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