Calculate the volume of a 30 % ($\rho_S=$ 1.1801 $\frac{g}{mL}$) solution $S$ of nitric acid
that must be diluted in a volumetric flask $j$ of 500 $mL$ in order to obtain a solution with $pH$ = 2.83
$HNO_3$: strong acid
Final solution:
$c$ $=$ $10^{-pH}$ =
$10^{-2.83}$ =
0.13 $\frac{mol}{L}$
$n_{HNO_3}$ =
$c_{HNO_3}\cdot V_{j}$ =
0.13$\cdot$0.5 =
0.065$\;mol$
The initial concentrated solution had the same number of moles:
Initial solution $S$:
$m_{HNO_3}$ =
$n_{HNO_3}\cdot M_{HNO_3}$ =
0.065$\cdot$63.02 =
4.096$\;g$
$m_S$ =
$\frac{m_{HNO_3}\cdot 100}{\%_S}$ =
$\frac{4.096\cdot 100}{30}$ =
13.653$\;g$
$V_S$ =
$\frac{m_S}{\rho_S}$ =
$\frac{4.096}{1.1801}$ =
3.471$\;mL$
