Question

Aniline behaves as a weak base. When 0.1 M, 50 mL solution of aniline was mixed with 0.1 M, 25 mL solution of $HCl$, the pH of the resulting solution was 8. Then the $pH$ of 0.01 M solution of anilinium chloride will be ($K_w={10}^{-14}$).

Answer 1

Anilinium chloride is the salt of a weak base and a strong acid. So, the $pH$ of this kind of solution is:
$pH=\frac{1}{2}p{K}_w-\frac{1}{2}(p{K}_b+logC)$
where C is concentration of salt.
To find $p{K}_b$ we can use the following buffer given the question,
$C_6{H}_{5}N{H}_2\left(aq\right)+H^{(}aq)+\leftrightharpoons C_6{H}_{5}N{H}_3^{+}(aq)$
$t=052.50$
$t=02.502.5$
$p{K}_b=pOH=14-pH=6$

$pH=7-\frac{1}{2}(6+log(0.01)=5$