Question

0.1 M aqueous solution of a weak monobasic acid has pH 2.0. The pKa of the monobasic acid is _______. (rounded off to one decimal place)

Answer 1

Correct Answer: 2.9 - 3

We are given that the concentration of the acid is [HA] = 0.1M and the pH of the solution is 2.0. We need to find the pK_a of the acid.

From the pH, we can calculate the concentration of H⁺ ions using:

\( pH = 2.0 \Rightarrow [H^+] = 10^{-2} = 0.01M \)

Next, we apply the acid dissociation expression for a weak acid HA:

HA \rightleftharpoons H^+ + A^- ,and use the formula for the acid dissociation constant K_a:

\( K_a = \frac{[H^+][A^-]}{[HA]} \).

Assuming that the concentration of A⁻ formed is equal to the concentration of H⁺, we have:

\( K_a = \frac{(0.01)(0.01)}{0.1 - 0.01} = \frac{0.0001}{0.09} \approx 1.11 \times 10^{-3} \)

Now, to find the pK_a:

\( pK_a = - \log(K_a) = - \log(1.11 \times 10^{-3}) \approx 2.9 \)

Thus, the pK_a of the acid is approximately 2.9.