By mixing $500\; mL$ $0.2\;M$ $NH_3$ with $500\; mL$ $NH_4Cl$(aq), one obtains a $pH$ of $9$. Calculate the mass of ammonium chloride in the late solution.
$pH$ $=$ $pKa$ $+$ $log \frac{n_{NH_3}}{n_{NH_4^+}}$ $9$ $=$ $9.2$ $+$ $log \frac{0.1}{n_{NH_4^+}}$ therefore: mass of ammonium chloride = $n_{NH_4Cl}\cdot M_{NH_4Cl}$ = $n_{NH_4^+}\cdot M_{NH_4Cl}$ = $10^{-0.8}\cdot 53.5$ = $8.48\; g$
