To what volume must a 250*mL volume of 10*mol*L^-1 HCl be diluted to give a 0.050*mol*L^-1 concentration?

1 Answer
Sep 11, 2017

Before we start remember the CARDINAL RULE:
"Acid to water not water to acid....." I get a final volume of 50*L of dilute acid.

Explanation:

"May her rest be long and placid, she added water to the acid."

"The other girl did as she oughter, she added acid to the water."

A small drop of water placed in a strong acid might spit as it heats up upon hydration (another old saw: "if you spit in acid it spits back at you").

For your problem, we use the relationship.....

"Concentration"="Moles of solute (moles)"/"Volume of solution (L)"

We have a starting volume of 250*mL of 10.0*mol*L^-1 HCl (the stuff you get out of the bottle is 10.6*mol*L^-1).

"Moles of HCl"=250*mLxx10^-3*L*mL^-1xx10*mol*L^-1

=2.50*mol with respect to HCl.

Now this molar quantity is diluted up to give a 0.050*mol*L^-1 (and remember acid to water!).

We solve for "volume" in "concentration"="moles of HCl"/"volume of solution"

i.e. V=(2.50*mol)/(0.050*mol*L^-1)=50*L, why? Because 1/L^-1=1/(1/L)=L as required.

AND WHEN YOU WORK WITH CONC. ACIDS YOU MUST WEAR (i) SAFETY SPECTACLES to protect your mince pies; and (ii) a LAB-COAT to protect your clothing.