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17.4 Solubility Equilibria

17.4 Solubility Equilibria. Hannah Nirav Joe. ↔. Big Ideas. You will be able to Understand what K sp is Find K sp from a reaction. Solubility Product Constant. K sp – equilibrium constant related to the equilibrium between a solid salt and its ions in a solution.

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17.4 Solubility Equilibria

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  1. 17.4 Solubility Equilibria Hannah Nirav Joe ↔

  2. Big Ideas • You will be able to • Understand what Ksp is • Find Ksp from a reaction

  3. Solubility Product Constant • Ksp– equilibrium constant related to the equilibrium between a solid salt and its ions in a solution. • Provides a quantitative measure of the solubility if a slightly soluble salt. • Smaller Ksp means that a small amount of the solid will dissolve in water • Appendix D contains a list of many Ksp values at 250C

  4. Solubility Product • Solubility product expression - the product of the concentration of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation • Same as all the other equilibrium expressions • Unitless number • Tells how much of the solid is dissolved to form a saturated solution

  5. Examples • BaSO4 (s) ↔ Ba2+(aq) + SO42- (aq) • Ksp = [Ba2+ ][SO42-] • CaF2 (s) ↔ Ca2+ (aq) + 2 F- (aq) • Ksp = [Ca2+ ][2 F-]2

  6. Solubility and Ksp • Ksponly has one value for a reaction at a given temperature • You can solve for Kspfrom solubility and solve for the solubility from Ksp

  7. Relationships between Solubility and Ksp Solubility of compound (g/L) Molar solubility of compound (mol/L) Molar concentrations of ions Ksp

  8. Practice Problems • 47, 53 in textbook • 47a) If the molar solubility of CaF2 is 1.24 x 10-3 mol/L, what is Ksp at this temperature?

  9. Solution • Step 1: Find the Dissociation Reaction 1 CaF2 (s) ↔ 1 Ca2+ (aq) + 2 F- (aq) • Step 2: Use coefficients to find the concentration of ions Solubility of 1 CaF2 = Concentration of 1 Ca2+ = 1.24 x 10-3 mol/L Solubility of 1 CaF2 = Concentration of 2 F- = 2.48 x 10-3 mol/L

  10. Solution • Step 3: Use the solubility product expression to solve for Ksp 1 CaF2 (s) ↔ 1 Ca2+ (aq) + 2 F- (aq) Ksp= [Ca2+ ][2 F-]2 Ksp= [1.24 x 10-3 mol/L ][2.48 x 10-3 mol/L]2 Ksp= 7.63 x 10-9

  11. More Problems • Calculate the Kspfor MgF2 if the molar solubility of this salt is 2.7 x 10-3 M. Ans. 7.9 x 10-8

  12. More Problems • The Kspfor BaCO3 is 5.0 x 10-9. What is the molarity of a saturated aqueous solution of BaCO3? Ans. 7.1 x 10-5 M

  13. More Problems • The Kspfor Fe(OH)2 is 4.1 x 10-15. What is the approximate pH of a saturated solution of Fe(OH)2? Ans. 9.00

  14. More Problems • 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? (Kspof AgCl = 1.8 x 10–10)

  15. Ksp= [x][.55+x] Ksp= .55x 1.8 * 10 -10 = .55x x = 3.3*10-10 = [Ag + ] = [AgCl]

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