Question

what is the PH of a 1.00 molar solution of sodium cyanide NaCN, The ionization constant for hydrocyanic acid is 6.2*10^-10

Answer 1

                NaCN(aq) -------------> Na^+(aq) + CN^- (aq)

                  1M                                                   1M

                   CN^- (aq) + H2O(l) ---------------> HCN(aq) + OH^- (aq)

          I        1                                                   0                  0

         C        -x                                                 +x                 +x

         E        1-x                                               +x                  +x

                        Kb    = Kw/Ka

                                   = 1*10^-14/(6.2*10^-10)   = 1.6*10^-5

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

                     1.6*10^-5   = x*x/1-x

                      1.6*10^-5 *(1-x) = x^2

                          x = 0.004

                       [OH^-]   = x   = 0.004M

                       POH   = -log[OH^-]

                                 = -log0.004

                                  = 2.3979

                           PH   = 14-POH

                                    = 14-2.3979

                                     = 11.6021 >>>>answer