Question

Given, ${{\text{K}}_{\text{a}}}$ for butyric acid is $\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}$. Calculate \[\text{pH}\] and hydroxyl ion concentration of $\text{0}\text{.2M}$ aqueous solution of sodium butyrate.

Answer 1

Hint: Salt hydrolysis is the process in which water reacts with cation and anion or both of a salt to change the concentration of ${{\text{H}}^{\text{+}}}$ and $\text{O}{{\text{H}}^{\text{-}}}$ ions of water.
Sodium butyrate is a weak acid and strong base type of salt.

Complete answer:
The calculation of $\text{pH}$ and hydroxyl ion takes place in four steps which are following –
In first step we will calculate the ${{\text{K}}_{\text{H}}}(\text{hydrolysis}\,\text{constant)}$
\[\begin{align}
  & {{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{CO}{{\text{O}}^{\text{-}}}\text{N}{{\text{a}}^{\text{+}}}\,\text{+}\,{{\text{H}}_{\text{2}}}\text{O}\,\,\rightleftharpoons \,{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{COOH}\,\,\text{+}\,\,\text{NaOH} \\
 & {{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{CO}{{\text{O}}^{\text{-}}}\,+\,\,{{\text{H}}_{\text{2}}}\text{O}\,\,\rightleftharpoons \,{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{COOH}\,\text{+}\,\text{O}{{\text{H}}^{\text{-}}} \\
 & {{\text{K}}_{\text{h}}}=\,\frac{[\,{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{COOH}\,][\text{O}{{\text{H}}^{\text{-}}}]}{[{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{CO}{{\text{O}}^{\text{-}}}\,]}.......(i)\,\,\,\text{ }\!\!\{\!\!\text{ From}\,\text{law}\,\text{of}\,\text{mass}\,\text{action }\!\!\}\!\!\text{ } \\
 & \text{For}\,\text{weak}\,\text{acid} \\
 & {{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{COOH}\rightleftharpoons {{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{CO}{{\text{O}}^{\text{-}}}\,+{{\text{H}}^{\text{+}}} \\
 & {{\text{K}}_{\text{a}}}=\,\frac{[{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{CO}{{\text{O}}^{\text{-}}}\,][{{\text{H}}^{\text{+}}}]}{[{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{COOH}]}......(ii) \\
 & \text{For}\,\text{water} \\
 & {{\text{H}}_{\text{2}}}\text{O}\rightleftharpoons \,\,{{\text{H}}^{\text{+}}}\,+\text{O}{{\text{H}}^{\text{-}}} \\
 & {{\text{K}}_{\text{w}}}=\,[{{\text{H}}^{\text{+}}}][\text{O}{{\text{H}}^{\text{-}}}]...........(iii) \\
\end{align}\]
From e.q (i), (ii) and (iii) we get
${{\text{K}}_{\text{h}}}\text{=}\frac{{{\text{K}}_{\text{w}}}}{{{\text{K}}_{\text{a}}}}......(iv)$
In second step we will calculate degree of hydrolysis (h)
\[\begin{align}
  & {{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{CO}{{\text{O}}^{\text{-}}}\,+\,\,{{\text{H}}_{\text{2}}}\text{O}\,\,\rightleftharpoons \,{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{COOH}\,\text{+}\,\text{O}{{\text{H}}^{\text{-}}} \\
 & \,\,\,\,\,\,\,\,\,\begin{matrix}
   \text{C}\,\, \\
   \text{C(1-h)}\,\, \\
\end{matrix}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{matrix}
   0 & 0 \\
   \text{Ch}\, & \text{Ch}\, \\
\end{matrix}\{\text{at}\,\text{initial}\,\text{concentration }\!\!\}\!\!\text{ } \\
 & \,\,{{\text{K}}_{\text{h}}}=\,\frac{[{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{COOH}\,\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ O}{{\text{H}}^{-}}]}{[{{\text{C}}_{\text{3}}}{{\text{H}}_{\text{7}}}\text{CO}{{\text{O}}^{-}}]} \\
 & \text{ }{{\text{K}}_{\text{h}}}\text{= C}{{\text{h}}^{\text{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ }\!\!\{\!\!\text{ at}\,\text{infinite}\,\text{dilution }\!\!\}\!\!\text{ } \\
 & \,\,\,\,\,\,\text{h}\,\text{=}\,\sqrt{\frac{{{\text{K}}_{\text{h}}}}{\text{C}}}\,......\text{(v)}\,\,\,\,\, \\
 & \text{so,}\,\text{after}\,\text{putting}\,\text{the}\,\text{value}\,\text{of}\,{{\text{K}}_{\text{h}}}\,\text{from}\,\text{equation}\,\text{(iv)} \\
 & \,\,\,\,\,\,\,\text{h}\,\text{=}\,\sqrt{\frac{{{\text{K}}_{\text{h}}}}{\text{C}}}=\,\sqrt{\frac{{{\text{K}}_{\text{w}}}}{{{\text{K}}_{\text{a}}}\text{C}}}.....(\text{vi}) \\
\end{align}\]
In the third step after putting the given value of ${{\text{K}}_{\text{a}}}$= $\text{2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}$ and $\text{C}$=$\text{0}\text{.2M}$ in e.q (VI) we get the value of hydroxyl ion concentration
\[\begin{align}
  & \text{ }\!\![\!\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\!\!]\!\!\text{ }\,=\,Ch\,=\,C\times \sqrt{\frac{{{K}_{w}}}{{{K}_{a}}\times \,C}} \\
 & \text{or }\!\![\!\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\!\!]\!\!\text{ }\,=\,\sqrt{\frac{{{K}_{w}}\times \,C}{{{K}_{a}}}} \\
 & \,\text{ }\!\![\!\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\!\!]\!\!\text{ }\,=\,\sqrt{\frac{{{10}^{-14}}\times \,0.2}{2\times {{10}^{-5}}}} \\
 & =\,\sqrt{\frac{2\times {{10}^{-15}}}{2\times {{10}^{-5}}}} \\
 & \text{ }\!\![\!\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\!\!]\!\!\text{ }\,=\,{{10}^{-5}}
\end{align}\]
In fourth step we will calculate the value of $\text{pH}$ by calculating the value of $\text{pOH}$
\[\begin{align}
  & \text{pOH =}\,\text{-log }\!\![\!\!\text{ O}{{\text{H}}^{\text{-}}}\text{ }\!\!]\!\!\text{ } \\
 & \text{pOH}\,=-\log [{{10}^{-5}}] \\
 & \text{pOH}\,=\,\,5 \\
\end{align}\]
So, $\text{pH}$ of the solution will be
\[\begin{align}
  & \text{pH}\,\,\text{+}\,\,\text{pOH}\,\,\text{=}\,\,\text{p}{{\text{K}}_{\text{w}}} \\
 & \text{pH}\,\,\text{+}\,\,\text{pOH =}\,\text{14} \\
 & \text{pH}\,\,\text{=}\,\,\text{14-}\,\text{5} \\
 & \text{pH}\,\,\text{=}\,\,\text{9} \\
\end{align}\]

Note:
Hydrolysis is the reverse process of neutralization.
-In this type of salt hydrolysis anionic part of the salt reacts with water therefore this reaction is also known as anionic hydrolysis. Finally the solution is basic in nature because of the increase in concentration of hydroxyl ions. So the final $\text{pH}$ of the solution will be more than seven.
Maximum hydrolysis occurs when the salt consists of weak acid and weak base. In this case both cation and anion both are reactive, so the solution is almost neutral but can be acidic and basic depending on the nature of acid and base.