Question

$0.1{\text{ mol}}$ of methyl amine $\left( {{K_b} = 5 \times {{10}^{ - 4}}} \right)$is mixed with $0.08{\text{ mol}}$ of $HCl$and the solution is diluted to $1{\text{ litre}}$. Determine the hydrogen ion concentration of the resulting solution.

Answer 1

Hint:To answer this question, you must recall the concept of basic buffer. A basic buffer is formed when a solution contains a weak base and its slat with a strong acid.
Formula used:
$pOH = p{K_b} + \log \dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$

Complete step by step answer:
In the question, we are given $0.1{\text{ mol}}$of methyl amine and $0.08{\text{ mol}}$of $HCl$. We know that methyl amine is a weak base and hydrochloric acid is a strong acid. The reaction taking place in the reaction mixture can be written as:
  $HCl + C{H_3}N{H_2}\xrightarrow[{}]{{}}C{H_3}N{H_3}Cl$

Before reaction	\[0.08\]	\[0.1\]	0
After reaction	0	\[0.02\]	\[0.08\]

 After the reaction occurs, the reaction mixture contains $0.08$moles of $C{H_3}N{H_3}Cl$and $0.02$moles of methyl amine. We can clearly see that this is a mixture of a weak base (methyl amine) and its salt with a strong acid (hydrochloric acid). So the solution is a basic buffer solution.
We know that the pOH of this solution can be given by the reaction
$pOH = p{K_b} + \log \dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$.
We can simplify the equation as
 $ - \log \left[ {OH} \right] = - \log {K_b} + \log \dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$
$ \Rightarrow \left[ {O{H^ - }} \right] = {K_b}\dfrac{{\left[ {base} \right]}}{{\left[ {salt} \right]}}$
Substituting the known values into the equation, we get,
$\left[ {O{H^ - }} \right] = \dfrac{{5 \times {{10}^{ - 4}} \times 0.02}}{{0.08}}$
Therefore, the hydroxide ion concentration in the solution is $1.25 \times {10^{ - 4}}$
We know that the ionization constant of water is given as ${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] = {10^{ - 14}}$
Using this equation, we can find the hydrogen ion concentration as $\left[ {{H^ + }} \right] = \dfrac{{{K_w}}}{{\left[ {O{H^ - }} \right]}}$
$\therefore \left[ {{H^ + }} \right] = 8 \times {10^{ - 11}}$

Note:
A buffer is a solution that can maintain its pH or the hydrogen on concentration with addition of small amounts of acid or base or on dilution. Since the pH of the solution does not change with minor changes, thus it can be used in various applications namely, fermentation, drug delivery, electroplating, activity of enzymes, etc.