Answer
The percentage of pyridine that forms pyridium ion is $0.01304 \%$.
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[OH^-] = [C_5H_5NH^+] = x$
-$[C_5H_5N] = [C_5H_5N]_{initial} - x = 0.1 - x$
For approximation, we consider: $[C_5H_5N] = 0.1M$
2. Now, use the Kb value and equation to find the 'x' value.
$Ka = \frac{[OH^-][C_5H_5NH^+]}{ [C_5H_5N]}$
$Ka = 1.7 \times 10^{- 9}= \frac{x * x}{ 0.1}$
$Ka = 1.7 \times 10^{- 9}= \frac{x^2}{ 0.1}$
$ 1.7 \times 10^{- 10} = x^2$
$x = 1.304 \times 10^{- 5}$
Percent ionization: $\frac{ 1.304 \times 10^{- 5}}{ 0.1} \times 100\% = 0.01304\%$
%ionization < 5% : Right approximation.
