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.
I got it!
The initial concentration of CN- would be 0.065 M rather than 0 M.
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.
To determine the molar solubility of Ni(CN)2 in a 0.065 M KCN solution, you need to consider the common ion effect. The presence of KCN in solution will increase the concentration of cyanide ions (CN^-), which can potentially react with Ni(CN)2 and decrease its solubility.
Using the Ksp expression for Ni(CN)2:
Ksp = [Ni^2+][CN^-]^2
Assuming x represents the molar solubility of Ni(CN)2 in the KCN solution, the concentration of Ni^2+ will also be x. Since each Ni(CN)2 molecule produces two CN^- ions, the concentration of CN^- will be (2x + 0.065) M (0.065 M from the KCN solution).
Substituting these values into the Ksp expression:
3.0 × 10^-23 = x * (2x + 0.065)^2
You can now solve this equation to find the molar solubility (x) of Ni(CN)2 in the 0.065 M KCN solution.
To determine the molar solubility of Ni(CN)2 in a 0.065 M KCN solution, we need to consider the common ion effect. The addition of KCN will provide extra CN- ions to the solution, which can affect the solubility of Ni(CN)2.
To solve this problem, let's assume the molar solubility of Ni(CN)2 in pure water is represented by "x" mol/L.
The balanced equation for the dissolution of Ni(CN)2 can be written as:
Ni(CN)2 (s) ⇌ Ni2+ (aq) + 2CN- (aq)
From the balanced equation, we can determine the expression for the solubility product constant (Ksp):
Ksp = [Ni2+][CN-]^2
Since [Ni2+] = x and [CN-] = 0.065 M (from the KCN solution), we can substitute the values into the Ksp equation:
Ksp = (x)(0.065)^2
Given that the value of Ksp for Ni(CN)2 is 3.0 × 10^(-23), we can set up the equation:
3.0 × 10^(-23) = (x)(0.065)^2
Simplifying the equation, we can solve for x:
x = (3.0 × 10^(-23)) / (0.065)^2
x = 1.957 × 10^(-8) M
Therefore, the molar solubility of Ni(CN)2 in the 0.065 M KCN solution is 1.957 × 10^(-8) mol/L.