MHT CET · Chemistry · Chemical Equilibrium
An acid dissociated to \(1.5 \%\) in its \(0.1 \mathrm{M}\) solution. Calculate its dissociation constant.
- A \(1.2 \times 10^{-5}\)
- B \(2.25 \times 10^{-5}\)
- C \(1.1 \times 10^{-5}\)
- D \(1.5 \times 10^{-5}\)
Answer & Solution
Correct Answer
(B) \(2.25 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & K_a=\alpha^2 C \\ & =\left(\frac{1.5}{100}\right)^2 \times 0.1 \\ & =2.25 \times 10^{-5}\end{aligned}\)
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