Question

Find the pH of 0.190M NaCN solution. For HCN, Ka=4.9?10?10.

Answer 1

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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