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

Answer:

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.