Question

What is the percentage of pyridine $$\text{(}{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N)}$$ that forms pyridinium ion $$\text{(}{\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N}}^{\text{ + }}\text{H)}$$ in a 0.10 M aqueous pyridine solution ( $${\text{K}}_{\text{b}}$$ for $${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}$$$$=1.7\times {10}^{-9}$$)?
(A) 0.77%
(B) 1.6%
(C) 0.0060%
(D) 0.013%

Answer 1

Hint: The law of dilution gives us a relation between dissociation constant and the degree of dissociation. This relation can be explained as: the degree of dissociation of a weak electrolyte ($$\alpha$$) is directly proportional to dilution constant $${\text{K}}_b$$, and it is inversely proportional to the concentration $${\text{C}}_{\text{0}}$$.

Complete step by step solution:
For the reaction,
Dilution constant can be written as:
$$K_b={\displaystyle \frac{\left[A^{+}\right]\left[B^{-}\right]}{\left[AB\right]}}={\displaystyle \frac{\left(\alpha C_0\right)\left(\alpha C_0\right)}{\left(1-\alpha \right)C_0}}={\displaystyle \frac{{\alpha }^2}{1-\alpha }}\times C_0$$
Where $$\alpha$$is the degree of dissociation of a weak electrolyte. And $${\text{C}}_{\text{0}}$$ is the concentration.
For weak electrolyte, $$\alpha <<0$$ so $$\left(1-\alpha \right)$$ can be neglected and the resulting equation is:
So, $$\alpha =\sqrt{{\displaystyle \frac{K_b}{C_0}}}$$
So, for pyridine on dilution with water results in pyridinium ion. In question, we are given that molarity of pyridine solution is 0.10M and $${\text{K}}_{\text{b}}$$ for $$C_5{H}_{5}N$$$$=1.7\times {10}^{-9}$$.
$$\alpha =\sqrt{{\displaystyle \frac{K_b}{C_0}}}=\sqrt{{\displaystyle \frac{1.7\times {10}^{-9}}{0.10}}=}1.30\times {10}^{-4}$$
So the degree of dissociation of pyridinium ion $$\alpha =$$$$1.30\times {10}^{-4}$$.
Therefore, percentage of pyridine that forms pyridinium ion is $$1.30\times 1{\text{0}}^{\text{ - 4}}\times 100=0\text{.013\% }$$.

Hence the correct option is (D).

Note: The Degree of dissociation of any solute within a solvent is basically the ratio of molar conductivity at C concentration and limiting molar conductivity at zero concentration or infinite dilution. This can be mathematically represented as $$\alpha ={\displaystyle \frac{{\mathrm{\Lambda }}_C}{{\mathrm{\Lambda }}_0}}$$.