pH of acids, bases and salts

Exercise 3

   

Calculate the $pH$ of a 3.982 % ($d$ = 1.02) nitric acid solution named $S$

Strong acid If $d$ = 1.02, then $\rho$ = 1.02 $\frac{g}{mL}$ Let's take $1\;L$ of solution $S$: $m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ = 1020 g $m_{HNO_3}$ = $\frac{\%_{HNO_3}\cdot m_S}{100} $ = $\frac{3.982 \cdot1020}{100} $ = 40.62$\; g$; $n_{HNO_3}=\frac{m_{HNO_3}}{M_{HNO_3}}$ = $\frac{40.62}{63.01}$ = 0.645$\; mol$ $c_{HNO_3}$ = $\frac{n_{HNO_3}}{V_S}$ = $\frac{0.645}{1}$ = 0.645 $\frac{mol}{L}$ $pH$ $=$ $-log\;c_{HNO_3}$ = $-log \;$0.645 = 0.191