WBJEE · Chemistry · Ionic Equilibrium
\(2.5 \mathrm{ml}\) of \(0.4(\mathrm{M})\) weak monoacidic base \(\left(\mathrm{k}_{\mathrm{b}}=1 \times 10^{-12}\right.\) at \(\left.25^{\circ} \mathrm{C}\right)\) is titrated with \(\frac{2}{15}(\mathrm{M}) \mathrm{HCl}\) in water at \(25^{\circ} \mathrm{C}\). The concentration of \(\mathrm{H}^{+}\)at equivalence point is \(\left(\mathrm{k}_{\mathrm{w}}=1 \times 10^{-14}\right.\), at \(\left.25^{\circ} \mathrm{C}\right)\),
- A \(3.7 \times 10^{-13}\) (M)
- B \(3.2 \times 10^{-7}(\mathrm{M})\)
- C \(3.2 \times 10^{-2}(\mathrm{M})\)
- D \(2.7 \times 10^{-2}(\mathrm{M})\)
Answer & Solution
Correct Answer
(C) \(3.2 \times 10^{-2}(\mathrm{M})\)
Step-by-step Solution
Detailed explanation
\(\mathrm{N}_{1} \mathrm{~V}_{1}=\mathrm{N}_{2} \mathrm{~V}_{2}\) (At equivalence point) \(0.4 \times 2.5=\frac{2}{15} \times \mathrm{V}_{2}\) \(\mathrm{BOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{B} \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(I)\)…
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