Question

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 2.160. The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10.

Hint: Use the systematic treatment of equilibrium to solve this problem.

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

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 2.16 then (H^+) = about 4E-3
Then (4E-3)(CN^-)/(HCN) = 6.2E-10=(4E-3)(CN^-)/(HCN) and solve for (HCN) = 6.56E6(CN^-)

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