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Calculate the pH of a 0.021 M NaCN solution, given
Ka(HCN) = 4.9 x 10-10
Answer
Given data
Ka of HCN = 4.9 x 10-10
Molarity of NaCN, C = 0.021 M
  ? Kb of NaCN = Kw / Ka
                             =1.0
*10-14 / 4.9 x 10-10
                            =
2.04*10-5
  Here Kb is small so
The concentration of OH- ion,[OH-]= (?Kb *
C)
                                                         =
( 2.04*10-5 * 0.021 M)0.5
                                                         =
6.546*10-4 M
                                                   pOH=
- log (6.546*10-4 M)
                                                           =
3.1839
                                                     pH
= 14 - 3.1839
                                                           =
10.816
