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
Aug 22, 2017

Answer:

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."#