If the Ksp of Zn (CN)2 (s) is 3×10^-16, what percentage of Zn2+ is precipitated when we know that

Question

[Zn 2+]=0.36M

[CN-]=5×10^-3M

Ksp=3×10^-16

Answer 1

I looked at this two ways. If Ksp is

Zn(CN)2 ==> Zn^2+ + 2CN^- then the reverse of that is

Zn^2+ + 2CN^- ==> Zn(CN)2 \ 1/Ksp = 3.333E15

That is such a HUGE number that we can make an assumption that ALL of the CN^- will be used which would leave 0.36 - 0.0025 = about 0.357 or approx (0.357/0.36)*100 = about 99.3% pptd.

OR we cn do it this way.

............Zn^2+ + 2CN^- ==> Zn(CN)2

I..........0.36........0.005.............0

C...........-x..........2x.................x

E.......0.36-x.....0.005-2x..........x

Then K=3.333E15 = [Zn(CN)2]/(Zn^2+)(CN^-)^2

Plug in the E line and solve for x.It's a cubic equation but you can find solutions on Google. I didn't check my answer but i obtained 0.0012 for x which is (Zn^2+). Then 0.36-0.0012 = 0.359, then

(0.359/0.36)*100 = 99.7%