MHT CET · Chemistry · Ionic Equilibrium
Calculate dissociation constant of a weak monobasic acid if it is \(0.05 \%\) dissociated in 0.02 M solution.
- A \(2.0 \times 10^{-9}\)
- B \(3.0 \times 10^{-9}\)
- C \(4.0 \times 10^{-9}\)
- D \(5.0 \times 10^{-9}\)
Answer & Solution
Correct Answer
(D) \(5.0 \times 10^{-9}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{HA} \rightleftharpoons \mathrm{H}_{\text {(aq) }}^{\oplus}+\mathrm{A}_{(\mathrm{aq})}^{\Theta} \\ & \begin{aligned} & \begin{aligned} & =0.05 \%\end{aligned}=0.05 \times 10^{-2} \\ & \mathrm{~K}_{\mathrm{a}}= \alpha^2 \mathrm{C} \\ &=\left(0.05 \times 10^{-2}\right)^2 \times 0.02 \mathrm{M} \\ &=5.0 \times 10^{-9}\end{aligned}\end{aligned}\)
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