Question

Aniline behaves as a weak base. When $0.1M,50mL$ solution of aniline was mixed with $0.1M,25mL$ solution of $HCl$ the $pH$ of resulting solution was 8. The $pH$ of $0.01M$ solution of anilinium chloride will be:

• A. $5$
• B. $4.5$
• C. $6$
• D. $6.5$

Answer 1

The correct option is A $5$

Given:
Milimoles of Aniline$=0.1\times 50=5$
Milimoles of $HCl=0.1\times 25=2.5$
Reaction taking place will be
$C_6{H}_{5}N{H}_2+HCl\rightleftharpoons C_6{H}_{5}N{H}_3^{+}C{l}^{-}$
$5+2.5\rightleftharpoons 0$
$2.5+0\rightleftharpoons 2.5$

Given $pH=8$
$\Rightarrow pOH=14-8=6=p{K}_b$


$pH$ of $0.01M$ solution of anilinium chloride:

Using the formula,
$pH=7-\frac{1}{2}P{K}_b-\frac{1}{2}logC$

putting the values,

$pH=7-\frac{1}{2}\left(6\right)-\frac{1}{2}log\left(0.01\right)$

$pH=5$