Question

Calculate the percentage ionisation of $0.01M$ acetic acid in $0.1M$$HCl$.${{K}_{a}}$of acetic acid is $1.8x{{10}^{-5}}$ .
A. $0.18%$
B. $0.018%$
C. $1.8%$
D. $18%$

Answer 1

Hint: An ionisation constant denoted by the symbol $\left( K \right)$ depends upon the equilibrium between ions and molecules that do not undergo complete ionisation in solution. It can be used to calculate the degree of dissociation or ionisation. For acids the ionisation constant is defined as ${{K}_{a}}$ . More the value of ionisation constant more is the dissociation of acid.

Formula Used:
$\alpha =\dfrac{D.M}{I.M}$
Where, $\alpha $ is the degree of dissociation
$D.M$ is the dissociated moles
$I.M$ is the initial moles
${{K}_{a}}=\dfrac{\left[ P \right]}{\left[ R \right]}$
${{K}_{a}}$ is the dissociation constant of acid
$P$ is concentration of the product
$R$ is the concentration of the reactant

Complete step by step answer:
Here, it is given that the concentration of acetic acid is $0.01M$ and the concentration of $HCl$ is $0.1M$
Dissociation constant of an acetic acid is $1.8\times {{10}^{-5}}$
$HCl$ is a strong acid so it will completely dissociates to give the product that is equal to the reactant
$HCl\to {{H}^{+}}+C{{l}^{-}}$
The concentration of $\left[ {{H}^{+}} \right]={{10}^{-1}}$
Now, the acetic acid is a weak acid that will not completely dissociate to give ions.
$C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$

	\[C{{H}_{3}}COOH\]	\[C{{H}_{3}}CO{{O}^{-}}\]	${{H}^{+}}$
Initial concentration	$0.01M$	$\_$	$\_$
Final concentration	$0.01-x$	$x$	$x+{{10}^{-1}}$.

${{K}_{a}}=\dfrac{\left[ P \right]}{\left[ R \right]}$
Where, ${{K}_{a}}$ is the dissociation constant of an acid
$P$ is concentration of product
$R$ is the concentration of reactant
Now, substituting the values in the above formula we get,
${{K}_{a}}=\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}^{+}} \right]}{\left[ C{{H}_{3}}COOH \right]}$
${{K}_{a}}=\dfrac{x\left( x+{{10}^{-1}} \right)}{0.01-x}$
Now substituting the value of ${{K}_{a}}$
$1.8\times {{10}^{-5}}=\dfrac{x\left( x+{{10}^{-1}} \right)}{0.01-x}$
Acetic acid is a weak acid, therefore $0.01-x\approx 0.01$ and $x+0.1\approx 0.1$
Now, on substituting the value we get,
$1.8\times {{10}^{-5}}=\dfrac{x\times 0.1}{0.01}$
$x=1.8\times {{10}^{-6}}$
The degree of dissociation
$\alpha =\dfrac{D.M}{I.M}$
$\alpha $ is the degree of dissociation
$D.M$ is the dissociated moles
$I.M$ is the initial moles
Now, substituting the value we get,
$\alpha =\dfrac{x}{0.01}$
On further solving,
$\alpha =\dfrac{1.8\times {{10}^{-6}}}{0.01}$
$\alpha =1.8\times {{10}^{-4}}$
The percentage ionization is $1.8\times {{10}^{-4}}\times 100=0.018%$

So, the correct answer is “Option B”.

Note: The formula is used for calculation of the degree of dissociation is applicable for weak electrolytes only. The law given is also known as Ostwald’s dilution law. In case of strong electrolytes, we can’t apply this law.