Calculate the ionisation constant of $0.08 \mathrm{~mol~} \mathrm{dm}^{-3}$ of a monobasic acid having…

Ionization constant calculation

Given a monobasic acid solution with concentration $C = 0.08 \text{ mol dm}^{-3}$ and pH = 2. The hydrogen ion concentration is determined from the definition:

$[\mathrm{H}^+] = 10^{-\text{pH}} = 10^{-2} = 0.01 \text{ mol dm}^{-3}$

For the dissociation $\mathrm{HA} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^-$, the ionization constant is expressed as:

$K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}$

Using the equilibrium concentrations: $[\mathrm{H}^+] = [\mathrm{A}^-] = 0.01$ and applying the approximation $[\mathrm{HA}] \approx C = 0.08$ (commonly used when dissociation is small), we compute:

$K_a = \frac{(0.01)(0.01)}{0.08} = \frac{0.0001}{0.08} = 1.25 \times 10^{-3}$

This corresponds to option D.