Calculate the solubility product

From the following half-cell reactions, calculate the solubility product (Ksp) of Mg(OH)2 (s):

Mg2+(aq) + 2e- <-------> Mg(s) E = -2.360 V
Mg(OH)2(s) + 2e- <--------> Mg(s) + 2OH-(aq) E = -2.690 V

a) 1.21 x 10^12 b) 2.37 x 10^-5 c) 6.98 x 10^7 d) 6.32 x 10^7 e) 4.21 x 10^-21


Please help me! Thank you :)

We want this reaction:

Mg(OH)2(s) <===> Mg2+(aq) + 2OH-(aq)

We can get that by reversing the first equation and adding the two equations together:

Mg(s) <-------> Mg2+(aq) + 2e- <-------> E = +2.360 V
Mg(OH)2(s) + 2e- <--------> Mg(s) + 2OH-(aq) E = -2.690 V

When we add the two voltages together, we obtain -0.33 V

Next, we use this relationship:

nFE = RT ln K

ln K = nFE / RT

ln K = [(2) (96,485) (-0.33)] / [(8.314) (298)]

ln K = -25.7

K = 6.9 x 10^-12

That's not any of your answer choices and I can't find any mistakes I may have made. You might want to look at this:

https://docs.google.com/viewer?a=v&q=cac…

Take a look at problem #2