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Solubility Equilibria

solubility product - Ksp  

  • equals product of ion concentrations raised to the power of their coefficients
  • can be used to calculate solubility (g/L)

Find the Ksp for a saturated solution of Mg(OH)2 if the pH is 10.17  

  • assume that Mg(OH)2, a strong base, dissociates completely
  • Mg(OH)2 >> Mg2+ + 2OH-
  • pOH = 14 - pH = 14 - 10.17 = 3.83
  • [OH-] = 10-pOH = 10-3.83 = 1.48 x 10-4
  • [Mg2+] = [OH-] / 2 = 7.4 x 10-5
  • Ksp = [Mg2+] [OH-]2 = 1.62 x 10-12

Find the molar solubility of Mn(OH)2 if the Ksp is 1.6 x 10-13  

  • Mn(OH)2 >> Mn2+ + 2OH-
  • Ksp = [Mn2+] [OH-]2
  • let c = molar solubility, concentration of Mn
  • Ksp = c x (2c)2 = 4c3
  • 1.6 x 10-13 = 4c3
  • c = 3.4 x 10-5

Find the molar solubility of Ba(IO3)2 in a solution of 0.010 M NaIO3  

  • Given:
    • Ksp = 6 x 10-10
  • [IO3-] = [NaIO3] = 0.010
  • Ksp = [Ba2+] [IO3-]2
  • 6 x 10-10 = [Ba2+] (0.010)2
  • [Ba2+] = molar solubility = 6 x 10-6

Find the molar solubility of Ba(IO3)2 in a solution of 0.01 M NaNO3  

  • Given:
    • Ksp = 6 x 10-10
  • NaNO3 has no effect on the solubility of Ba(IO3)2
  • Ksp = [Ba2+] [IO3-]2
  • let c = molar solubility, concentration of Ba
  • Ksp = c x (2c)2 = 4c3
  • 6 x 10-10 = 4c3
  • c = 5.3 x 10-4
Subject: 
Chemistry
Subject X2: 
Chemistry

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