# How would you prepare 500.0 mL of a 2.00 mol/L solution of HCI from a concentrated solution of 12.0 mol/L?

Aug 22, 2017

Carefully, and we adds acid to water. This is an important practical difficulty......

#### Explanation:

In fact the most concentrated hydrochloric acid you can buy is approx. 32% $\text{w/w}$, and this is about $10.6 \cdot m o l \cdot {L}^{-} 1$. Anyway we will go with the terms of the question.

Here we use the ratio, $\text{concentration"="moles of solute"/"volume of solution}$.

And thus $\text{moles of solute"="concentration"xx"volume}$.

We want $500.0 \times {10}^{-} 3 \cdot L \times 2.00 \cdot m o l \cdot {L}^{-} 1 = 1 \cdot m o l$ with respect to $H C l$.

And so we take the quotient, $\frac{1.00 \cdot m o l}{12.0 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1$

$= 83.3 \cdot m L$.

And so we take $83.3 \cdot m L$ of conc. acid, and we add approx. $420 \cdot m L$ of water. The order of addition is VERY important.

Remember what I said about diluting conc. acid: $\text{If you spit in acid it spits back.}$