MHT CET · Chemistry · Ionic Equilibrium
Find \(\left[\mathrm{OH}^{-}\right]\)if a monoacidic base is \(3 \%\) ionised in its \(0.04 \mathrm{M}\) solution.
- A \(3.1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}\)
- B \(4.5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
- C \(9.0 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}\)
- D \(1.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
Answer & Solution
Correct Answer
(D) \(1.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\)
Step-by-step Solution
Detailed explanation
For a monoacidic base,
\(
\begin{aligned}
& {\left[\mathrm{OH}^{-}\right]=\mathrm{c} \times \alpha} \\
& {\left[\mathrm{OH}^{-}\right]=0.04 \times \frac{3}{100}} \\
& {\left[\mathrm{OH}^{-}\right]=1.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}}
\end{aligned}
\)
\(
\begin{aligned}
& {\left[\mathrm{OH}^{-}\right]=\mathrm{c} \times \alpha} \\
& {\left[\mathrm{OH}^{-}\right]=0.04 \times \frac{3}{100}} \\
& {\left[\mathrm{OH}^{-}\right]=1.2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}}
\end{aligned}
\)
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