Question

2. If a solution contained 0.045 M of NaCN, what would be the pH of the solution? (HCN Ka = 4.9x10-10)

Answer 1

Ka HCN = 4.9 x 10^-10. Let us find out Kb of It's conjugate base CN^-.

Ka x Kb = Kw = 1.0 x 10^-14.

(4.9 x 10^-10) x Kb = 1.0 x 10^-14.

Kb = 1.0 x 10^-14 / 4.9 x 10^-10.

Kb = 2.04 x 10^-5.

NaCN is a strong electrolyte and ionizes fully as,

NaCN (aq) ---------> Na⁺ (aq) + CN^- (aq).

[CN-] = {NaCN] = 0.045 M

But CN- (aq) being strong base hydrlyzes back to HCN as,

CN^- (aq) + H-OH <------> HCN + HO^- (aq)

Kb = [HCN][HO^-] / [CN^-] = 2.04 x 10^-5. ----------------- (1)

Initially, [CN^-] = [NaCN] = 0.045 M

Let at equlibrium 'X' M of CN- hydrolyzed and hence we write ICE table as,

CN^- (aq) <------> HCN + HO^- (aq)

Initially 0.045 0 0

Change -X +X +X

At Eqm. (0.045-X) X X

Using these euilibrium concenrations in eq.(1) we get,

(X)(X) / (0.045-X) = 2.04 x 10^-5.

X² / (0.045-X) = 2.04 x 10^-5.

SMall concentration assumption says 0.045 >> X so 0.045-X = 0.045 (apprx)

So we get,

X² / 0.045 = 2.04 x 10^-5.

X² = 0.045 x 2.04 x 10^-5.

X² = 91.80 x 10^-8.

X = 9.58 x 10^-4.

By ICE table,

[HO^-] = X = 9.58 x 10^-4 M.

By definition of pOH,

pOH = -log[HO^-]

pOH = -log(9.58 x 10^-4)

pOH = 3.02

Then,

pH + pOH = 14

pH + 3.02 = 14

pH = 14 - 3.02

pH = 10.98

pH of 0.045 M NaCN solution is 10.98.

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