Question

The dissociation constant of acetic acid at a given temperature is $1.69\times {{10}^{-5}}$ at any given temperature. The degree of dissociation of $0.01M$ acetic acid in the presence of $0.01M\text{ }HCl$ is equal to:
A. 0.41
B. 0.13
C. $1.69\times {{10}^{-3}}$
D. 0.013

Answer 1

Hint: Think about the degree of dissociation and the formula for the dissociation constants when a weak acid comes into picture. Add the protons formed by the hydrochloric acid to the protons formed by the acetic acid and then carry out further calculations.

Complete step by step answer:
We know that acetic acid is a weak acid and hydrochloric acid is a strong acid. Acetic acid will dissociate according to the degree of dissociation and hydrochloric acid will dissociate completely. We know that the formula to find the dissociation considers the concentration of the ion species formed divided by the concentration of the undissociated molecules. So, for acetic acid, the formula of the dissociation constant will be:
\[{{K}_{a}}=\dfrac{[C{{H}_{3}}CO{{O}^{=}}][{{H}^{+}}]}{[C{{H}_{3}}COOH]}\]
Here, ${{K}_{a}}$ is the dissociation constant whose value is $1.69\times {{10}^{-5}}$ and let us assume that $\alpha $ is the degree of dissociation. Let, $C$ be the concentration of acetic acid.
The value of the $C{{H}_{3}}CO{{O}^{-}}$ ions will be the concentration of the acetic acid multiplied by the degree of dissociation i.e. $C\alpha $. We know that the concentration of the undissociated acetic acid will be $C(1-\alpha )$. Now, to find the concentration of the ${{H}^{+}}$ we need to add the concentration of the ions obtained from acetic acid as well as from hydrochloric acid
Hydrochloric acid dissociates completely, so the concentration of ${{H}^{+}}$ ions contributed by it will be 0.01M. The ${{H}^{+}}$ contributed by the acetic acid will be $C\alpha $.
Now, putting these values in the equation, we will get:
\[{{K}_{a}}=\dfrac{(C\alpha )(C\alpha +0.01)}{C(1-\alpha )}\]
Cancelling $C$ from the first bracket and denominator and taking 0.01 common from the second bracket, we get:
\[{{K}_{a}}=\dfrac{0.01\alpha (1+\alpha )}{(1-\alpha )}\]
As compared to 1, we know that the value of the degree of dissociation is very low, so we can ignore it in case of $(1+\alpha ),(1-\alpha )$ and take their values as 1. Putting the value of the dissociation constant and solving for $\alpha $ we get:
\[\begin{align}
  & 1.69\times {{10}^{-5}}=0.01\alpha \\
 & \alpha =1.69\times {{10}^{-3}} \\
\end{align}\]

Hence, the answer to this question is ‘C. $1.69\times {{10}^{-3}}$’.

Note: Please note that the dissociation constant will always be less than the degree of dissociation in the presence of the same ions as the weak acid. It shifts the equilibrium by creating an excess of ions and making it more difficult for the molecules to dissociate.