Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN⁻(aq)
+ Na⁺(aq).
Chemical reaction 2: CN⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻)
= 0.021 M.
Ka(HCN) = 4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻]
/ [CN⁻].
[HCN] · [OH⁻] =
x.
[CN⁻] = 0.021 M - x..
2.04·10⁻⁵ = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH⁻] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.
