Question

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?

Answer 1

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............