Question

What range of pH would permit the selective precipitation of Cu2+ as Cu(OH)2 from a solution that contains 0.088 M Cu2+ and 0.088 M Mn2+? For Mn(OH)2, Ksp = 1.6 × 10-13, and for Cu(OH)2, Ksp = 4.8 × 10-20.

When pH = _____ – _____ Mn(OH)2 will be soluble, but some Cu(OH)2 will precipitate out of solution.

Answer 1

Cu(OH)2 <--< Cu+2 + 2OH-

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

[Cu+2] = 0.088 M

[Mn+2] = 0.088 M

Ksp Cu = 4.8*10^-20

Ksp Mn = 1.6*10^-13

since Ksp fo Cu is smaller, then it will precipitate first

assume this will be saturated:

4.8*10^-20 = (0.088)(OH)^2

[OH-] = sqrt(4.8*10^-20))/(0.088))

[OH-] = 2.489647*10^-9 M

pOH = -log(2.489647*10^-9) = 8.60386

now.. calculate until Mn(OH)2 precipitates:

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

1.6*10^-13 = (0.088)(OH-)^2

[OH-] = sqrt((1.6*10^-13)/(0.088)) = 0.0000013483

pOH = -log(0.0000013483) = 5.870

pH = 14-5.870 = 8.13

pH range:

8.13< pH <8.60