Question

When a solution of acetic acid was titrated with $NaOH$ the $pH$ of the solution when half the acid neutralized was $4.74$. Dissociation constant of the acid is:

• A. $1.8\times {10}^{-5}$
• B. $3.2\times {10}^{-5}$
• C. $8.7\times {10}^{-8}$
• D. $6.42\times {10}^{-4}$

Answer 1

The correct option is A $1.8\times {10}^{-5}$

$pH=p{K}_a+log\left[\frac{[-A]}{\left[HA\right]}\right]$

When half of the acid is neutralized.

Ratio of $\left[A^{-}\right]/\left[HA\right]=1$

This is half equivalence point then $pH=p{K}_a$

Thus, $p{K}_a$ of weak acid is $4.74$

Dissociation Constant$⟶{10}^{-4.74}=1.8\times {10}^{-5}$