Question

What is the Ka for a weak acid, HA, if a .2 M HA solution has a pH of 4.0?

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 Concentration , c = 0.2 M

also pH = 4.0

- log[H⁺] = 4.0

       [H⁺] = 10 ^-4.0

              = 1.0x10^-4 M

       ca = 1.0x10^-4 M

        a = (1.0x10^-4 M ) / 0.2

          = 5.0x10^-4

Therefore K_a = ca²

                  = 0.2 x (5.0x10^-4 )²

                  = 5.0x10^-8

Therefore the Ka of this weak acid is 5.0x10^-8