Question

A weak acid $HA$ after treating with $12\text{mL}$ of $0.1M$ strong base $BOH$ has a $pH$ of 5. At the end point the volume of same base required is $26.6\text{mL}$. Calculate the $K_a$ of an acid.

• A. $1.8\times {10}^{-5}$
• B. $8.22\times {10}^{-6}$
• C. $1.8\times {10}^{-6}$
• D. $8.2\times {10}^{-5}$

Answer 1

The correct option is B $8.22\times {10}^{-6}$

For neutralisation:
Total $M_{\text{eq.}}$ of acid $=M_{\text{eq.}}$ of base $=26.6\times 0.1=2.66$
Now for partial neutralisation of acid
$\begin{array}{cccccccc}& HA& +& BOH& \to & BA& +& H_{2}O\\ M_{\text{eq.}}\text{before reaction}& 2.66& & 1.2& & 0& & 0\\ M_{\text{eq.}}\text{after reaction}& 1.46& & 0& & 1.2& & 1.2\end{array}$
The resultant mixture acts as a buffer solution.
Hence, $\left[HA\right]$ and $\left[BA\right]$ may be placed in terms of $M_{\text{eq}}$ since volume of mixture is constant.
$pH=-\text{log}{K}_a+\text{log}\frac{\left[Salt\right]}{\left[Acid\right]}$
or $5=-\text{log}{K}_a+\text{log}\frac{1.2}{1.46}$
$K_a=8.22\times {10}^{-6}$