How do you find molar solubility, given pKsp and pH?

Question

For ferrous hydroxide, pKsp = 15.097. What is the base-10 logarithm of the molar solubility of this compound when pH = 7.968?

The answer is -3.033, I just need to know how to work the problem.

Thanks for any help!!!

Answer 1

If s is the solubility of Fe(OH)2 then from the eqilibrium:

Fe(OH)2(s) <==> Fe2+(aq) + 2OH-(aq)

the concentration of Fe2+ ions will be also s mol/L. You can find the molarity of OH- ions from the pH of the solution:

pOH = 14 - pH, pOH = 14 - 7.968, pOH = 6.032. Now:

Ksp = [Fe2+]*[OH-]^2. Taking logarithms and multiplying by -1 both sides we get:

-log(Ksp) = -log[Fe2+] + 2*(-log[OH-]). But:

-log(Ksp) = pKsp, -log[Fe2+] = -log(s), -log[OH-] = pOH, so

pKsp = -log(s) + 2pOH, => -log(s) = pKsp - 2pOH,

-log(s) = 15.097 - 2*6.032 = 3.033, so

log(s) = 3.033

Answer 2

ph=10.a million then pOH = 14 - 10.a million = 3.9 pOH = - log[OH-] = 3.9 to locate conc. [OH-] ksp = [Fe+2][OH-]^2 Use the ksp fee and [OH-] above to locate [Fe+2]. this might inform you procedures plenty Fe(OH)2 has dissolved. Do the comparable for b) yet [OH-] would be distinctive through fact pH is distinctive