# How many liters of 8.0M HCl are needed to prepare 1.50 L of 2.5M HCl?

Oct 10, 2016

0.47 L of 8.0 M HCl are needed to prepare 1.50 L of 2.5 M HCl.

#### Explanation:

For dilution problems: ${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$, where ${M}_{1}$ and ${V}_{1}$ are the concentration and volume of the original solution and ${M}_{2}$ and ${V}_{2}$ are the concentration and volume of the diluted solution.

We are calculating ${V}_{1}$.

First rearrange the dilution equation:
${V}_{1} = \frac{{M}_{2} {V}_{2}}{M} _ 1$

Plug in the values:
${V}_{1} = \frac{\left(2.5 M\right) \left(1.50 L\right)}{8.0 M}$

Calculate:
${V}_{1} = 0.47 L$