Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 2.390. The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10.

Question

I know the following is what needs to be done, but im unsure of how to go about it.

1. Write the pertinent reactions. 2. Write the charge balance equation. 3. Write the mass balance equations. 4. Write the equilibrium constant expressions for each reaction. 5. Count the equations and unknowns. 6. Solve for all unknowns.

Answer 1

Yes but there is an easier way to do it.
Let S = solubility
Zn(CN)2==> Zn^2+ + 2CN^-,then + H^+ =>HCN
..solid....S.......2S
Ksp = (Zn^2+)(CN^-)^2

For HCN ==> H^+ + CN^- You should confirm all of this.
Ka = 6.2E-10 = (H^+)(CN^-)/(HCN)
If pH 2.39 then (H^+) = about 4E-3 but you can do it more accurately than that.
Then (4E-3)(CN^-)/(HCN) = 6.2E-10=(4E-3)(CN^-)/(HCN) and solve for (HCN) = 6.56E6(CN^-) and substitute this into the below equation.

You know (Zn^2+) = S and you know
2S = (CN^-) + (HCN)
2S = (CN^-) + 6.56E6 = essentially 6.56E6.
S = 3.28E6(CN^-) and
(CN^-) = S/3.28E6
Now substitute S for Zn and S/3.28E6 for CN and solve for S.

Answer 2

I am a bit confused at the substitution part at the end. Can you please break it down further so I can find the concentrations?