Calculate the volume of a 32 % ($\rho_S=$ 1.1594 $\frac{g}{mL}$) solution $S$ of chlorhydric acid
that must be diluted in a volumetric flask $j$ of 500 $mL$ in order to obtain a solution with $pH$ = 0.6
$HCl$: strong acid
Final solution:
$c$ $=$ $10^{-pH}$ =
$10^{-0.6}$ =
0.02 $\frac{mol}{L}$
$n_{HCl}$ =
$c_{HCl}\cdot V_{j}$ =
0.02$\cdot$0.5 =
0.01$\;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$
$m_S$ =
$\frac{m_{HCl}\cdot 100}{\%_S}$ =
$\frac{0.365\cdot 100}{32}$ =
1.141$\;g$
$V_S$ =
$\frac{m_S}{\rho_S}$ =
$\frac{0.365}{1.1594}$ =
0.315$\;mL$
