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$\;mL$
