pH des acides, bases et sels

Exercice 3

   

Calculer le pH d'une solution S d'acide chlorhydrique  à  4,388 % (d = 1,02)

Acide fort Si d = 1,02, alors $\rho$ = 1,02 $\frac{g}{mL}$ Prenons 1L de cette solution S: $m_S=\rho \cdot V_S = \rho \cdot 1000 $ = 1020 g $m_{HCl}$ = $\frac{\%_{HCl}\cdot m_S}{100} $ = $\frac{4,388 \cdot1020}{100} $ = 44,76 g; $n_{HCl}=\frac{m_{HCl}}{M_{HCl}}$ = $\frac{44,76}{36,46}$ = 1,228  mol $c_{HCl}$ = $\frac{n_{HCl}}{V_S}$ = $\frac{1,228}{1}$ = 1,228 $\frac{mol}{L}$ $pH=-log\;c_{HCl}$ = $-log \;$1,228 = -0,089