What is the molarity of the solution prepared when #13.50*mL# of #8.00*mol*L^-1# #"hydrochloric acid"# is add to #225.0*mL# of water?

1 Answer
May 1, 2017

Answer:

The new concentration is approx. #0.5*mol*L^-1# with respect to sulfuric acid.

Explanation:

#"Concentration (molarity)"-="Moles of solute"/"Volume of solution"#.

And thus here, we have.................

#(13.50xx10^-3*Lxx8.00*mol*L^-1)/(225.0xx10^-3*L)=0.480*mol*L^-1#.

This value is the formal concentration with respect to #H_2SO_4#; of course we know that sulfuric acid speciates in aqueous solution (to what?).

Just to add that I approve of your order of addition. Conc. acid is ALWAYS added to water, and NEVER the reverse. Why not? Because if you spit in acid it spits back at you............