Question

Calculate the pH and the equilibrium concentrations of H2C6H5O7-, HC6H5O72- and C6H5O73- in a 0.1840 M aqueous citric acid solution. For H3C6H5O7, Ka1 = 7.4×10-3, Ka2 = 1.7×10-5, and Ka3 = 4.0×10-7

pH =

[H2C6H5O7-] = __M

[HC6H5O72-] =__ M

[C6H5O73-] = __M

Answer 1

           H3C6H5O7 (aq) ----------------> H^+ (aq) + H2C6H5O7^- (aq)

I          0.184                                          0                      0

C           -x                                             +x                     +x

E         0.184-x                                       +x                      +x

               ka1    = [H^+][H2C6H5O7^-]/[H3C6H5O7]

               7.4*10^-3   = x*x/(0.184-x)

             7.4*10^-3*(0.184-x) = x^2

                   x   = 0.0334

            [H^+]   = x    = 0.0334M

            [H2C6H5O7^-]   = x   = 0.0334M

           PH    = -log[H^+]

                   = -log0.0334

                    = 1.4762

             H2C6H5O7^-(aq) ---------------> H^+ (aq) + HC6H5O7^2-(aq)

                   Ka2      =   [H^+][HC6H5O7^2-]/[H2C6H5O7^-]

                    1.7*10^-5   = 0.0334[HC6H5O7^2-]/0.0334

                    [HC6H5O7^2-]    = 1.7*10^-5 M

        HC6H5O7^2-(aq) ---------------> H^+ (aq) + C6H5O7^3-(aq)

                   Ka3      =   [H^+][C6H5O7^3-]/[HC6H5O7^2-]

                  4*10^-7     = 0.0334[C6H5O7^3-]/1.7*10^-5

                 [C6H5O7^3-]      = 4*10^-7*1.7*10^-5/0.0334    = 2*10^-10M

           

pH =1.4762

[H2C6H5O7-] = 0.0334M

[HC6H5O72-] =1.7*10^-5M

[C6H5O73-] =    2*10^-10 M