I got it!
The initial concentration of CN- would be 0.065 M rather than 0 M.
The Ksp of Ni(CN)2 is 3.0 × 10−23. What is
the molar solubility of Ni(CN)2 in a 0.065 M KCN solution?
Answer in units of mol/L.
I found that the normal molar solubility is 1.957x10^-8, but I don't know how the KCN solution affects it.
Sig figs aren't accounted for.
2 answers
Technically the CN- really is 0 initially and 2x + 0.065 at equilibrium.
......Ni(CN)2 ==> Ni^2+ + 2CN^-
I.....solid........0.......0
C.....solid........x.......2x
E.....solid........x.......2x
Then ......KCN --> K^+ + CN^-
I........0.065.....0......0
C.......-0.065....0.065..0.065
E..........0..... 0.065..0.065
Then Ksp = (Ni^2+)(CN^-)^2
Substitute Ni^2+ = x from above
Substitute CN = 0.065 from KCN and 2x from Ni(CN)2 so it looks like this
Ksp = (x)(2x+ 0.065)^2
Then you make the simplifying assumption that 0.065+ 2x = 0.065 and that assumes 2x is too small to make much difference. The answer comes out to be the same answer you obtained but with a little funny stuff in between. It helps to do it this way so that when 2x + 0.065 does NOT allow the simplifying assumption, you have the set up to go through with the quadratic assumption.
......Ni(CN)2 ==> Ni^2+ + 2CN^-
I.....solid........0.......0
C.....solid........x.......2x
E.....solid........x.......2x
Then ......KCN --> K^+ + CN^-
I........0.065.....0......0
C.......-0.065....0.065..0.065
E..........0..... 0.065..0.065
Then Ksp = (Ni^2+)(CN^-)^2
Substitute Ni^2+ = x from above
Substitute CN = 0.065 from KCN and 2x from Ni(CN)2 so it looks like this
Ksp = (x)(2x+ 0.065)^2
Then you make the simplifying assumption that 0.065+ 2x = 0.065 and that assumes 2x is too small to make much difference. The answer comes out to be the same answer you obtained but with a little funny stuff in between. It helps to do it this way so that when 2x + 0.065 does NOT allow the simplifying assumption, you have the set up to go through with the quadratic assumption.
3 answers