Question

What is the pH of a 0.025 M aqueous solution of sodium propionate?
$Na{{C}_{3}}{{H}_{5}}{{O}_{2}}\left( aq \right)+{{H}_{2}}O\left( l \right)\text{ }\rightleftharpoons \text{ }H{{C}_{3}}{{H}_{5}}{{O}_{2}}\left( aq \right)\text{ }+\text{ }N{{a}^{+}}\left( aq \right)\text{ }+\text{ }O{{H}^{-}}(aq)$ . The $N{{a}^{+}}$ is a spectator ion, so it can be eliminated from the equation to give: ${{C}_{3}}{{H}_{5}}O_{2}^{-}\left( aq \right)+{{H}_{2}}O\left( l \right)\text{ }\rightleftharpoons \text{ }H{{C}_{3}}{{H}_{5}}O_{2}^{-}\left( aq \right)\text{ }+\text{ }O{{H}^{-}}$
The ${{K}_{a}}$ for propionic acid is $1.3{{e}^{-5}}$

Answer 1

There is a relation between acid dissociation constant and base dissociation constant and it is as follows.
${{K}_{b}}=\dfrac{{{K}_{w}}}{{{K}_{a}}}$
Here ${{K}_{b}}$ = base dissociation constant
${{K}_{a}}$ = acid dissociation constant
${{K}_{w}}$ = water ionization constant

Complete answer:
- In the question it is asked to calculate the pH of the 0.025 M aqueous solution of sodium propionate by using the given data in the question.
- First, we have to calculate the base dissociation constant by using the given data with the following formula.
${{K}_{b}}=\dfrac{{{K}_{w}}}{{{K}_{a}}}$
Here ${{K}_{b}}$ = base dissociation constant
${{K}_{a}}$ = acid dissociation constant = $1.3\times {{10}^{-5}}$
${{K}_{w}}$ = water ionization constant = $1\times {{10}^{-14}}$
- Substitute the above known values in the above formula to get the base dissociation constant.

& {{K}_{b}}=\dfrac{{{K}_{w}}}{{{K}_{a}}} \\\ & {{K}_{b}}=\dfrac{1\times {{10}^{-14}}}{1.3\times {{10}^{-5}}} \\\ & {{K}_{b}}=7.7\times {{10}^{-10}} \\\ \end{aligned}$$ \- By using the base dissociation constant, we can solve the change in concentration using algebra and it is as follows. $${{\text{K}}_{\text{b}}}\text{=}\dfrac{\text{Equilibrium concentration of the products}}{\text{Equilibrium concentration of the reactants}}$$ \- From the given chemical reaction, we can easily write the equilibrium concentrations and it is as follows. $$\begin{aligned} & \underset{\operatorname{int}ial\text{ }concentration}{\mathop{{}}}\,\underset{x}{\mathop{\text{ }{{C}_{3}}{{H}_{5}}O_{2}^{-}\left( aq \right)}}\,+{{H}_{2}}O\left( l \right)\text{ }\rightleftharpoons \text{ }\underset{0}{\mathop{H{{C}_{3}}{{H}_{5}}O_{2}^{-}\left( aq \right)\text{ }}}\,+\text{ }\underset{0}{\mathop{O{{H}^{-}}}}\, \\\ & \underset{\text{At equilibrium}}{\mathop{{}}}\,\underset{0.025-x}{\mathop{\text{ }{{C}_{3}}{{H}_{5}}O_{2}^{-}\left( aq \right)}}\,+{{H}_{2}}O\left( l \right)\text{ }\rightleftharpoons \text{ }\underset{x}{\mathop{H{{C}_{3}}{{H}_{5}}O_{2}^{-}\left( aq \right)\text{ }}}\,+\text{ }\underset{x}{\mathop{O{{H}^{-}}}}\, \\\ \end{aligned}$$ At equilibrium we can write the base constant of the above chemical reaction is as follows. $$\begin{aligned} & {{\text{K}}_{\text{b}}}\text{=}\dfrac{\text{Equilibrium concentration of the products}}{\text{Equilibrium concentration of the reactants}} \\\ & 7.7\times {{10}^{-10}}=\dfrac{{{x}^{2}}}{0.025-x} \\\ & {{x}^{2}}+(7.7\times {{10}^{-10}})x-1.925\times {{10}^{-11}} \\\ \end{aligned}$$ \- We can compare the above result with a quadratic equation and we will get the pH of the solution and it is as follows. $$\begin{aligned} & {{x}^{2}}+(7.7\times {{10}^{-10}})x-1.925\times {{10}^{-11}} \\\ & a{{x}^{2}}+bx+c \\\ \end{aligned}$$ Here a = 1, b = $7.7\times {{10}^{-10}}$ , c = $-1.925\times {{10}^{-11}}$ \- Substitute the above values in the below formula to get the pH of the solution. $$\begin{aligned} & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\\ & ~9.1\text{ }=\text{ }14\text{ }+\text{ }log\text{ }(1.4\times {{10}^{-5}}) \\\ & pH=9.1 \\\ \end{aligned}$$ \- Therefore, the pH of the aqueous solution of 0.025 M sodium propionate is 9.1. **Note:** Initially we have to find the base dissociation constant of the given chemical to solve the remaining problem. Without solving the base dissociation constant of the respective chemical, we cannot find the pH of the solution.