Question

An aqueous solution of aspirin (HA) is prepared at pH 7.4. The ratio of concentration of \( \text{A}^- \) and \( \text{HA} \) at equilibrium is ________. (round off to the nearest integer).

Hint:

For weak acid dissociation, the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \) can be calculated using the dissociation constant and the concentration of \( \text{H}^+ \) from the pH.

Answer 1

Correct Answer: 9700

For a weak acid like aspirin, the dissociation equilibrium is given by: \[ \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \] The dissociation constant \( K_a \) is: \[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}. \] We are given \( K_a = 3.98 \times 10^{-4} \), and we are asked to find the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \) at equilibrium. At pH 7.4, the concentration of \( \text{H}^+ \) is: \[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-7.4} \approx 4.0 \times 10^{-8} \, \text{M}. \] Now, using the expression for \( K_a \): \[ 3.98 \times 10^{-4} = \frac{(4.0 \times 10^{-8}) [\text{A}^-]}{[\text{HA}]}. \] Since the concentration of \( \text{H}^+ \) and \( \text{A}^- \) are approximately equal (because \( \text{HA} \) dissociates to form \( \text{A}^- \) in a 1:1 ratio), we can write: \[ [\text{A}^-] \approx [\text{H}^+] = 4.0 \times 10^{-8} \, \text{M}. \] So, the ratio of concentrations is approximately: \[ \frac{[\text{A}^-]}{[\text{HA}]} \approx \frac{4.0 \times 10^{-8}}{1.0 \times 10^{-3}} = 0.04. \] Thus, the ratio of \( \text{A}^- \) to \( \text{HA} \) at equilibrium is approximately \( \boxed{0.04} \).