MHT CET · Chemistry · Ionic Equilibrium
The degree of ionization of \(0.4 \mathrm{M}\) acetic acid will be \(\left(K_{a}=1.8 \times 10^{-5}\right)\)
- A \(6.71 \times 10^{-3}\)
- B \(1.6 \times 10^{-3}\)
- C \(0.4 \times 1.8 \times 10^{-5}\)
- D \(1.8 \times 10^{-5}\)
Answer & Solution
Correct Answer
(A) \(6.71 \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
For a weak acid (acetic acid),
\(
\begin{aligned}
K_{a} &=C \alpha^{2} \\
1.8 \times 10^{-5} &=0.4 \times \alpha^{2} \\
\alpha &=\sqrt{\frac{1.8 \times 10^{-5}}{0.4}} \\
&=6.71 \times 10^{-3}
\end{aligned}
\)
\(
\begin{aligned}
K_{a} &=C \alpha^{2} \\
1.8 \times 10^{-5} &=0.4 \times \alpha^{2} \\
\alpha &=\sqrt{\frac{1.8 \times 10^{-5}}{0.4}} \\
&=6.71 \times 10^{-3}
\end{aligned}
\)
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