| [A] The pH of a mixture containing 400 mL of \[0.1\,M\,{{H}_{2}}S{{O}_{4}}\]and 400 mL of 0.1 M NaOH will be approximately 1.3. |
| [B] Ionic product of water is temperature dependent. |
| [C] A monobasic acid with \[{{K}_{a}}={{10}^{-5}}\] has a pH = 5. The degree of dissociation of this acid is 50%. |
| [D] The Le Chatelier's principle is not applicable to common-ion effect. the correct statement are : |
A) [B] and [D]
B) [A], [B] and [C]
C) [A] and [B]
D) [B] and [C]
Correct Answer: B
Solution :
[A]\[\underset{20}{\mathop{\underset{400\times .1=40}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,}}\,+\underset{0}{\mathop{\underset{400\times .1=40}{\mathop{2NaOH}}\,}}\,\xrightarrow[{}]{{}}N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O\] \[\therefore \]\[[{{H}^{+}}]=\frac{20\times 2}{800}=\frac{1}{20}\Rightarrow pH=-\log \left( \frac{1}{20} \right)\] \[\therefore \]\[pH=1.3\]so [a] is correct [B] \[\log \left( \frac{K{{\text{w}}_{2}}}{K{{\text{w}}_{1}}} \right)=\frac{\Delta H}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\] so ionic product of water is temp. dependent hence is correct. [C] \[{{K}_{a}}={{10}^{-5}},pH=5\Rightarrow [{{H}^{+}}]={{10}^{-5}}\] \[{{K}_{a}}=\frac{c{{\alpha }^{2}}}{(1-\alpha )}\Rightarrow {{K}_{a}}=\frac{[{{H}^{+}}].\alpha }{(1-\alpha )}\] \[\therefore \]\[{{10}^{-5}}=\frac{{{10}^{-5}}.\alpha }{(1-\alpha )}\Rightarrow 1-\alpha =\alpha \Rightarrow \alpha =\frac{1}{2}=50%\] so [c] is correct. [D] Le-chatelier's principle is applicable to common -Ion effect so option is wrong \[\therefore \] correct answer [b]
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