pH of acids, bases and salts
Exercise 3

Calculate the $pH$ of a 7 % ($d$ = 1.049) bromhydric acid solution named $S$
Strong acid
If $d$ = 1.049, then $\rho$ = 1.049 $\frac{g}{mL}$
Let's take $1\;L$ of 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