Question

# A 0.01 M solution of acetic acid is 5% ionised at ${{25}^{o}}C$. Calculate its dissociation constant.

Hint: Acetic acid is a weak electrolyte. Weak electrolytes follow the Ostwald’s dilution law -
${{K}_{a}}=\dfrac{C{{\alpha }^{2}}}{1-\alpha }$
5 % ionisation of acetic acid means that out of 100 acetic acid molecules, only 5 of them get dissociated. These data can be used to calculate the dissociation constant of acid.

- The dissociation constant of a compound can be described as the equilibrium constant that measures the natural tendency of a large compound to separate reversibly into smaller components.
- Acetic acid is a weak monoprotic acid (it has one H atom). Hence it gets dissociated as -
$C{{H}_{3}}COOH\rightleftarrows {{H}^{+}}+C{{H}_{3}}CO{{O}^{-}}$

- Weak electrolytes follow the Ostwald’s dilution law. The dissociation constant of acetic acid (which is a weak electrolyte) can be calculated using the equation of Ostwald’s dilution law. This law describes the relation of dissociation constant of a weak electrolyte with the concentration of the electrolyte and its degree of dissociation.

${{K}_{a}}=\dfrac{C{{\alpha }^{2}}}{1-\alpha }$-> Equation (1)
Here, ${{K}_{a}}$ is the dissociation constant that we have to find. C is the concentration of the given compound. And $\alpha$ is the degree of dissociation.
- We have,
$C=0.01M,$
$\alpha =5%=\dfrac{5}{100}=0.05$

- We need to find the value of the dissociation constant of acetic acid. For that, we have to substitute the values of C and $\alpha$ in Equation (1) to obtain the value.
${{K}_{a}}=\dfrac{(0.01\times {{0.05}^{2}})}{(1-0.05)}=2.63\times {{10}^{-5}}M$
Therefore, the dissociation constant of acetic acid is $2.63\times {{10}^{-5}}$ M.

Note: - It is important to take care that while substituting the value of $\alpha$, you should not substitute the value given in percentage as such. You have to divide it by 100 to get the actual value. Take care to write the unit of the final answer. The unit will be the same as that of molarity (mol/L).