pH des acides, bases et sels
Exercice 3

Calculer le pH d'une solution S d'acide chlorhydrique à 2,364 % (d = 1,01)
Acide fort
Si d = 1,01, alors $\rho$ = 1,01 $\frac{g}{mL}$
Prenons 1L de cette solution S:
$m_S=\rho \cdot V_S = \rho \cdot 1000 $ =
1010 g
$m_{HCl}$ =
$\frac{\%_{HCl}\cdot m_S}{100} $ =
$\frac{2,364 \cdot1010}{100} $ =
23,88 g;
$n_{HCl}=\frac{m_{HCl}}{M_{HCl}}$ =
$\frac{23,88}{36,46}$ =
0,655 mol
$c_{HCl}$ = $\frac{n_{HCl}}{V_S}$ =
$\frac{0,655}{1}$ =
0,655 $\frac{mol}{L}$
$pH=-log\;c_{HCl}$ =
$-log \;$0,655 =
0,184