When pKb for CN- (CN ion) at 25 degree C is 4.7 , then pH of 0.5 M NaCN aqueous solution will be?

1) 11
2) 11.5
3) 2.5
4) 12

Answer:
CN- + H2O <> HCN + OH-
Initial concentration
0.5 . . . . . . .. . .. .0 .. . . . 0
at equilibrium
0.5-x .. . . . . . . . . x . .. . .x
Kb = 10^-4.7 = 1.99x 10 ^-5 = x^2 / 0.5 - x
x = [OH-] = 0.00315 M
pOH = 2.59
pH = 14 - 2.5 = 11.5
The answer is 2)
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