Question

The pH of an aqueous monoprotic weak acid solution is 6.20 at 25 C. Calculate the Ka for the acid if the initial concentration is 0.010 M.

Answer 1

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

initial conc.            c               0         0

change                -cα            +cα      +cα

Equb. conc.         c(1-α)          cα    cα

Dissociation constant , K_a = cα x cα / ( c(1-α)

                                         = c α² / (1-α)

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

So K_a = cα²

==> α = √ ( K_a / c )

Given c = concentration = 0.010 M

pH = 6.20

- log[H⁺] = 6.20

[H⁺] = 10^-6.20 = 6.31x10^-7M

cα = 6.31x10^-7M

α = 6.31x10^-5

K_a = cα² = 0.010 x (6.31x10^-5 )²

= 3.98x10^-11

Therefore Ka for the acid is  3.98x10^-11