Question

A weak acid $HA$ after treatment with $12mL$ of $0.1M$ strong base has a pH of $5$. At the end point, the volume of the same base required is $26.6mL$. Calculate the $K_a$ of the acid.

Answer 1

At the end point milliequivalents of acid=milliequivalents of base

$N_1{V}_1=N_2{V}_2$

So milliequivalents of an acid=$26.6\times 0.1=2.66meq.$

$HA+BOH\to BA+H_{2}O$

$2.66$ $1.2$

$1.46$ $0$ $1.2$ $1.2$

As they form buffer so

$pH=p{K}_a+\log \frac{\left[Salt\right]}{\left[Acid\right]}$

$5=p{K}_a+\log \frac{1.2}{1.46}$

$p{K}_a=5.86$

$p{K}_a=-log\left(K_a\right)=5.86$

$K_a=8.2\times {10}^{-6}$