What is the pH of 50.0ml of 1.150M MgCl2 added to 50.0ml of 0.300M NaOH? ksp of Mg(OH)2 is 1.2E-1?

Question

I got the answer 13.18, but I didn't use the ksp anywhere in my problem solving.

Answer 1

The final volume of the solution is 50+50=100 ml = double the volume of each of the solutions. So the concentration of each compound is going to be half in the final solution: 0.575 M MgCl2 and 0.150 M NaOH

MgCl2 -> Mg(+2) + 2Cl- so the initial concentration of Mg+2 is [Mg+2]0=[MgCl2]=0.575
similarly [OH-]0=[NaOH]=0.150

Mg(OH)2 <=> Mg(+2) + 2OH-

so Ksp=[Mg2+][OH-]^2=0.12
If P=[Mg2+]0[OH-]0^2=Ksp or PKsp then there will be precipitation which will sequester some OH- and the pH will be smaller.

P=[Mg2+]0[OH-]0^2 =0.575*(0.15^2) =0.0129 < Ksp so the solution is below saturation, there will be no precipitation and thus pH=13.18.

HOWEVER I found on the web that for Mg(OH2) Ksp=5.16*10^-12. Then the answer would be completely different. So please check that the value Ksp=1.2*10^-1 is what they truly gave you.

Answer 2

what??