Question

A solution of $0.1M$ weak base $B$ is titrated with $0.1M$ of a strong acid $\left(HA\right)$. The variation of $pH$ of the solution with the volume of $HA$ added is shown in the figure below.

What is the $p{k}_b$ of the base? The neutralization reaction is given by

$B+HA\to B{H}^{+}+A^{-}$

Answer 1

$B+HA\to B{H}^{\oplus }+A^{-}$
From the figure, it is clear that there is a sharp change in the curve when $6ml$ of $HA$ has been added. So, this is the equivalence point. As the concentration of $B\text{and}HA$ are same, volume of $B$ = volume of $HA$ = $6ml$
So, total volume =$12ml$
At equivalence point, mili mol of $B$ = mili mol of $HA$
=$\text{molarity}\times \text{volume(ml)}=0.1\times 6=0.6mmol$
Hence, concentration of $A^{-}$$=\frac{0.6}{12}=\frac{1}{20}M$
As $pH=6,[H{]}^{+}={10}^{-6}M$
But, $[H{]}^{+}=\sqrt{\frac{k_w}{k_b}\times c}$
$or,{10}^{-6}=\sqrt{\frac{{10}^{-14}\times 0.6}{k_b\times 12}}$
$k_b=5\times {10}^{-4}$
$p{k}_b=4-log5=4-0.7=3.3$