Question

\[{\text{2}}{\text{.5 mL}}\] of \[\dfrac{2}{5}{\text{M}}\] weak monoacidic base ( \[{K_b} = 1 \times {10^{ - 12}}\] at \[{\text{2}}{{\text{5}}^o}{\text{C}}\]) is titrated with \[\dfrac{2}{15}\] M HCl in water at \[{\text{2}}{{\text{5}}^o}{\text{C}}\] .
The concentration of $H^+$ at equivalence point is:
\[k_w=1 \times10^{-14} \;at \;25^0C\]
(A) \[3.7 \times {10^{ - 13}}M\]
(B) \[3.2 \times {10^{ - 7}}M\]
(C) \[3.2 \times {10^{ - 2}}M\]
(D) \[2.7 \times {10^{ - 2}}M\]

Answer 1

Hint: Write the expression for the hydrolysis of cation of the salt at the equivalence point. Use the expression for the acid dissociation constant of conjugate acid of base and calculate the degree of hydrolysis. Calculate the hydrogen ion concentration from the degree of hydrolysis and salt concentration.

Complete step by step answer:
Let \[{\text{BOH}}\] represent the weak monoacidic base.
Write balanced acid base neutralization reaction.
\[{\text{BOH +HCl }} \to {\text{ BCl + }}{{\text{H}}_2}{\text{O}}\]
Calculate the volume of \[{\text{HCl}}\] needed to reach the equivalence point.
\[{{\text{V}}_{{\text{HCl}}}}{\text{ = }}\dfrac{{{{\text{M}}_{{\text{BOH}}}}{\text{ }} \times {\text{ }}{{\text{V}}_{{\text{BOH}}}}}}{{{{\text{M}}_{{\text{HCl}}}}}}{\text{ }} \\
{{\text{V}}_{{\text{HCl}}}}{\text{ = }}\dfrac{{\dfrac{2}{5}{\text{M}} \times {\text{2}}{\text{.5 mL}}}}{{\dfrac{2}{{15}}{\text{M}}}} \\
{{\text{V}}_{{\text{HCl}}}} = 7.5{\text{ mL}} \\\]
Calculate the concentration of the salt at the equivalence point:
\[\left[ {{\text{BOH}}} \right]{\text{ = }}\dfrac{{\dfrac{2}{5}{\text{M}} \times {\text{2}}{\text{.5 mL}}}}{{7.5{\text{ mL}}}}{\text{ }} \\
\left[ {{\text{BOH}}} \right]{\text{ = 0}}{\text{.1 M}} \\\]
Consider hydrolysis of salt at the equivalence point:
\[{{\text{B}}^ + }{\text{ + }}{{\text{H}}_2}{\text{O }} \rightleftharpoons {\text{ BOH + }}{{\text{H}}^ + } \\
{\text{C}}\left( {1 - {\text{h}}} \right){\;\;\;\;\;\;\;\;\;\;\text{ Ch}\;\;\;\;\;\;\;Ch} \\\]
\[{K_b} = 1 \times {10^{ - 12}}\]
\[{K_w} = 1 \times {10^{ - 14}}\]
Calculate \[{K_a}\]
\[{K_a}{\text{ = }}\dfrac{{{K_w}}}{{{K_b}}} = \dfrac{{1 \times {{10}^{ - 14}}}}{{1 \times {{10}^{ - 12}}}} = 1 \times {10^{ - 2}}\]
Calculate the degree of hydrolysis
\[{K_h}{\text{ = }}\dfrac{{\left[ {{\text{BOH}}} \right] \times \left[ {{{\text{H}}^ + }} \right]}}{{\left[ {{{\text{B}}^ + }} \right]}} \\
1 \times {10^{ - 2}} = \dfrac{{{{\left( {{\text{Ch}}} \right)}^2}}}{{{\text{C}}\left( {1 - {\text{h}}} \right)}} \approx\dfrac{{{{\left( {{\text{Ch}}} \right)}^2}}}{{\text{C}}}{\text{ = C}}{{\text{h}}^2}{\text{ as h <<< 1}} \\
1 \times {10^{ - 2}}{\text{ = }}\left( {{\text{0}}{\text{.1 M}}} \right){{\text{h}}^2} \\
{\text{h}} = 0.27\;{\text{M}} \\\]
Calculate hydronium ion concentration
\[\left[ {{{\text{H}}^ + }} \right]{\text{ = Ch}} = {\text{0}}{\text{.1 }}\times0.27 = 2.7{\kern 1pt} \times {10^{ - 2}}{\text{M}}\]
Hence, the hydrogen ion concentration at the equivalence point is \[2.7 \times {10^{ - 2}}M\]

Hence, the correct option is the option (D).

Note: The approximation \[\dfrac{{{{\left( {{\text{Ch}}} \right)}^2}}}{{{\text{C}}\left( {1 - {\text{h}}} \right)}} \approx\dfrac{{{{\left( {{\text{Ch}}} \right)}^2}}}{{\text{C}}}\] can be used because \[{\text{h <<< 1}}\] . This is because \[{K_h}\] has small value.