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Solubility Product Constant 6-5

Solubility Product Constant 6-5. Ksp. is a variation on the equilibrium constant for a solute-solution equilibrium. remember that the solubility equilibrium is based on a saturated solution. Solids are not written in the equilibrium expression. CaF 2 (s) Ca 2+ (aq) + 2F - (aq).

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Solubility Product Constant 6-5

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  1. Solubility Product Constant6-5 Ksp

  2. is a variation on the equilibrium constant for a solute-solution equilibrium. • remember that the solubility equilibrium is based on a saturated solution. • Solids are not written in the equilibrium expression.

  3. CaF2 (s) Ca 2+ (aq) + 2F -(aq) Ksp = [Ca 2+] [F -]2 Where Ksp is the solubility product constant, or simply the solubility product.

  4. CaCl2 (s) Ca 2+ (aq) + 2 Cl -(aq) Ksp = [Ca 2+] [Cl -]2

  5. Given the solubility, calculate the Ksp of the substance. 1).Write the chemical equation for the substance dissolving and dissociating. 2) Write the Ksp expression. 3) Insert the concentration of each ion and multiply out.

  6. Calculating the solubility product: CuBr has a measured solubility of 2.0 10-4 M at 25 C. Calculate the Ksp value. CuBr (s) Cu + (aq) + Br -(aq) Ksp = [Cu +] [Br -] = (2.0 10-4 )(2.0 10-4 ) = 4.0 10-8

  7. x EX: Calculate the Ksp for Bi2 S3 , which has a solubility of 1.36 x 10-15 mol/L at 25°C. Bi2S3 (s) 2 Bi3+ (aq) + 3 S2- (aq) 2(1.36 x 10-15 ) 3(1.36 x 10-15) 2.72 x 10-15 4.08 x 10-15 x 3x 2x

  8. Ksp = [Bi 3+]2 [S 2-]3 = (2.72 x 10-15)2 (4.08 x 10-15)3 = 5.02 x 10-73

  9. Determine the Ksp of calcium fluoride (CaF2 ), given that its molar solubility is 2.14 x 10-2 M. • CaF2 (s) → Ca2+ (aq) + 2 F¯ (aq) • The Ksp expression is: Ksp = [Ca2+ ] [F¯]2 2x x

  10. Ksp = [Ca2+ ] [F¯]2 • Ksp = (2.14 x 10-2 ) (4.28 x 10-2 )2 • Ksp = 3.92 x 10-5

  11. Convert g/l to mol/l The solubility of Fe(OH)2 is found to be 1.4 x 10-3 g/l. What is the Ksp value? Change the g/l to mol/l. Molar mass of Fe(OH)2 is 89.87g/mol Solubility is = 1.56 x 10-5 mol/l Fe(OH)2(s) → Fe2+(aq) + 2OH-(aq) x x 2x Ksp = [Fe2+ ][OH- ]2 =(1.56 x 10-5 ) (2[1.56 x 10-5 ]) 2 =1.5 x 10-14

  12. Calculate the Ksp of silver bromide: ( AgBr ), if the solubility is 1.3 x 10 -4 g / L .

  13. We can also use a known value of Ksp to find out the solubility of a substance. • Write out the chemical equation for the substance • Write the Ksp expression • Solve for x

  14. If the Ksp is 1.8 x 10 -8 of silver chloride, find the solubility. • write Ksp expression for silver chloride. • AgCl (s) ------> Ag+ (aq) + Cl- (aq) • x x

  15. Ksp = [Ag + (aq) ] [Cl - (aq)] • 1.8 x 10 -8 = x2 • 1.3 x 10-4 = x

  16. The Ksp value for Cu(IO3)2 is 1.4 x 10-7 at 25°C. Calculate its solubility at this temperature. Cu(IO3)2 (s) Cu2+ (aq) + 2 IO31- (aq) x x 2x

  17. Ksp = [Cu 2+] [IO31-]2 • 1.47 x 10-7 = (x) (2x)2 • 1.47 x 10-7 = 4x3 • 1.47 x 10-7 = 4x3 4 4 3.3 x 10-3 = x Solubility of Cu(IO3)2

  18. If Ksp of calcium hydroxide is 5.5 x 10-6, find the solubility. Ca(OH)2 (s)→ Ca+2 (aq) + 2OH-(aq) x=0.011 M

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