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