Calculate the volume of a 38 % ($\rho_S=$ 1.1886 $\frac{g}{mL}$) solution $S$ of chlorhydric acid
that must be diluted in a volumetric flask $j$ of 1000 $mL$ in order to obtain a solution with $pH$ = 1.38
$HCl$: strong acid
Final solution:
$c$ $=$ $10^{-pH}$ =
$10^{-1.38}$ =
0.01 $\frac{mol}{L}$
$n_{HCl}$ =
$c_{HCl}\cdot V_{j}$ =
0.01$\cdot$1 =
0.01$\;mol$
The initial concentrated solution had the same number of moles:
Initial solution $S$:
$m_{HCl}$ =
$n_{HCl}\cdot M_{HCl}$ =
0.01$\cdot$36.47 =
0.365$\;g$
$m_S$ =
$\frac{m_{HCl}\cdot 100}{\%_S}$ =
$\frac{0.365\cdot 100}{38}$ =
0.961$\;g$
$V_S$ =
$\frac{m_S}{\rho_S}$ =
$\frac{0.365}{1.1886}$ =
0.307$\;mL$
