Question

Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows: C5H5N+H2O⇌C5H5NH++OH− The pKb of pyridine is 8.75. What is the pH of a 0.335 M solution of pyridine?

Answer 1

Answer : The pH of the solution is, 9.4

Solution :  Given,

Concentration (c) = 0.025 M

Base dissociation constant = $pK_b=8.75$

First we have to calculate the value of $K_b$.

The expression used for the calculation of $pK_b$ is,

$pK_b=-\log (K_b)$

Now put the value of $pK_b$ in this expression, we get:

$8.75=-\log (K_b)$

$K_b=1.78\times 10^{-9}$

The given equilibrium reaction is,

                         $C_5H_5N+H_2O\rightleftharpoons C_5H_5NH^++OH^-$

initially conc.     0.335                    0                0

At eqm.           (0.335-x)                  x                x

Formula used :

$k_b=\frac{[C_5H_5NH^+][OH^-]}{[C_5H_5N]}$

Now put all the given values in this formula ,we get:

$1.78\times 10^{-9}=\frac{(x)(x)}{(0.335-x)}$

By solving the terms, we get:

$x=2.4\timees 10^{-5}$

Now we have to calculate the concentration of hydroxide ion.

$[OH^-]=x=2.4\times 10^{-5}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (2.4\times 10^{-5})$

$pOH=4.6$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.6\\\\pH=9.4$

Therefore, the pH of the solution is, 9.4