# To make 500 mL of 3.0 M HCl, how many liters of concentrated 18.0 M HCl does one need?

Sep 16, 2016

Approx. $0.0833 \cdot L$ are required. Note that the question was not set by a chemist. $18.0 \cdot m o l \cdot {L}^{-} 1$ $H C l$ is not available.

#### Explanation:

To begin, I don't think you can buy $18.0$ $m o l \cdot {L}^{-} 1$ $H C l$, but for now I will ignore this fact.

We need $500 \cdot m L$, $3.0 \cdot m o l \cdot {L}^{-} 1$ $H C l$. This is a molar quantity of $500 \times {10}^{-} 3 \cdot L \times 3.0 \cdot m o l \cdot {L}^{-} 1 = 1.50 \cdot m o l$.

There is $18.0 \cdot m o l \cdot {L}^{-} 1$ $H C l$ available according to the question.

Thus $\frac{1.50 \cdot m o l}{18.0 \cdot m o l \cdot {L}^{-} 1}$ $=$ $0.0833 \cdot L$, and you would add the acid to the distilled water in a volumetric flask.

The strongest $H C l \left(a q\right)$ you can buy is 36% $\frac{w}{w}$. This has a concentration of under $12.0 \cdot m o l \cdot {L}^{-} 1$.

This question is a real dog's breakfast, in that the examiner was clearly not a chemist. I have written before here (on and on and on), that when you work with concentrated acids, the order of addition is important. Acid is always added to water, not water to acid, Why not? Because water to acid WILL spit at you. The question was not sound from this practical viewpoint either.

When we work with acids (with real concentrations), we use the relationship:

$\text{Concentration}$ $=$ $\text{Moles of Acid"/"Volume of solution}$