A) 50 ml.
B) 100 ml.
C) 200 ml.
D) None of these
Correct Answer: B
Solution :
Let V ml of \[RN{{H}_{3}}Cl\]added into \[RN{{H}_{2}}\]solution\[[RN{{H}_{2}}Cl]\]in resultant solution \[=\frac{0.2\times V}{100+V}\] \[[RN{{H}_{2}}]=\frac{100\times 0.1}{100+V};pOH=p{{K}_{b}}+\log \frac{[RNH_{3}^{+}]}{[RN{{H}_{2}}]}\]\[5.3=5+\log \left[ \frac{\left( \frac{0.2\times V}{100+V} \right)}{\frac{100\times 0.1}{100+V}} \right]\therefore 2\]\[=\frac{0.2\times V}{10}\Rightarrow V=100ml.\]
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