pH of acids, bases and salts

Exercise 3

   

Calculate the $pH$ of a 6 % ($d$ = 1.041) bromhydric acid solution named $S$

Strong acid If $d$ = 1.041, then $\rho$ = 1.041 $\frac{g}{mL}$ Let's take $1\;L$ of solution $S$: $m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ = 1041 g $m_{HBr}$ = $\frac{\%_{HBr}\cdot m_S}{100} $ = $\frac{6 \cdot1041}{100} $ = 62.46$\; g$; $n_{HBr}=\frac{m_{HBr}}{M_{HBr}}$ = $\frac{62.46}{80.91}$ = 0.772$\; mol$ $c_{HBr}$ = $\frac{n_{HBr}}{V_S}$ = $\frac{0.772}{1}$ = 0.772 $\frac{mol}{L}$ $pH$ $=$ $-log\;c_{HBr}$ = $-log \;$0.772 = 0.112