AP EAMCET · Chemistry · Ionic Equilibrium
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
- A \(0.005,0.005,0.495\)
- B \(0.05,0.05,0.45\)
- C \(0.01,0.01,0.49\)
- D \(0.005,0.495,0.005\)
Answer & Solution
Correct Answer
(A) \(0.005,0.005,0.495\)
Step-by-step Solution
Detailed explanation
\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…
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