A) \[3.7\times {{10}^{-13}}M\]
B) \[3.2\times {{10}^{-7}}M\]
C) \[3.2\times {{10}^{-2}}M\]
D) \[2.7\times {{10}^{-2}}M\]
Correct Answer: D
Solution :
Weak monoacidic base, e.g., B OH is neutralised as follows \[BOH+HCl\xrightarrow{{}}BCl+{{H}_{2}}O\] At equivalence point, all B OH get converted into salt. Remember! the concentration of \[{{H}^{+}}\] (or pH of solution) is due to hydrolysis of the resultant salt (B \[Cl\], cationic hydrolysis here.) \[\underset{C\,\,(1-h)}{\mathop{{{B}^{+}}}}\,\,+{{H}_{12}}O\underset{Ch}{\mathop{BOH}}\,+\underset{Ch}{\mathop{{{H}^{+}}}}\,\] Volume of \[HCl\] used up, \[{{V}_{a}}=\frac{{{N}_{b}}{{V}_{b}}}{{{N}_{a}}}\] \[=\frac{2.5\times 2\times 15}{2\times 5}\] = 7.5 mL Concentration of salt, \[[BCl]=\frac{cone.\text{ }of\,base}{total\text{ }volume}\] \[=\frac{2\times 2.5}{5\,(7.5+2.5)}\] \[=\frac{1}{10}=0.1\] \[{{K}_{h}}=\frac{C{{h}^{2}}}{1-h}=\frac{{{K}_{w}}}{{{K}_{b}}}\] (h 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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