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?

1 Answer
anor277
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%32% "w/w"w/w, and this is about 10.6*mol*L^-110.6⋅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"concentration=moles of solutevolume of solution.

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

We want 500.0xx10^-3*Lxx2.00*mol*L^-1=1*mol500.0×10−3⋅L×2.00⋅mol⋅L−1=1⋅mol with respect to HClHCl.

And so we take the quotient, (1.00*mol)/(12.0*mol*L^-1)xx10^3*mL*L^-11.00⋅mol12.0⋅mol⋅L−1×103⋅mL⋅L−1

=83.3*mL=83.3⋅mL.

And so we take 83.3*mL83.3⋅mL of conc. acid, and we add approx. 420*mL420⋅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."If you spit in acid it spits back.

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