pH des acides, bases et sels
Exercice 3

Calculer le pH d'une solution S d'acide chlorhydrique à 3,374 % (d = 1,015)
Acide fort
Si d = 1,015, alors $\rho$ = 1,015 $\frac{g}{mL}$
Prenons 1L de cette solution S:
$m_S=\rho \cdot V_S = \rho \cdot 1000 $ =
1015 g
$m_{HCl}$ =
$\frac{\%_{HCl}\cdot m_S}{100} $ =
$\frac{3,374 \cdot1015}{100} $ =
34,25 g;
$n_{HCl}=\frac{m_{HCl}}{M_{HCl}}$ =
$\frac{34,25}{36,46}$ =
0,939 mol
$c_{HCl}$ = $\frac{n_{HCl}}{V_S}$ =
$\frac{0,939}{1}$ =
0,939 $\frac{mol}{L}$
$pH=-log\;c_{HCl}$ =
$-log \;$0,939 =
0,027