MHT CET · Chemistry · Ionic Equilibrium
In a solution, \(0.02 \mathrm{M}\) acetic acid is \(4 \%\) dissociated. The \(\left[\mathrm{OH}^{-}\right]\) in the solution is
- A \(8 \times 10^{-4}\)
- B \(2 \times 10^{-14}\)
- C \(8 \times 10^{10}\)
- D \(1.25 \times 10^{-11}\)
Answer & Solution
Correct Answer
(D) \(1.25 \times 10^{-11}\)
Step-by-step Solution
Detailed explanation
\(\left[\mathrm{H}^{+}\right]=\sqrt{\alpha^{2} C^{2}}=\sqrt{(0.02)^{2} \cdot(0.04)^{2}}=8 \times 10^{-4}\)
\(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}\)
\(\therefore\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{8 \times 10^{-4}}=1.25 \times 10^{-11} \mathrm{M}\)
\(\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}\)
\(\therefore\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{8 \times 10^{-4}}=1.25 \times 10^{-11} \mathrm{M}\)
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