Question

When a $40mL$ of a $0.1M$ weak base is titrated with $0.16M$ $HCl$, the $pH$ of the solution at the end point is $5.23$. What will be the $pH$ if $15mL$ of $0.12M$ $NaOH$ is added to the resulting solution?

Answer 1

$\underset{4-}{BOH}+\underset{4-}{HCl}⟶\underset{4}{BCl}+H_{2}O$
At end point $m$ moles of $BOH=m$ moles of $HCl$
$0.16\times V=4$ $V=25mL$
Total volume$=40+25=65mL$
$\left[BCl\right]=\frac{4}{65}$
since $BCl$ is SAWB
$pH=7-\frac{1}{2}p{k}_b-\frac{1}{2}\log C$
$5.23=7-\frac{1}{2}p{k}_b-\frac{1}{2}\log 465$
$p{k}_b=4.75$
Now on further adding $NaOH$
$\underset{42.2}{BCl}+\underset{1.8-}{NaOH}⟶\underset{1.8}{BOH}+NaCl$
$pOH=p{k}_b+\log \frac{2.2}{1.8}=4.837$
$\Rightarrow$ $pH=9.1628$