Question

The pK_a of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous solution of chloroacetic acid at pH 2.4, what percentage of molecules/ions will be in the basic form?

Answer 1

Let's denote the acid by HA and proceed to set up an ICE table with equilibrium dissociation of α. Also at pH = 2.4, there will be an initial H⁺ concentration of 3.98 * 10^-3 M

                       HA →              H⁺         +      A^-

initial conc:      0.1                            3.98 * 10^-3                   0

equilibrium conc:      0.1(1-α)           3.98 * 10^-3 + 0.1α              0.1α

We have pK_a = 2.9

-log(K_a) = 2.9

K_a = 1.26 * 10^-3

From the equation, K_a = [H⁺] * [A^-] / [HA]

1.26 * 10^-3 = (3.98 * 10^-3 + 0.1α )* 0.1α / 0.1(1-α)     

Since α<<1, we assume 1-α = 1

Solving the equation, we have: α = 0.094

Since this is the fraction of acid that has dissociated, we can say that % of Base form = 100 * α= 9.4%