Question

Calculate the pH at 25°C of a 0.73M solution of potassium cyanide KCN . Note that hydrocyanic acid HCN is a weak acid with a pKa of 9.21 .

Answer 1

find out the Kb of HCN

pKa + pKb = 14

pKb = 14-pKa = 14 - 9.21 = 4.79

Kb = 10^-pKb = 10^-4.79 = 1.6 x 10^-5

now construct the ICE table

           KCN + H2O <---> HCN + KOH

I          0.73                             0         0

C        -x                                 +x        +x

E        0.73-x                             +x           +x

Kb = [HCN][KOH] / [KCN]

1.6 x 10^-5 = [x][x] /[0.73-x]

x² + x 1.6 x 10^-5 - 1.168*10^-5 = 0

solve the quadratic equation

x = 0.0034 = [KOH]

pOH = -log(0.0034) = 2.47

pH = 14-POH = 14-2.47 = 11.53