For a weak acid $\mathrm{HA}$, the percentage of dissociation is nearly $1 \%$ at equilibrium. If the…

For a weak acid $\mathrm{HA}$, the percentage of dissociation is nearly $1 \%$ at equilibrium. If the concentration of acid is $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$, then the correct option for its $\mathrm{K}_{\mathrm{a}}$ at the same temperature is
  1. $1 \times 10^{-5}$
  2. $1 \times 10^{-3}$
  3. $1 \times 10^{-4}$
  4. $1 \times 10^{-6}$

Solution

$\begin{aligned} & \mathrm{HA} \rightleftharpoons \mathrm{H}^{+}+\mathrm{A}^{-} \\ & \text {At } \mathrm{t}=0 \quad \mathrm{C} \quad- \\ & \text { At eq. } \mathrm{C}-\mathrm{C} \alpha \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \\ & \mathrm{K}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}=\frac{(\mathrm{C} \alpha)(\mathrm{C} \alpha)}{\mathrm{C}(1-\alpha)}=\frac{\mathrm{C} \alpha^2}{1-\alpha}=\mathrm{C} \alpha^2 \quad(\because \alpha \text { is very small }) \\ & =0.1(0.01)^2=1 \times 10^{-5} \\ & \end{aligned}$

Asked in: NEET 2023 (Manipur)

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