pH of acids, bases and salts

Exercise 15

    

Calculate the volume of a 26 % ($\rho_S=$ 1.1536 $\frac{g}{mL}$) solution $S$ of nitric acid that must be diluted in a volumetric flask $j$ of 1000 $mL$ in order to obtain a solution with $pH$ = 0.15

$HNO_3$: strong acid Final solution: $c$ $=$ $10^{-pH}$ = $10^{-0.15}$ = 0.09 $\frac{mol}{L}$ $n_{HNO_3}$ = $c_{HNO_3}\cdot V_{j}$ = 0.09$\cdot$1 = 0.09$\;mol$ The initial concentrated solution had the same number of moles: Initial solution $S$: $m_{HNO_3}$ = $n_{HNO_3}\cdot M_{HNO_3}$ = 0.09$\cdot$63.02 = 5.672$\;g$ $m_S$ = $\frac{m_{HNO_3}\cdot 100}{\%_S}$ = $\frac{5.672\cdot 100}{26}$ = 21.815$\;g$ $V_S$ = $\frac{m_S}{\rho_S}$ = $\frac{5.672}{1.1536}$ = 4.917$\;mL$