Calculate the volume of a 32 % ($\rho_S=$ 1.1934 $\frac{g}{mL}$) solution $S$ of nitric acid
that must be diluted in a volumetric flask $j$ of 500 $mL$ in order to obtain a solution with $pH$ = 2.98
$HNO_3$: strong acid
Final solution:
$c$ $=$ $10^{-pH}$ =
$10^{-2.98}$ =
0.03 $\frac{mol}{L}$
$n_{HNO_3}$ =
$c_{HNO_3}\cdot V_{j}$ =
0.03$\cdot$0.5 =
0.015$\;mol$
The initial concentrated solution had the same number of moles:
Initial solution $S$:
$m_{HNO_3}$ =
$n_{HNO_3}\cdot M_{HNO_3}$ =
0.015$\cdot$63.02 =
0.945$\;g$
$m_S$ =
$\frac{m_{HNO_3}\cdot 100}{\%_S}$ =
$\frac{0.945\cdot 100}{32}$ =
2.953$\;g$
$V_S$ =
$\frac{m_S}{\rho_S}$ =
$\frac{0.945}{1.1934}$ =
0.792$\;mL$
