question_answer

The percentage of pyridine \[({{C}_{5}}{{H}_{5}}N)\] that forms pyridinium ion \[({{C}_{5}}{{H}_{5}}{{N}^{+}}H)\] in a 0.10 M aqueous pyridine solution \[({{K}_{b}}\,for\text{ }{{C}_{5}}{{H}_{5}}N=1.7\times {{10}^{-9}}\,\]is 

A)  0.0060%                      

B)  0.013%

C)  0.77%             

D)  1.6%

Correct Answer: B

Solution :

\[\therefore {{K}_{b}}=\frac{C{{\alpha }^{2}}}{1-\alpha }\approx C{{\alpha }^{2}}\] \[\therefore \alpha =\sqrt{\frac{{{K}_{b}}}{C}}=\sqrt{\frac{1.7\times {{10}^{-9}}}{0.1}}\] \[\therefore %\alpha =1.3\times {{10}^{-4}}\times 100=0.013%\]