Question

Aniline is an amine that is used to manufacture dyes. It is isolated as aniline hydrochloride \[\left[ {{C_{6}}{H_{5}}N{H_3}} \right]Cl\], a salt of aniline and \[HCl\] . Calculate the \[pH{\text{ }}of\;0.233M\;\] solution of aniline. \[{K_b}\left( {aniline} \right) = 4.6 \times {10^{ - 10}}\].

Answer 1

Hint: For conjugate-acid base pairs, the acid dissociation constant \[{K_a}\], and base ionization constant \[{K_b}\] are related by the following equations:
\[\Rightarrow {K_a} \cdot {K_b} = {K_w}\]
where \[{K_w}\]​ is the autoionization constant
\[\Rightarrow {\text{p}}{K_{\text{a}}} + {\text{p}}{K_{\text{b}}} = 14\;\;{\text{at }}25{{\mkern 1mu} ^\circ }{\text{C}}\]

Complete step by step answer:
Weak acids, generically abbreviated as HA donate H+(or proton) to water to form the conjugate base \[\Rightarrow {A^-}and{\text{ }}{H_3}O + \]
\[\Rightarrow {\text{HA}}(aq) + {{\text{H}}_2}{\text{O}}(l) \rightleftharpoons {{\text{H}}_3}{{\text{O}}^ + }(aq) + {{\text{A}}^ - }(aq)\]
Similarly, a base B accept a proton in water to form the conjugate acid, \[H{B^ + }{\text{ }} and {\text{ }}O{H^ - }\]
\[\Rightarrow {\text{B}}(aq) + {{\text{H}}_2}{\text{O}}(l) \rightleftharpoons {\text{H}}{{\text{B}}^ + }(aq) + {\text{O}}{{\text{H}}^ - }(aq)\]
For a weak acid or base, the equilibrium constant for the ionization reaction quantifies the relative amounts of each species. In this article, we will discuss the relationship between the equilibrium constants Ka and Kb, a conjugate acid-base pair.

Let's look more closely at the dissociation reaction for a monoprotic weak acid [HA]
\[\Rightarrow {\text{HA}}(aq) + {{\text{H}}_2}{\text{O}}(l) \rightleftharpoons {{\text{H}}_3}{{\text{O}}^ + }(aq) + {{\text{A}}^ - }(aq)\]
The products of this reversible reaction are \[{A^ - }\], the conjugate base of \[\;HA{\text{ }} and {\text{ }}{H_3}{O^ + }\]. We can write the following expression for the equilibrium constant Ka,
\[{K_{\text{a}}} = \dfrac{{[{{\text{H}}_3}{{\text{O}}^ + }][{{\text{A}}^ - }]}}{{[{\text{HA}}]}}\]

Since $A^-$ is a base, we can also write the reversible reaction for $A^−$ acting as a base by accepting a proton from water:
\[\Rightarrow {{\text{A}}^ - }(aq) + {{\text{H}}_2}{\text{O}}(l) \rightleftharpoons {\text{HA}}(aq) + {\text{O}}{{\text{H}}^ - }(aq)\]
The products of this reaction are \[HA{\text{ }}and{\text{ }}O{H^ - }\]. we can write out the equilibrium constant Kb for the reaction where $A^-$ acts as a base.
\[\Rightarrow {K_{\text{b}}} = \dfrac{{[{\text{HA}}][{\text{O}}{{\text{H}}^ - }]}}{{[{{\text{A}}^ - }]}}\]

If we multiply Ka for HA with the Kb of its conjugate base $A^-$ , that gives:
$\Rightarrow {K_a}{K_b}\;=\;({\dfrac{[{{H_3}{O^+}}][A^-]}{[HA]}})(\dfrac{[HA][OH^-]}{{[A^-]}})\;=\;{K_w}$
where Kw ​is the water dissociation constant. This relationship is very useful for relating Ka and Kb conjugate acid-base pairs!! We can also use the value of Kw​ at $25^\circ$ To derive other handy equations:
\[\Rightarrow {K_{\text{a}}} \cdot {K_{\text{b}}} = {K_{\text{w}}} = 1.0 \times {10^{ - 14}}\;{\text{at }}25{{\mkern 1mu} ^\circ }{\text{C}}\quad \]

If we take the negative {\log _{10}}\] of both sides of the equation, we get:
\[\Rightarrow p{K_{a}} + p{K_b} = 14\;\;at\;{25^ \circ }C\]
Aniline hydrochloride is conjugate acid of aniline. To calculate pKa​ of \[\left[ {{C_{6}}{H_{5}}N{H_3}} \right]Cl\] and treat it like weak acid.

So using above formula we can write;
\[\Rightarrow p{K_a} + p{K_{b}} = p{K_w}\]
\[ \Rightarrow p{K_a} = 14 - ( - log4.6 \times {10^{ - 10}}) = 4.66\]
Now pH, pKa and concentration of a solution is related by below formula:
\[\Rightarrow pH = \dfrac{1}{2}(p{K_a} - logC)\]\[\; = \dfrac{{4.66}}{2} + \dfrac{{0.63}}{2} = 2.64\]

Note: If the pH is lower than the pKa, then the compound will be protonated. If the pH is higher than the pKa, then the compound will be deprotonated. Acids are neutral when protonated and negatively charged (ionized) when deprotonated. Bases are neutral when deprotonated and positively charged (ionized) when protonated.