Calculate the volume of a 38 % ($\rho_S=$ 1.2335 $\frac{g}{mL}$) solution $S$ of nitric acid
that must be diluted in a volumetric flask $j$ of 500 $mL$ in order to obtain a solution with $pH$ = 2.81
$HNO_3$: strong acid
Final solution:
$c$ $=$ $10^{-pH}$ =
$10^{-2.81}$ =
0.15 $\frac{mol}{L}$
$n_{HNO_3}$ =
$c_{HNO_3}\cdot V_{j}$ =
0.15$\cdot$0.5 =
0.075$\;mol$
The initial concentrated solution had the same number of moles:
Initial solution $S$:
$m_{HNO_3}$ =
$n_{HNO_3}\cdot M_{HNO_3}$ =
0.075$\cdot$63.02 =
4.727$\;g$
$m_S$ =
$\frac{m_{HNO_3}\cdot 100}{\%_S}$ =
$\frac{4.727\cdot 100}{38}$ =
12.439$\;g$
$V_S$ =
$\frac{m_S}{\rho_S}$ =
$\frac{4.727}{1.2335}$ =
3.832$\;mL$
