Four hydroxide precipitates have the Ksp vlaues tablulated below: Assume no meta

Question

Four hydroxide precipitates have the Ksp vlaues tablulated below: Assume no metal hydrolysis occurs at any pH and assume ionic strength effects are equal for all the ions. Assume all four metal ions, Ba^2+ Ca^2+ Cu^2+ and Mn^2+ are together in an acidic solution, and each is 1 mM (0.001 M). KOH is then added slowly with vigorous stirring. Which metal hydroxide will precipitate first? What is the order of precipitation after that? At what pH will the first hydroxide precipitate form, assuming equilibrium conditions? Does the solution need to be basic (pH > 7) to precipitate hydroxides?

Explanation / Answer

1) Ba(OH)2                  Ksp = 5.0 x 10^-3

2) Ca(OH)2                   Ksp = 1.3 x 10^-6

3) Cu(OH)2                   Ksp = 1.6 x 10^-19

4) Mn(OH)2                  Ksp = 2 x 10^-13

a) among all precipitates Cu(OH)2 has low solubility product value. so Cu(OH)2 precipitates first.

the order is

Cu(OH)2

Mn(OH)2

Ca(OH)2

Ba(OH)2

2) Cu(OH)2   ------------------>   Cu+2   + 2 OH-

                                                    S             2S

Ksp = [Cu+2][OH-]^2

1.6 x 10^-19 = S x (2S)^2

1.6 x 10^-19 = 4S^3

S = 3.42 x 10^-7

[OH-] = 2S = 6.84 x 10^-7 M

pOH = -log[OH-]

        = - log (6.84 x 10^-7)

        = 6.16

pH + pOH = 14

pH = 14--pOH

      = 14 - 6.16

      = 7.84

pH = 7.84

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