pH of acids, bases and salts

Exercise 3

   

Calculate the $pH$ of a 1 % ($d$ = 1.005) bromhydric acid solution named $S$

Strong acid If $d$ = 1.005, then $\rho$ = 1.005 $\frac{g}{mL}$ Let's take $1\;L$ of solution $S$: $m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ = 1005 g $m_{HBr}$ = $\frac{\%_{HBr}\cdot m_S}{100} $ = $\frac{1 \cdot1005}{100} $ = 10.05$\; g$; $n_{HBr}=\frac{m_{HBr}}{M_{HBr}}$ = $\frac{10.05}{80.91}$ = 0.124$\; mol$ $c_{HBr}$ = $\frac{n_{HBr}}{V_S}$ = $\frac{0.124}{1}$ = 0.124 $\frac{mol}{L}$ $pH$ $=$ $-log\;c_{HBr}$ = $-log \;$0.124 = 0.906