At $27^{\circ} \mathrm{C}$. the degree of dissociation of HA (weak acid) in 0.5 M of its solution is $1 \%$.…
- $0.005,0.005,0.495$
- $0.05,0.05,0.45$
- $0.01,0.01,0.49$
- $0.005,0.495,0.005$
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)