Question

Equilibrium constant for the following reaction is $1 \times {10^{ - 9}}$
${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N(aq}}{\text{.) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^{\text{ + }}}{\text{(aq}}{\text{.) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq}}{\text{.)}}$
Determine the mole of pyridinium chloride (${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}{\text{.HCl}}$) that should be added to 500ml solution of 0.4M pyridine (${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$) to obtain buffer solution of ${\text{pH = 5}}$ .
A.${\text{0}}{\text{.1mole}}$
B.${\text{0}}{\text{.2mole}}$
C.${\text{0}}{\text{.3mole}}$
D.${\text{0}}{\text{.4mole}}$

Answer 1

The constant defined for the equations of chemical reactions at equilibrium is known as equilibrium constant. It is basically the ratio of forward and backward rate of reactions at the equilibrium. The solution which resists the change in pH of a solution upon the addition of a small amount of weak acid or base is known as buffer solution.

Complete step by step answer:
Equilibrium constant is a constant defined for the chemical reactions which is basically the ratio of pressure or concentration units of products to reactants when reaction is at equilibrium stage. Only aqueous phase units are included in this term equilibrium constant, solids and excess reagents or products are excluded during the calculation of equilibrium constant.
Given: Equilibrium constant =$1 \times {10^{ - 9}}$ and ${\text{pH = 5}}$
Reaction: ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N(aq}}{\text{.) + }}{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^{\text{ + }}}{\text{(aq}}{\text{.) + O}}{{\text{H}}^{\text{ - }}}{\text{(aq}}{\text{.)}}$
A water solvent based solution which basically contains a mixture of weak acid and its conjugate base is known as buffer solution. Mostly their work is to resist the change in pH of the solution if a small amount of weak acid or weak bases are added to it.
So here we have a buffer solution, and according to chemical equilibrium we can say that the relation between pH and acid or base concentrations for buffer solutions is as follows:
${\text{pH = pKa}}{\text{ - log}}\dfrac{{{\text{[base]}}}}{{{\text{[acid]}}}}$
$\Rightarrow {\text{5 = 14 - pKb}}{\text{ + log}}\dfrac{{{\text{[pyride]}}}}{{{\text{[C6}}{\text{H5}}{\text{NHCl]}}}}$
let us suppose the moles of pyridinium chloride is $x$ .
\Rightarrow 5 = 10 - 9 + log\dfrac{{0.4}}{x} \times 0.5 \\\ \Rightarrow log\dfrac{{0.2}}{x} = 0 \\\ \Rightarrow \dfrac{{0.2}}{x} = 1 \\\ \Rightarrow x = 0.2 \\\
Therefore the required moles pyridinium chloride is 0.2 mol.

Hence, option (B) is correct.

Note:
Pyridinium is a weak acid. When pyridinium chloride is introduced to water it splits itself into ions, chloride ion and pyridinium ion. This makes the solution a little acidic (due to presence of ${{\text{H}}^{\text{ + }}}$).