pH of acids, bases and salts

Exercise 3

   

Calculate the $pH$ of a 2.364 % ($d$ = 1.01) chlorhydric acid solution named $S$

Strong acid If $d$ = 1.01, then $\rho$ = 1.01 $\frac{g}{mL}$ Let's take $1\;L$ of solution $S$: $m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ = 1010 g $m_{HCl}$ = $\frac{\%_{HCl}\cdot m_S}{100} $ = $\frac{2.364 \cdot1010}{100} $ = 23.88$\; g$; $n_{HCl}=\frac{m_{HCl}}{M_{HCl}}$ = $\frac{23.88}{36.46}$ = 0.655$\; mol$ $c_{HCl}$ = $\frac{n_{HCl}}{V_S}$ = $\frac{0.655}{1}$ = 0.655 $\frac{mol}{L}$ $pH$ $=$ $-log\;c_{HCl}$ = $-log \;$0.655 = 0.184