Question

The acid-dissociation constant of hydrocyanic acid (HCN) at 25.0°C is 4.9 × 10^-10. What is the pH of an aqueous solution of 0.030 M sodium cyanide (NaCN)?

Answer 1

We Know that Kb x Ka = 10^-14

then Kb =  10^-14 /Ka =  10^-14 / 4.9 x  10^-10 = 2.04 x 10^-5

The reaction is

CN^- + H₂O HCN + OH^-

Now lets see ICE table for above reaction

CN^- + H₂O HCN + OH^-

Initial 0.030 0 0

Change -x +x +x

Equilibrium 0.03-x x x

Kb for above reaction at equilibrium is

Kb = [HCN][ OH^- ]/[CN^-]

So

2.04 x 10^-5 = (x)(x) / (0.03-x) = x² / 0.03-x)

x² + 2.04x x 10-5 - 0.0612 x 10^-5 = 0

by solving above quadratic equation we get

x = 0.00077 M = [OH^-]

So , pOH = -log[OH^-]

pOH = -log(0.00077) = 3.11

And

pH = 14 - pOH

= 14 - 3.11

= 10.89