MHTCET 11 May 2024 Shift 1 Paper

Concept:For a weak monobasic acid, the dissociation constant is calculated from equilibrium concentrations of ions and undissociated acid.Explanation:Let the acid be $\mathrm{HA}$.Its dissociation is:$\mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-}$Given: initial concentration $C = 0.02\ \mathrm{M}$ and degree of dissociation $\alpha = 5\% = 0.05$.At equilibrium:$[\mathrm{H}^{+}] = C\alpha = 0.02 \times 0.05 = 0.001\ \mathrm{M}$$[\mathrm{A}^{-}] = C\alpha = 0.001\ \mathrm{M}$$[\mathrm{HA}] = C(1-\alpha) = 0.02(1-0.05) = 0.019\ \mathrm{M}$The dissociation constant is given by:$$K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}$$Substituting the values:$$K_a = \frac{(0.001)(0.001)}{0.019} = \frac{1 \times 10^{-6}}{0.019}$$$$K_a = 5.26 \times 10^{-5} \approx 5 \times 10^{-5}$$Answer:The dissociation constant of the acid is $5 \times 10^{-5}$, which matches option C.