Question

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?

Answer 1

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%` `"w/w"`, and this is about `10.6*mol*L^-1`. Anyway we will go with the terms of the question.

Here we use the ratio, `"concentration"="moles of solute"/"volume of solution"`.

And thus `"moles of solute"="concentration"xx"volume"`.

We want `500.0xx10^-3*Lxx2.00*mol*L^-1=1*mol` with respect to `HCl`.

And so we take the quotient, `(1.00*mol)/(12.0*mol*L^-1)xx10^3*mL*L^-1`

`=83.3*mL`.

And so we take `83.3*mL` of conc. acid, and we add approx. `420*mL` of water. The order of addition is VERY important.

Remember what I said about diluting conc. acid: `"If you spit in acid it spits back."`