Question

The percentage of pyridine $\left(C_5{H}_{5}N\right)$ that forms pyridium ion $\left(C_5{H}_5{N}^{+}H\right)$ in a $0.10M$ aqueous pyridine solution is:

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

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

Answer 1

The correct option is C $0.77\%$

Let $x$ be the concentration of $C_5{H}_5{N}^{+}H$ formed or $C_5{H}_{5}N$ reacted.

$\therefore$ $K_b=\frac{K_w}{K_a}$

$\therefore K_a=\frac{K_w}{K_b}$

$p{K}_w=14$

$\therefore -\log k_w=14$

$\therefore k_w={10}^{-14}$

$K_a=\frac{{10}^{-14}}{1.7\times {10}^{-9}}=5.8\times {10}^{-6}$

$K_a=\frac{x^2}{(0.1-x)}=5.8\times {10}^{-6}$

$x=0.00077$ M

$\therefore$ % of $C_5{H}_{5}N$ that forms $C_5{H}_5{N}^{+}H=\frac{0.00077}{0.1}\times 100$

$=0.77$ %

Hence, the correct option is $\text{C}$