Question

The ionisation constant of acetic acid is $1.8\times {10}^{-5}$.The concentration at which it will be dissociated to $2$% is:

• A. $1M$
• B. $0.045M$
• C. $0.018M$
• D. $0.45M$

Answer 1

The correct option is B $0.045M$

When acetic acid is dissolved in water, it partially dissociates (2%).

Ionisation constant $K_a=1.87\times {10}^{-5}$

$K_a=\frac{C\alpha \times C\alpha }{C(1-\alpha )}$

We assume $\alpha \sim 0\Rightarrow 1-\alpha \sim 1$

$\Rightarrow 1.87\times {10}^{-5}=\frac{C{\alpha }^2}{1}=C\times 0.02\times 0.02$

$\Rightarrow C=\frac{1.87\times {10}^{-5}}{{0.02}^2}$

$=0.0467M$

Correct answer is option-B.