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

• Correct answer
  A

  $0.005,0.005,0.495$
• B

  $0.05,0.05,0.45$
• C

  $0.01,0.01,0.49$
• D

  $0.005,0.495,0.005$

Solution

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

Hence, the correct answer is option: 'A'.