Answer
is: pH of solution of sodium cyanide is 11.07.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l)
⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.068 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⁻]
= 0.068 M - x..
2.04·10⁻⁵ = x² / (0.068 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00116 M.
pOH = -log(0.00116 M) = 2.93.
pH = 14 - 2.93 = 11.07.
The pH of the sodium cyanide solution is 11.56.
Let the cyanide ion be X
We have to set up the ICE table for the problem as follows;
X^-(aq) + H2O(l) ⇄ HX(aq) + OH^-(aq)
I 0.068 x x
C -x + x +x
E 0.068 - x x x
But Kb = Kw/Ka = 1 × 10^-14/4.9 × 10-10
Kb = 2 × 10^-4
So;
Kb = [HX] [OH^-]/[X^-]
2 × 10^-4 = x^2/ 0.068 - x
2 × 10^-4(0.068 - x) = x^2
1.36 × 10^-5 - 2 × 10^-4x = x^2
x^2 + 2 × 10^-4x - 1.36 × 10^-5 = 0
x= 0.0036 M
Since x = [OH^-] = 0.0036 M
pOH = - log(0.0036 M)
pOH = 2.44
pH = 14 - 2.44 = 11.56
The pH of the sodium cyanide solution is 11.56.
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