MHT CET · Chemistry · Chemical Equilibrium
A weak base is \(1.42 \%\) dissociated in its \(0.05 \mathrm{M}\) solution. Calculate its dissociation constant.
- A \(5.5 \times 10^{-5}\)
- B \(4.0 \times 10^{-5}\)
- C \(1.8 \times 10^{-5}\)
- D \(1.0 \times 10^{-5}\)
Answer & Solution
Correct Answer
(D) \(1.0 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
Percent dissociation \(=1.42 \%\)
\(\therefore \quad \alpha=0.0142\)
For a weak monoacidic base,
\(\begin{aligned}
\mathrm{K}_{\mathrm{b}} & =\alpha^2 \mathrm{C} \\
& =(0.0142)^2 \times(0.05) \\
& =1.0082 \times 10^{-5}
\end{aligned}\)
\(\therefore \quad \alpha=0.0142\)
For a weak monoacidic base,
\(\begin{aligned}
\mathrm{K}_{\mathrm{b}} & =\alpha^2 \mathrm{C} \\
& =(0.0142)^2 \times(0.05) \\
& =1.0082 \times 10^{-5}
\end{aligned}\)
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