Question

The percentage of pyridine $\left(C_5{H}_{5}N\right)$ that forms pyridinium ion $\left(C_5{H}_{5}NH\right)$ in a $0.10$M aqueous pyridine solution is:

[$K_b$ for $C_5{H}_{5}N=1.7\times {10}^{-9}$]

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

Answer 1

The correct option is A $0.77\%$

We know that

$K_a=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.7\times {10}^{-9}}=5.88\times {10}^{-6}$

$C_5{H}_{5}N+H_{2}O\rightleftharpoons C_5{H}_{5}NH+C_5{H}_{5}N{H}^{+}+O{H}^{-}$

Initial conc: $0.1$ $0$ $0$ $0$

Final Conc: $0.1-x$ $x$ $x$

$K_a=\frac{x^2}{0.1-x}$ [Since $x<<0.1$ it can be neglected]

$5.88\times 0.1\times {10}^{-6}=x^2$

$x=7.6\times {10}^{-4}$

%$Pyridine=\frac{7.6\times {10}^{-4}}{0.1}\times 100=0.77\%$

$\because (\frac{dissociatedions}{concentrationofsolution}\times 100)$

Hence, the correct option is $A$