Question

Calculate the degree of ionisation of \[0.01\text{ M}\] acetic acid. The dissociation constant of acetic acid is $1.8\times {{10}^{-5}}$.

Answer 1

Hint: We can solve these type of questions by remembering the following formula that is for a weak acid, ${{K}_{a}}=c{{\alpha }^{2}}$
Where, c is the concentration of the acid and $\alpha $ is the dissociation constant.

Complete answer:
Given,
$\left[ C{{H}_{3}}COOH \right]=0.01M$
${{K}_{a}}~=1.8\times {{10}^{-5}}$
Now, consider the reaction
$C{{H}_{3}}COOH\rightleftharpoons ~C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$
\[~~\left( 0.01-0.01\alpha \right)\left( 0.01\alpha \right)\left( 0.01\alpha \right)\]
These are the concentrations at equilibrium, so we can write the following
${{K}_{a}}~=~\dfrac{\left( 0.01\alpha \right)\left( 0.01\alpha \right)}{\left( 0.01-0.01\alpha \right)}=1.8\times {{10}^{-5}}$
As is a weak acid we can take the following approximations,
$0.01-~x~\approx ~0.01$and $x~+~0.1~\approx ~0.1$
Now substituting these values in above equation we get,
${{K}_{a}}~=~\dfrac{\left( 0.01\alpha \right)\left( 0.01\alpha \right)}{\left( 0.01 \right)}=1.8\times {{10}^{-5}}$
$\alpha =\sqrt{18\times {{10}^{-6}}}$
$\alpha =4.2\times {{10}^{-2}}$
Hence, the degree of ionization is \[0.042\].

Note: The degree of ionization (also known as ionization yield) is defined as the proportion of neutral particles, such as those in a gas or aqueous solution, that are ionized to charged particles. The strong electrolytes generally have degree of ionisation as 1 and for weak electrolytes it is less than 1.