Answer
The percentage of pyridine that forms pyiridinium ion is equal to 0.013%.
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$C_5H_5N(aq) + H_2O(l) \lt -- \gt C_5H_5NH^+(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $C_5H_5NH^+$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M
-$[OH^-] = [C_5H_5NH^+] = 0 + x = x$
-$[C_5H_5N] = [C_5H_5N]_{initial} - x $
For approximation, we are going to consider $[C_5H_5N]_{initial} = [C_5H_5N]$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][C_5H_5NH^+]}{ [C_5H_5N]}$
$Kb = 1.7 \times 10^{- 9}= \frac{x * x}{ 0.10}$
$Kb = 1.7 \times 10^{- 9}= \frac{x^2}{ 0.10}$
$x^2 = 1.7 \times 10^{-9} \times 0.10$
$x = \sqrt { 1.7 \times 10^{-9} \times 0.10} = 1.3 \times 10^{-5}$
Percent ionization: $\frac{ 1.3 \times 10^{- 5}}{ 0.10} \times 100\% = 0.013\%$
%ionization < 5% : Right approximation.
