Question

Derive Ostwald’s dilution law for weak acid. Derive $pH + pOH = 14$ at ${25^ \circ }C$.

Answer 1

Hint: This question gives the knowledge about the Ostwald’s dilution law. Ostwald’s dilution law is the application of law of mass action. The degree of dissociation of any electrolyte increases as the concentration decreases.

Complete step by step answer:
According to the Ostwald’s dilution law, the degree of dissociation of any electrolyte increases as the concentration decreases.
Consider a reaction of a weak acid as follows:
$C{H_3}COOH + {H_2}O \rightleftharpoons C{H_3}CO{O^ - } + {H_3}{O^ + }$
In the above reaction, we observe that the initial concentration of acetic acid is $c$ and after time $t$ the initial concentration becomes $c - c\alpha $ , the concentration of acetate ion is $c\alpha $ and the hydronium ion concentration is $c\alpha $ .
The dissociation constant of the above reaction is as follows:
$ \Rightarrow K = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]\left[ {{H_2}O} \right]}}$
On rearranging we get,
$ \Rightarrow K \times \left[ {{H_2}O} \right] = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}$
The dissociation constant of the acid is as follows:
$ \Rightarrow {K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}$
Substitute the values in above equation as follows:
$ \Rightarrow {K_a} = \dfrac{{\left( {c\alpha } \right)\left( {c\alpha } \right)}}{{c - c\alpha }}$
On simplifying, we get
$ \Rightarrow {K_a} = \dfrac{{c{\alpha ^2}}}{{1 - \alpha }}$
If the degree of dissociation is less than or equal to $5\% $ then consider $1 - \alpha \approx 1$, the above equation will be expressed as:
$ \Rightarrow {K_a} = c{\alpha ^2}$
On rearranging, we get
$\alpha = \sqrt {\dfrac{{{K_a}}}{c}} $
Consider a reaction to derive the relation $pH + pOH = 14$ as follows:
${H_2}O \rightleftharpoons H{O^ - } + {H^ + }$
In the above reaction, we observe that the initial concentration of water is $c$ and after time $t$ the concentration of hydroxide ion is $c\alpha $ and the hydronium ion concentration is $c\alpha $ .
The dissociation constant of the above reaction is as follows:
$ \Rightarrow {K_d} = \dfrac{{\left[ {H{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {{H_2}O} \right]}}$
On rearranging we get,
$ \Rightarrow {K_d} \times \left[ {{H_2}O} \right] = \left[ {H{O^ - }} \right]\left[ {{H^ + }} \right]$
The dissociation constant of the water is as follows:
$ \Rightarrow {K_w} = \left[ {H{O^ - }} \right]\left[ {{H^ + }} \right]$
The dissociation constant of water is also known as ionic product of water or ionic constant of water or autoprotolysis of water or equilibrium constant of water.
Multiply both sides with –log on both sides as follows:
$ \Rightarrow - \log {K_w} = - \log \left[ {H{O^ - }} \right] - \log \left[ {{H^ + }} \right]$
Using $pH$ formulas we conclude that the above equation is equal to,
$ \Rightarrow p{K_w} = pOH + pH$
As we know, the $p{K_w}$ of water at ${25^ \circ }C$ is $14$. Substitute the value of $p{K_w}$ as $14$ in the above equation as follows:
$ \Rightarrow 14 = pOH + pH$

Note: Always remember that the degree of dissociation of any electrolyte increases as the concentration decreases. Ostwald’s dilution law is the application of law of mass action. It is valid for weak acids only.