pH of acids, bases and salts
Exercise 7

Using the table of acid-base couples, calculate the $pH$ of a 2 % ($d$ = 1.0045) ammonium chloride solution named $S$.
$NH_4^+: $ weak acid
If $d$ = 1.0045, then $\rho$ = 1.0045 $\frac{g}{mL}$
Let's take $1\;L$ of solution $S$:
$m_S$ $=$ $\rho \cdot V_S $ $=$ $ \rho \cdot 1000 $ =
1004.5$\; g$
$m_{NH_4Cl}$ =
$\frac{\%_{NH_4Cl}\cdot m_S}{100} $ =
$\frac{2 \cdot1004.5}{100} $ =
20.09$\; g$;
$n_{NH_4Cl}=\frac{m_{NH_4Cl}}{M_{NH_4Cl}}$ =
$\frac{20.09}{53.5}$ =
0.376$\; mol$
$c_{NH_4^+}$ =
$c_{NH_4Cl}$ =
$\frac{n_{NH_4Cl}}{V_S}$ =
$\frac{0.376}{1}$ =
0.376 $\frac{mol}{L}$
Given $y=[H_3O^+]$
The equation
$y^2$ $+$ $K_a\;y$ $-$ $K_a\;c$ $=$ $0$
becomes:
$y^2$ $+$ $10^{-9.2}y$ $-$ $10^{-9.2} 0.376$ $=$ $0$
and produces:
$y=$ 1.54 10-5
and so:
$pH$ $=$ $-log\; y$ $ =$ 4.812