Question

Calculate the pH of 1.0L of 0.10 M pyridine solution to which 0.3 mol of pyridinium chloride ${C_5}{H_5}N{H^ + }C{l^ - }$ , has been added, assuming no change in volume.

Answer 1

Note that we are not given the value of $p{K_b}$ hence we’ll solve this question without using the $p{K_b}$ . The pH of a solution containing a weak acid (pyridine) and its conjugate base (Pyridinium Chloride) can be given by the Henderson-Hasselbalch equation. This equation is given as:
$pOH = p{K_b} + \log \left( {\dfrac{{[{\text{conjugate acid]}}}}{{[{\text{weak base]}}}}} \right)$ .

Complete answer:
When a weak base (pyridine) is dissolved with the conjugate acid (pyridinium chloride) to form a basic buffer solution. We have been given the concentrations of the weak base. The concentration of the base (pyridine) is 0.10M.
The no. of moles of the salt formed is 0.3moles. the concentration of the salt pyridinium chloride in 1L solution can be given as: $c = \dfrac{n}{V} = \dfrac{{0.3}}{1} = 0.3M$ (where n is the no. of moles and V is the volume)
Therefore, $[{C_3}{H_5}N] = [Weak{\text{ }}Base] = 0.1M$ and $[{C_3}{H_5}N{H^ + }C{l^ - }] = 0.3M$
The value of ${K_b}$ is considered the standard value which is ${K_b} = 1.5 \times {10^{ - 9}}$ . The value of $p{K_b} = - \log ({K_b}) = - \log (1.5 \times {10^{ - 9}})$
$p{K_b} = 8.8239$
Substitute the values in the Henderson-Hasselbalch equation, to get the value of pOH
$pOH = 8.8239 + \log \left( {\dfrac{{0.3}}{{0.1}}} \right)$
$pOH = 8.8239 + 0.47712$
$pOH = 9.30$
The relationship between pH and pOH can be given as: $pH + pOH = 14$
The pH of the solution of pyridine and pyridinium chloride is: $pH = 14 - 9.30 = 4.69$ . The pH of 1.0L of 0.10 M pyridine solution to which 0.3 mol of pyridinium chloride ${C_5}{H_5}N{H^ + }C{l^ - }$ is 4.69.
This is the required answer.

Note:
Basic buffer is the solution which is formed by the mixture of weak base and its conjugate acid. Buffer solution is the one that doesn’t promote the change in pH of the solution. Acidic buffer is the solution of weak acid and its conjugate base. The pH of the acidic buffer solution by the Henderson-Hasselbalch can be given as: $pH = p{K_a} + \log \left( {\dfrac{{[{\text{conjugate base]}}}}{{[{\text{weak acid]}}}}} \right)$ .