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
IÂ Â Â Â Â Â Â Â Â
0.73Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â 0
CÂ Â Â Â Â Â Â
-x                               Â
+x       +x
EÂ Â Â Â Â Â Â
0.73-x                           Â
+x         Â
+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
