Question

A 0.160 M weak acid solution has a pH of 3.90. Find Ka for the acid.

Answer 1

1) Write the dissociation equation for the acid:

HA <===> H⁺ + A¯

2) Write the equilibrium expression:

K_a = ( [H⁺] [A¯] ) / [HA]

3) Our task now is to determine the three concentrations on the right-hand side of the equilibrium expression since the K_a is our unknown.

a) We will use the pH to calculate the [H⁺]. We know pH = -log [H⁺], therefore [H⁺] = 10¯^pH

[H⁺] = 10¯^3.26 = 1.2589 x 10¯⁴ M

b) From the dissociation equation, we know there is a 1:1 molar ratio between [H⁺] and [A¯]. Therefore:

[A¯] = 1.2589 x 10¯⁴ M

c) the final value, [HA] is given in the problem. In the example being discussed, 0.120 M is the value we want. Some teachers will use 0.120, while others would say to subtract the 5.4954 x 10¯⁴value from 0.120 first. Let's do both.

c1) K_a = [(1.2589 x 10¯⁴ M) (1.2589 x 10¯⁴ M)] / 0.160

K_a = 9.90 x 10¯⁸