At $27^{\circ} \mathrm{C}$. the degree of dissociation of HA (weak acid) in 0.5 M of its solution is $1 \%$.…

At $27^{\circ} \mathrm{C}$. the degree of dissociation of HA (weak acid) in 0.5 M of its solution is $1 \%$. The concentrations of $\mathrm{H}_3 \mathrm{O}^{-}$, $\mathrm{A}^{-}$and HA at equilibrium (in $\mathrm{mol} \mathrm{L}^{-1}$ ) are respectively
  1. $0.005,0.005,0.495$
  2. $0.05,0.05,0.45$
  3. $0.01,0.01,0.49$
  4. $0.005,0.495,0.005$

Solution


$\begin{aligned} & {\left[\mathrm{H}^{+}\right]=\mathrm{c} \alpha } \\ & =0.5 \times\left(10^{-2}\right) \\ & =0.005 \\ \therefore & {\left[\mathrm{H}_3 \mathrm{O}\right]^{+}=0.005 } \\ & {\left[\mathrm{~A}^{-}\right]=0.005 } \\ & {[\mathrm{HA}]=0.5-\mathrm{x} } \\ & =0.5-0.005 \\ & =0.495\end{aligned}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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