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Question: Calculate the degree of ionisation of \[0.01\text{ M}\] acetic acid. The dissociation constant of ac...

Calculate the degree of ionisation of 0.01 M0.01\text{ M} acetic acid. The dissociation constant of acetic acid is 1.8×10−51.8\times {{10}^{-5}}.

Explanation

Solution

Hint: We can solve these type of questions by remembering the following formula that is for a weak acid, Ka=cα2{{K}_{a}}=c{{\alpha }^{2}}
Where, c is the concentration of the acid and α\alpha is the dissociation constant.

Complete answer:
Given,
[CH3COOH]=0.01M\left[ C{{H}_{3}}COOH \right]=0.01M
Ka =1.8×10−5{{K}_{a}}~=1.8\times {{10}^{-5}}
Now, consider the reaction
CH3COOH⇌ CH3COO−+H+C{{H}_{3}}COOH\rightleftharpoons ~C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}
  (0.01−0.01α)(0.01α)(0.01α)~~\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
Ka = (0.01α)(0.01α)(0.01−0.01α)=1.8×10−5{{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 ≈ 0.010.01-~x~\approx ~0.01and x + 0.1 ≈ 0.1x~+~0.1~\approx ~0.1
Now substituting these values in above equation we get,
Ka = (0.01α)(0.01α)(0.01)=1.8×10−5{{K}_{a}}~=~\dfrac{\left( 0.01\alpha \right)\left( 0.01\alpha \right)}{\left( 0.01 \right)}=1.8\times {{10}^{-5}}
α=18×10−6\alpha =\sqrt{18\times {{10}^{-6}}}
α=4.2×10−2\alpha =4.2\times {{10}^{-2}}
Hence, the degree of ionization is 0.0420.042.

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.