AP EAMCET · Chemistry · Ionic Equilibrium
At \(27^{\circ} \mathrm{C}\), the degree of dissociation of weak acid (HA) in its 0.5 M aqueous solution is \(1 \%\). Its \(\mathrm{K}_{\mathrm{a}}\) value is approximately
- A \(5 \times 10^{-4}\)
- B \(5 \times 10^{-5}\)
- C \(5 \times 10^{-6}\)
- D \(5 \times 10^{-8}\)
Answer & Solution
Correct Answer
(A) \(5 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
degree of dissociation of weak acid (HA) \(=1 \%\) \(\alpha=1 \%\) \(\Rightarrow \frac{1}{100}=0.01\) Dissociation constant \(\left(\mathrm{K}_{\mathrm{a}}\right)=\mathrm{C} \alpha^2\) where, \(C(\text { concentration })=0.5 \mathrm{M}\)…
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