Question

Determine the Ka of an acid whose 0.364M solution has a pH of 1.80.

Answer 1

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

initial conc.            c               0         0

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 α is very small so 1-a is taken as 1

So K_a = ca²

Given pH = 1.80

- log[H⁺] = 1.80

        [H⁺] = 10 -1.80

               = 0.016M

[H⁺] = ca = 0.016 M

Where c = acid concentration = 0.364 M ( given)

So a = 0.016 / c

       = 0.016 / 0.364

       = 0.0435

So K_a = ca²

     K_a = 0.364 x 0.0435²

          = 6.90x10^-4

Therefore the K_a of the acid is 6.90x10^-4