pH of acids, bases and salts
Exercise 3

Calculate the $pH$ of a 1.36 % ($d$ = 1.005) chlorhydric acid solution named $S$
Strong acid
If $d$ = 1.005, then $\rho$ = 1.005 $\frac{g}{mL}$
Let's take $1\;L$ of solution $S$:
$m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ =
1005 g
$m_{HCl}$ =
$\frac{\%_{HCl}\cdot m_S}{100} $ =
$\frac{1.36 \cdot1005}{100} $ =
13.67$\; g$;
$n_{HCl}=\frac{m_{HCl}}{M_{HCl}}$ =
$\frac{13.67}{36.46}$ =
0.375$\; mol$
$c_{HCl}$ = $\frac{n_{HCl}}{V_S}$ =
$\frac{0.375}{1}$ =
0.375 $\frac{mol}{L}$
$pH$ $=$ $-log\;c_{HCl}$ =
$-log \;$0.375 =
0.426