Question

An aqueous solution of 0.24M aniline (${{K}_{b}}=4.166\times {{10}^{-10}}$) is mixed with NaOH solution to maintain anilinium ion concentration to $1\times {{10}^{-8}}M$.
The pOH of NaOH solution used was:

Answer 1

We know that when the rate of forward reaction in a system is equal to the rate of backward reaction, the system is said to be in equilibrium. Also, the ionization of a base in an aqueous solution is given by the base equilibrium constant, ${{K}_{b}}$.

Complete step-by-step answer: Now, we know that in an aqueous reaction, the dissociation of NaOH takes place as follows
$NaOH(aq)\rightleftarrows N{{a}^{+}}(aq)+O{{H}^{-}}(aq)$
The hydrolysis of aniline in presence of a water molecule takes place as follows
${{C}_{6}}{{H}_{5}}N{{H}_{2}}+HOH\rightleftharpoons {{C}_{6}}{{H}_{5}}NH_{3}^{+}+O{{H}^{-}}$
So the equilibrium constant for the ionization of the base ${{K}_{b}}$ of the reaction can be given as
${{K}_{b}}=\dfrac{[{{C}_{6}}{{H}_{5}}NH_{3}^{+}][O{{H}^{-}}]}{{{C}_{6}}{{H}_{5}}N{{H}_{2}}}$
Now, we can see that the hydrolysis of aniline produces aniline ion and hydroxide. To maintain the concentration of the aniline ions produced, we need to introduce a sufficient amount to hydroxide ion to maintain the equilibrium (Le Chatlier's Principle)
These hydroxide ions can be provided by NaOH as it also dissociates in an aqueous solution to give sodium ions and hydroxide ions.
Now it is given to us that the concentration of the reactant aniline is 0.24M and the concentration of the product aniline ion is $1\times {{10}^{-8}}M$.
Also, it is given to us that ${{K}_{b}}=4.166\times {{10}^{-10}}$.
So, by substituting these values in the base ionization constant equation, we get
$

{{K}_{b}}=4.166\times {{10}^{-10}}=\dfrac{1\times {{10}^{-8}}\times [O{{H}^{-}}]}{0.24} \\
[O{{H}^{-}}]=\dfrac{0.24\times 4.166\times {{10}^{-10}}}{{{10}^{-8}}} \\
[O{{H}^{-}}]=0.99984\times {{10}^{-2}}M \\
$Hence, 0.0099984M solution of NaOH is required to be mixed with 0.24M of aniline aqueous solution to maintain the number of aniline ions to$1\times {{10}^{-8}}M$. Now, we know that$p[OH]=-\log [O{{H}^{-}}]$.
So,

$$p[OH]=-\log (0.99984\times {{10}^{-2}}) \\\ p[OH]=-\log {{10}^{-2}}-\log 0.99984 \\\ p[OH]=2+6.95\times {{10}^{-5}} \\\ p[OH]\cong 2 \\\$$

So, the p[OH] of the NaOH solution used is approximately 2.

Note: It should be noted that the relation between the acid ionization and the base ionization reaction is given by
${{K}_{w}}={{K}_{a}}{{K}_{b}}$
Where ${{K}_{w}}$ is the dissociation constant of water and has a value $1\times {{10}^{-14}}$.