Question

Calculate the mass of NaCN that must be added to 250.0 mL of water in order to obtain a solution having a pH of 11.25? [Ka of HCN = 4.9 × 10–10] Also, calculate the % ionization of the of the solution.

Answer 1

Kb = Kw / Ka = 1.0 x 10^-14 / 4.9 x 10^-10

      = 2.04 x 10^-5

pH = 11.25

pOH = 2.75

[OH-] = x = 1.78 x 10^-3 M = x

CN -   + H2O ---------------> HCN + OH-

C                                              0         0 ------------> initial

C -x                                          x          x-----------------> equilibrium

Kb = [HCN][OH-]/[CN-]

2.04 x 10^-5 = x^2 / C -x

2.04 x 10^-5 = (1.78 x 10^-3 )^2 / (C -1.78 x 10^-3 )

C = 0.0157 M

% ionisation = (x / C) x 100

                      = (1.78 x10^-3 / 0.0157) x 100

                      = 11.33%

C = 0.0157 M

volume = 250 mL = 0.25 L

moles = 0.0157 x 0.25 = 3.925 x 10^-3

NaCN mass = moles x molar mass

                    = 3.925 x 10^-3 x 49

                    = 0.19 g