Question

Calculate the apparent solubility of benzoic acid at pH 5. Benzoic acid is a weak acid, with pK_a = 4.2 and solubility of nonionized species S₀ = 0.028 mol/L.
Please show steps

Answer 1

Let a be the dissociation of the weak acid
                            HA <---> H ⁺ + A^-

initial conc.            c               0         0

change                -ca            +ca      +ca

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids a is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given pK_a = 4.2

   - log K_a = = 4.2

          K_a = 10^-4.2

               = 6.31x10^-5

pH = - log[H⁺] = 5

               [H⁺] = 10^-5 M

                 ca = 10^-5 M ----> c = 10^-5 /a

solubility of nonionized species = c(1-a) = 0.028

                                     (10^-5 /a ) (1-a) = 0.028

                                                   1-a = 0.028xa / 10^-5

                                                   1-a = 2800 a

                                                     a = 1 / 2801

                                                        = 3.57x10^-4

So c = 10^-5 /a = 10^-5 /(3.57x10^-4) = 0.028 M

∴ Apparent solubility of benzoic acid is ca = 0.028x 3.57x10^-4 = 9.99x10^-6 M