Question

A saturated solution of Mg(OH)2 is prepared having a large excess of Mg(OH)2. Sn(NO3)2 is added to the solution. Ksp = 1.8 10-11 for Mg(OH)2 and Ksp = 5.1 10-26 for Sn(OH)2.
(a) What [Sn2+] is required to start the precipitation of Sn(OH)2?
(b) What [Sn2+] is required so that the [Mg2+] in solution will be 0.14 M?

Answer 1

First determine the OH^- already in the solution from Mg(OH)2.
.......Mg(OH)2 ==> Mg2+ + 2OH^-
I.......solid.......0......0
C.......solid.......x......2x
E.......solid.......x......2x

Ksp = (Mg^2+)(OH^-)^2
Solve for x = Mg and OH = 2x.

Then go through the Sn(OH)2 the same way, subtitute OH from the Mg(OH)2 calculation and solve for (Sn^2+).

b. Go back to Mg(OH)2, substitute 0.14M for (Mg^2+) and solve for OH^-. Substitute that OH^- into the Sn(OH)2 and solve for (Sn^2+)

Answer 2

Thank you! I got this one right.