Question

Prove that degree of dissociation of a weak monoprotic acid is given by $\alpha = \dfrac{1}{{1 + {{10}^{\left( {p{K_a} - pH} \right)}}}}$
Where ${K_a}$ is its dissociation constant.

Answer 1

It is related to the formation of cation and anion when an electrolyte has been dissociated into its respective ions. When an acid is dissolved in water, the covalent bond between the electronegative atom and the hydrogen atom is broken by heterolytic fission. As a result, it gives a proton and a negative ion. Dissociation takes place and here “degree” means “extent”.

Complete answer: The degree of dissociation is represented by the Greek symbol $\alpha$and is defined as the fraction of solute molecules that dissociate.
The extent to which an electrolyte is dissociated can be determined by this degree of dissociation and this degree of dissociation depends on whether the electrolyte is a strong or weak electrolyte.
Now let us prove that an equation for degree of dissociation $\alpha$ of a weak monoprotic acid is $\alpha = \dfrac{1}{{1 + {{10}^{\left( {p{K_a} - pH} \right)}}}}$.
Consider $HA \to {H^ + } + {A^ - }$
Now ${K_a}$ for this dissociation is given by
${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}$
Let us consider the concentration of the dissociated ions as $\alpha$and the concentration of the reactant become $\left( {1 - \alpha } \right)$.
So, ${K_a} = \dfrac{{[{H^ + }][c\alpha ]}}{{c(1 - \alpha )}}$
Where c is the concentration of weak acid.
The c gets cancelled and we have,
${K_a} = \dfrac{{[{H^ + }][\alpha ]}}{{(1 - \alpha )}}$.
On rearrangement,
$\dfrac{{(1 - \alpha )}}{\alpha } = \dfrac{{\left[ {{H^ + }} \right]}}{{{K_a}}}$
Applying log an both sides,
$\log \left( {\dfrac{{1 - \alpha }}{\alpha }} \right) = \log \left[ {{H^ + }} \right] - \log {K_a}$
We know that

$$pH = - \log \left[ {{H^ + }} \right] \\\ p{K_a} = - \log {K_a} \\\$$

Substituting this we get,
$\log \left( {\dfrac{{1 - \alpha }}{\alpha }} \right) = - pH + p{K_a}$
on simplification we have,
$\Rightarrow \left( {\dfrac{{1 - \alpha }}{\alpha }} \right) = {10^{\left( {p{K_a} - pH} \right)}}$
$\Rightarrow \dfrac{1}{\alpha } - 1 = {10^{\left( {p{K_a} - pH} \right)}}$
$\Rightarrow \dfrac{1}{\alpha } = 1 + {10^{\left( {p{K_a} - pH} \right)}}$
On reciprocation, we have
$\Rightarrow \alpha = \dfrac{1}{{1 + {{10}^{\left( {p{K_a} - pH} \right)}}}}$
Hence proved.

Note:
There is a difference between the degree of dissociation and the dissociation constant. The degree of dissociation is the dissociation of molecules into smaller ions, whereas dissociation constant is the ratio of the dissociated ions to the undissociated or the original molecules.