Question

Determine the pH of a 0.115M sodium citrate (NAC6H7O7) solution. The Ka for citric acid (HC6H7O7) is 7.4X10^-4

Answer 1

NaC6H7O7(aq) ---------------> Na^+ (aq) + C6H7O7^- (aq)

0.115M                                                        0.115M

              C6H7O7^- (aq) + H2O(l) -------------------> HC6H7O7(aq) + OH^- (aq)

I              0.115                                                              0                      0

C            -x                                                                    +x                     +x

E            0.115-x                                                            +x                     +x

                  Kb     = Kw/Ka

                             = 1*10^-14/7.4*10^-4   = 1.35*10^-11

                       Kb     = [HC6H7O7][OH^-]/[C6H7O7^-]

                      1.35*10^-11   = x*x/0.115-x

                      1.35*10^-11*(0.115-x) = x^2

                          x = 1.25*10^-6

                           [OH^-]   = x   = 1.25*10^-6M

                           POH   = -log[OH^-]

                                      = -log1.25*10^-6

                                       = 5.9030

                              PH   = 14-POH

                                       = 14-5.9030   = 8.097 >>>>answer