pH des acides, bases et sels
Exercice 3

Calculer le pH d'une solution S d'acide bromhydrique à 6 % (d = 1,041)
Acide fort
Si d = 1,041, alors $\rho$ = 1,041 $\frac{g}{mL}$
Prenons 1L de cette solution S:
$m_S=\rho \cdot V_S = \rho \cdot 1000 $ =
1041 g
$m_{HBr}$ =
$\frac{\%_{HBr}\cdot m_S}{100} $ =
$\frac{6 \cdot1041}{100} $ =
62,46 g;
$n_{HBr}=\frac{m_{HBr}}{M_{HBr}}$ =
$\frac{62,46}{80,91}$ =
0,772 mol
$c_{HBr}$ = $\frac{n_{HBr}}{V_S}$ =
$\frac{0,772}{1}$ =
0,772 $\frac{mol}{L}$
$pH=-log\;c_{HBr}$ =
$-log \;$0,772 =
0,112