# What volume of concentrated hydrochloric acid (12.0 M HCl) is require to make 2.0 liters of a 3.0 M HCl solution?

Jan 17, 2017

$500 \cdot m L$ of $12.0 \cdot m o l \cdot {L}^{-} 1$ $H C l \cdot \left(a q\right)$ are required.

#### Explanation:

The product ${C}_{1} {V}_{1}$ is $\text{concentration"xx"volume}$, and when we multiply the typical units we get $m o l \cdot \cancel{{L}^{-} 1} \times \cancel{L}$, i.e. units of $\text{moles}$ as required.

Anyway, as to your problem, ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, because both sides have the units of moles.

${V}_{1} = \frac{{C}_{2} {V}_{2}}{C} _ 1 = \frac{3.0 \cdot m o l \cdot {L}^{-} 1 \times 2.0 \cdot L}{12.0 \cdot m o l \cdot {L}^{-} 1}$

$= \frac{1}{2} \cdot L$

Note that we typically buy conc. $H C l$ as a 32%(w/w) solution, and this is about $10 \cdot m o l \cdot {L}^{-} 1$. This question was not well proposed.