Question

Calculate the pH of 0.14 M RbOCl solution

Answer 1

RbOCl(aq) ------------------> Rb^+ (aq) + OCl^- (aq)

0.14M                                                          0.14M

                        OCl^- (aq) + H2O ----------------> HOCl(aq) + OH^-(aq)

          I              0.14                                        0                 0

          C            -x                                                     +x               +x

           E           0.14-x                                               +x               +x

                           Kb    = Kw/Ka

                                    = 1*10^-14/3.5*10^-8

                                      = 2.86*10^-7

                          Kb    = [HOCl][OH^-]/[OCl^-]

                          2.86*10^-7   = x*x/0.14-x

                       2.86*10^-7 *(0.14-x) = x^2

                                x   = 0.0002

                      [OH^-]    =x   = 0.0002M

                       POH   = -log[OH^-]

                                 = -log0.0002

                                  =3.6989

                        PH    = 14-POH

                                  = 14-3.6989

                                  = 10.3011 >>>>answer