Question

A chemist need 250mL of 3 mol/L hydrochloric acid. What volume of 12mol/L solution would be needed to be diluted to get the needed solution?

Answer 1

During the dilution, the number of the moles of the hydrochloric acid HCl is conserved. This means that the number of the moles of HCl $n_{12M}$ in the needed volume of 12 M solution and the number of the moles of HCl $n_{3M}$ in 250 mL of 3 M solution are equal:

$n_{12M} = n_{3M}$ .

The number of the moles of HCl in 250 mL of 3M solution is volume times concentration. The volume must be converted in liters, so $V = 250·10^{-3}$ L:

$n=cV = 3·250·10^{-3} = 0.75$ mol.

Using the same relation, the volume of 12 M HCl needed is:

$V_{12M} = \frac{n}{c} = \frac{0.75}{12} = 0.0625$ L, or 62.5 mL.

Answer: the volume of 12 M HCl of solution needed to prepare 250 mL of 3 M HCl solution is 62.5 mL.