Calculate the volume of a 34 % ($\rho_S=$ 1.2068 $\frac{g}{mL}$) solution $S$ of nitric acid
that must be diluted in a volumetric flask $j$ of 250 $mL$ in order to obtain a solution with $pH$ = 1.06
$HNO_3$: strong acid
Final solution:
$c$ $=$ $10^{-pH}$ =
$10^{-1.06}$ =
0.09 $\frac{mol}{L}$
$n_{HNO_3}$ =
$c_{HNO_3}\cdot V_{j}$ =
0.09$\cdot$0.25 =
0.0225$\;mol$
The initial concentrated solution had the same number of moles:
Initial solution $S$:
$m_{HNO_3}$ =
$n_{HNO_3}\cdot M_{HNO_3}$ =
0.0225$\cdot$63.02 =
1.418$\;g$
$m_S$ =
$\frac{m_{HNO_3}\cdot 100}{\%_S}$ =
$\frac{1.418\cdot 100}{34}$ =
4.171$\;g$
$V_S$ =
$\frac{m_S}{\rho_S}$ =
$\frac{1.418}{1.2068}$ =
1.175$\;mL$
