pH des acides, bases et sels

Exercice 3

   

Calculer le pH d'une solution S d'acide bromhydrique  à  7 % (d = 1,049)

Acide fort Si d = 1,049, alors $\rho$ = 1,049 $\frac{g}{mL}$ Prenons 1L de cette solution S: $m_S=\rho \cdot V_S = \rho \cdot 1000 $ = 1049 g $m_{HBr}$ = $\frac{\%_{HBr}\cdot m_S}{100} $ = $\frac{7 \cdot1049}{100} $ = 73,43 g; $n_{HBr}=\frac{m_{HBr}}{M_{HBr}}$ = $\frac{73,43}{80,91}$ = 0,908  mol $c_{HBr}$ = $\frac{n_{HBr}}{V_S}$ = $\frac{0,908}{1}$ = 0,908 $\frac{mol}{L}$ $pH=-log\;c_{HBr}$ = $-log \;$0,908 = 0,042