# 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?

May 1, 2017

The new concentration is approx. $0.5 \cdot m o l \cdot {L}^{-} 1$ with respect to sulfuric acid.

#### Explanation:

$\text{Concentration (molarity)"-="Moles of solute"/"Volume of solution}$.

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

$\frac{13.50 \times {10}^{-} 3 \cdot L \times 8.00 \cdot m o l \cdot {L}^{-} 1}{225.0 \times {10}^{-} 3 \cdot L} = 0.480 \cdot m o l \cdot {L}^{-} 1$.

This value is the formal concentration with respect to ${H}_{2} S {O}_{4}$; 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............