Question

If the Ka of a monoprotic weak acid is 7.0 × 10^-6, what is the pH of a 0.24 M solution of this acid?

Answer 1

Let given monoprotic acid be designated by HA. The equilibrium established can be shown as follows:

HA    =   H⁺    +   A^-

Let 's' be the amount of HA dissociated at equilibrium. Thus, ICE chart can be shown as follows:

                                HA          =               H⁺         +                A^-

Initial                     0.24M                   0                              0

Change                 -s                            +s                           +s

Equilibrium         0.24-s                    s                              s

Expression for equilibrium constant (Ka) can be written as follows:

Ka = [H⁺][A^-] / [HA]

7.0 × 10^-6 = s x s / (0.24-s)

For weak acid, s can be assumed to be very small.

Therefore, s can be neglected from (0.24 – s) expression,

7.0 × 10^-6 = s x s / 0.24

s2 = (7.0 × 10^-6) x 0.24

s2 = 1.68 × 10^-6

s = 1.30 × 10^-3

Therefore, [H⁺] = 1.30 × 10^-3

pH = -log [H⁺]

pH = -log (1.30 × 10^-3)

pH = 2.89

Therefore, pH of the 0.24M solution of given monoprotic acid is 2.89