Concentrated hydrobromic acid has a density of #1.482*g*cm^-3#, and a concentration of #47.6%# #"w/w"#. What volume of such acid is required to prepare a #2.50*dm^3# volume of solution whose #pH=3.4#?

1 Answer
anor277
Nov 5, 2017

Constant boiling hydrobromic acid is #47.6%# BY MASS, and has a concentration of #8.89*mol*L^-1#. Note that this question is an intellectual exercise only.....

Explanation:

But of course we should establish the concentration from first principles FROM THE GIVEN DATA...

#"Concentration"="Moles of solute"/"Volume of solution"#...and thus has units of #mol*L^-1# as required... We work on a #1*cm^3# volume of the stuff....

#[HBr]=((1*cm^3xx47.6%xx1.482*g*cm^-3)/(80.91*g*mol^-1))/(1*cm^3xx10^-3*dm^3*cm^-3)=8.72*mol*dm^-3#, the which is near enuff to the result quoted...

We want a #2.50*dm^3# volume whose #pH=3.4#, i.e. #[H_3O^+]=10^(-3.4)=3.981xx10^-4*mol*L^-1#...

And thus........
#n_(HBr)="concentration"xx"volume"=3.981xx10^-4*mol*dm^-3xx2.50*dm^3=0.995xx10^-4*mol#

And so we take the quotient #"moles of solute"/"concentration"# to get to the required volume....whew, arithmetic....and it is up to you check whether I made an error....

#=(0.995xx10^-4*mol)/(8.72*mol*dm^-3)xx10^3*cm^3*dm^-3=0.114*cm^3# and good luck measuring that volume of conc. #HBr#; it would be a highly impractical undertaking. This is especially true given that when you work with conc. acids, YOU MUST ADD acid to water! Why?

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