Question

What is the $[O{H^ - }]$ in a 0.01M solution of aniline hydrochloride? ${K_b}$ for aniline is $4.0 \times {10^{ - 10}}$?

Answer 1

To get the concentration of hydroxide ions, pOH calculation can be done in the terms of dissociation constant and the molarity of the solution. Dissociation constant is the ratio of dissociated product ions to reactant ions.

Complete step-by-step solution:
Aniline hydrochloride is a salt that reacts with water and dissociates into aniline, which is a weak base. It is also a weak acid having aniline as its conjugate base. Concentration of hydroxide ion depends on the molarity of the solution. It is used in the titration process against strong bases like sodium hydroxide.
${C_6}{H_5}N{H_2} + {H_2}O \rightleftharpoons {C_6}{H_5}N{H_3}^ + + O{H^ - }$
The equilibrium constant for this reaction is known as base-dissociation constant for the amine, denoted by ${K_b}$. pOH is determined to get the concentration of hydroxide ions as it is the negative algorithm of hydroxide ion concentration. It can also be written in the form of dissociation constant as-
$pOH = \dfrac{1}{2}[p{K_b} - \log b]$
We are given the value of ${K_b}$, so using the below formula and putting its value in it, we get the value of $p{K_b}$.
$p{K_b} = - \log {K_b}$
$= - \log (4.0 \times {10^{ - 10}})$
$= 9.34$
Now we can put this value in the below formula and get the value of $pOH$
$pOH = \dfrac{1}{2}[p{K_b} - \log b]$
$pOH = \dfrac{1}{2}[9.34 - \log (0.01)]$
$\therefore pOH = 4.67 - ( - 2)$
$\therefore pOH = 6.67$
$\therefore - \log [O{H^ - }] = 6.67$
$\therefore [O{H^ - }] = 4.7 \times {10^6}$
From this we conclude that it is a weak base as its pH comes out to be 7.33 i.e. on subtracting pOH from 14.

Note: There is another method to solve this numerical. This method involves ionisation constant and change in concentration from initial to final for all the species using formulas like ${K_\alpha } = \dfrac{{{K_w}}}{{{K_b}}}$ and ${K_\alpha } = c{\alpha ^2}$, where $\alpha < < 1$.