ag2CrO4(s) 2 Ag+(aq) + CrO42-(aq) 1 1 E0 1.30 x 10 -4 M 6.50 x 10 -5 M [Ag2CrO4]i = 6.50 x 10 -5 M 330 g 6.5 x 10-5mol = 0.022 g/L 1 L 1 mol

  18. Precipitation

  19. Compare value of Q, with givenKsp to determine if an aqueous solution is saturated or unsaturated. Q = KspSaturated solution, no precipitate. Q >Ksp Precipitate forms (“oversaturated”) Q < KspSolution is unsaturated. Qsp = [A+]a[B¯]b

  20. PbF2(s) Pb2+(aq) + 2 F¯(aq) Ksp of lead (II) fluoride is 1.6 x 10-5. If 0.57 g are mixed with 1500 mL of water, is solution saturated? Pb – 207 + 2 (19) = 245g/mol 0.57 g 1 mol = 2.32 x 10-3 mol/L 1 L 245.2 g Ksp = [Pb2+][F-]2 Qsp = (2.32 x 10-3)(4.64 x 10-3)2 Qsp = 5.0 x 10-8 Q >Ksp Precipitate forms

  21. Predict if there is a precipitate of PbCl2 if 100 mL each of of and are added together. Ksp of PbCl2 = 1.7 x 10 -5 PbF2(s) Pb2+(aq) + 2 Cl¯(aq) 0.01 M NaCl 0.02 M Pb(NO3)2 0.01 M NaCl 0.02 M Pb(NO3)2 Qsp = [Pb2+][Cl-]2

  22. Because you are adding volumes, you must account for the new diluted concentrations. C1V1 = C2V2 (0.01 M)(0.1 L) = (C2)(0.2 L) = 0.005 M Cl- (0.02 M)(0.1 L) = (C2)(0.2 L) = 0.01 M Pb2+ Qsp = [0.01][0.005]2 = 2.5 x 10 -7 Qsp< Ksp A precipitate will not form.

  23. AgBr(s) Ag+(aq) + Br¯(aq) If 20.0 mL of 0.0010 M silver nitrate is mixed with 20.0 mL of 3.0 x 10-5 M potassium bromide, does silver bromide (Ksp = 5.0 x 10-13) precipitate? Assume the volumes are additive. Ksp = [Ag+][Br-] (0.001 M)(0.02 L) = (C2)(0.04 L) = 5.0 x 10-4 M Ag+ (2e-5 M)(0.02 L) = (C2)(0.04 L) = 1.5 x 10-5 M Br- Qsp = [5.0 x 10-4][1.5 x 10-5] = 7.5 x 10 -9 Qsp> Ksp A precipitate will form.

  24. Substances which are insoluble are actually slightly soluble. • The solubility product, Ksp, describes the product of ion concentrations in saturated solutions. • Solubility can be determined from the solubility product.

  25. CAN YOU? / HAVE YOU? • Write a balanced chemical equation to represent • equilibrium in a saturated solution. • Write a solubility product expression. • Answer questions about Ksp and various missing concentrations using I.C.E. tables.

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