pH of acids, bases and salts

Exercise 7

    

Using the table of acid-base couples, calculate the $pH$ of a 0.5 %  ($d$ = 0.9998)  ammonium chloride solution named $S$.

$NH_4^+: $ weak acid If $d$ = 0.9998, then $\rho$ = 0.9998 $\frac{g}{mL}$ Let's take $1\;L$ of solution $S$: $m_S$ $=$ $\rho \cdot V_S $ $=$ $ \rho \cdot 1000 $ = 999.8$\; g$ $m_{NH_4Cl}$ = $\frac{\%_{NH_4Cl}\cdot m_S}{100} $ = $\frac{0.5 \cdot999.8}{100} $ = 5$\; g$; $n_{NH_4Cl}=\frac{m_{NH_4Cl}}{M_{NH_4Cl}}$ = $\frac{5}{53.5}$ = 0.093$\; mol$ $c_{NH_4^+}$ = $c_{NH_4Cl}$ = $\frac{n_{NH_4Cl}}{V_S}$ = $\frac{0.093}{1}$ = 0.093 $\frac{mol}{L}$ Given $y=[H_3O^+]$ The equation $y^2$ $+$ $K_a\;y$ $-$ $K_a\;c$ $=$ $0$ becomes: $y^2$ $+$ $10^{-9.2}y$ $-$ $10^{-9.2} 0.093$ $=$ $0$ and produces: $y=$ 7.68 10^-6 and so: $pH$ $=$ $-log\; y$ $ =$ 5.115