Solve Ksp Equation for Sn^2+ When MnS Begins to Precipitate

[Newtime]

Homework Statement

For stannous sulfide, SnS, the Ksp = 1.0X10^-25
For manganese sulfide, MnS, the Ksp = 3.0 x 10^-15

Assume a solution contains .01 M Sn^2+ and .01 M Mn^2+. If the sulfide ion (S^2-) concentration is increased gradually without any dilution of the first solution, what will be the molar concentration of Sn^(2+) ion when manganese sulfide first starts to precipitate?

Homework Equations

Ksp = ?

The Attempt at a Solution

I've tried this question in what I, and a few of my friends who are stuck on it with me, consider a logical progression. I ended up with [Sn] = 1E-25, [Mn] = 3E-15, =1E-27 and =3E-17, respectively. [Sn][3E-17]=1E-15 and I end up with [Sn]=3.33E-9M. The answer is 3.33E-13 and I don't understand how it was attained. Any help is appreciated, thank you.

 

[Borek]

0.01x10^-17 is not 10^-15.

 

[Newtime]

0.01x10^-17 is not 10^-15.

Which part of my post are you referring to?

 

[Borek]

Anyone. You did the same mistake several times.