pH des acides, bases et sels

Exercice 3

   

Calculer le pH d'une solution S d'acide chlorhydrique  à  5,408 % (d = 1,025)

Acide fort Si d = 1,025, alors $\rho$ = 1,025 $\frac{g}{mL}$ Prenons 1L de cette solution S: $m_S=\rho \cdot V_S = \rho \cdot 1000 $ = 1025 g $m_{HCl}$ = $\frac{\%_{HCl}\cdot m_S}{100} $ = $\frac{5,408 \cdot1025}{100} $ = 55,43 g; $n_{HCl}=\frac{m_{HCl}}{M_{HCl}}$ = $\frac{55,43}{36,46}$ = 1,52  mol $c_{HCl}$ = $\frac{n_{HCl}}{V_S}$ = $\frac{1,52}{1}$ = 1,52 $\frac{mol}{L}$ $pH=-log\;c_{HCl}$ = $-log \;$1,52 = -0,182