Question

The percentage of pyridine $\left(C_5{H}_{5}N\right)$ that forms pyridinium ion $\left(C_5{H}_5{N}^{+}H\right)$ in a 0.10 M aqueous pyridine solution $(K_b$ for $C_5{H}_{5}N=1.7\times {10}^{-9}$ ) is

• A. $0.77\%$
• B. $1.6\%$
• C. $0.0060\%$
• D. $0.013\%$

Answer 1

The correct option is D $0.013\%$

$C_5{H}_{5}N+H_{2}O\to C_5{H}_{5}N{H}^{+}+O{H}^{-1}$

$K_b=C{\alpha }^2$

$\alpha =\sqrt{\frac{K_b}{C}}$

$\alpha =\sqrt{\frac{1.7\times {10}^{-9}}{0.1}}$

$\alpha =1.30\times {10}^{-4}$

$percentageof\alpha =1.30\times {10}^{-4}\times 100=0.013\%$

Hence option D is the correct answer