a.(
(HCN ka = 4.0 x 10-10)
Kb = 10^-14 / Ka = 10^-14 / 4 x 10^-10 = 2.5 x 10^-5
CN- + H2OÂ Â --------------> HCN + OH-
0.237Â 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 Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â 0
0.237 -
x                              Â
x         x
Kb = x^2 / 0.237 - x
2.5 x 10^-5 = x^2 / 0.237 - x
x = 2.42 x 10^-3
[OH-] = 2.42 x 10^-3 M
pOH = -log [OH-] = -log (2.42 x 10^-3)
       = 2.62
pH = 11.38
b.)
Kb = 1.8 x 10^-5
pH = 7 - 1/2 (pKb + log C)
     = 7 - 1/2 (4.74 + log 0.166)
     = 5.02
pH = 5.02
