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BCECE Engineering BCECE Engineering Solved Paper-2014

  • question_answer
    \[\text{2}.\text{5 mL}\] of \[\frac{2}{5}M\] weak monoacidic base\[({{K}_{b}}=1\times {{10}^{-12}}\text{ at 2}{{\text{5}}^{\circ }}C)\] is titrated with \[\frac{2}{15}M\] \[\text{HCl}\] in water at\[\text{25}{}^\circ \text{C}\]. The concentration of \[{{\text{H}}^{\text{+}}}\] at equivalence point is\[({{K}_{\omega }}=1\times {{10}^{-14}}\text{ at 2}{{\text{5}}^{\circ }}C)\]

    A) \[\text{3}.\text{7}\times \text{1}0{{-}^{\text{13}}}\text{M}\]                   

    B) \[3.2\times {{10}^{-7}}M\]

    C) \[3.2\times {{10}^{-2}}M\]                          

    D) \[2.2\times {{10}^{-2}}M\]

    Correct Answer: D

    Solution :

    Weak monoacidic base. e.g., B OH is neutralized as follows\[BOH+HCl\xrightarrow[{}]{{}}BCl+{{H}_{2}}O\] At equivalence point, all S OH get converted into salt. Remember! the concentration of \[{{H}^{+}}\] (or pH of solution) is due to hydrolysis of the resultant salt (B \[\text{Cl}\], cationic hydrolysis here.) \[\underset{C(1-h)}{\mathop{{{B}^{+}}}}\,+{{H}_{2}}O\rightleftharpoons \underset{Ch}{\mathop{BOH}}\,+H_{Ch}^{+}\] Volume of HCI used up, \[{{V}_{a}}=\frac{{{N}_{b}}{{V}_{b}}}{{{N}_{a}}}\] \[=\frac{2.5\times 2\times 15}{2\times 5}\] \[=7.5mL\] Concentration of salt, \[\text{ }\!\![\!\!\text{ BC }\!\!]\!\!\text{ }\frac{\text{conc}\text{.of base}}{\text{total volume}}\] \[=\frac{2\times 2.5}{5(7.5+2.5)}=\frac{1}{10}=0.1\]                 \[{{K}_{h}}=\frac{C{{h}^{2}}}{1-h}=\frac{{{K}_{w}}}{{{k}_{b}}}\] should be estimated whether that can 'be neglected or not) On calculating, h = 0.27 (significant, not negligible) \[[{{H}^{+}}]=Ch=0.1\times 0.27=2.7\times {{10}^{-2}}M\]


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