# What is the molarity of a sulfuric acid solution prepared by diluting 500.0 mL of 1.00 M H2SO4 to a total volume of 2.50 L?

Nov 17, 2016

$\text{Molarity}$ $=$ $2.00 \cdot m o l \cdot {L}^{-} 1$ with respect to $\text{sulfuric acid}$.

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$, and thus has the units of $m o l \cdot {L}^{-} 1$.

So we work out this quotient with respect to the given conditions:

$\frac{500 \times {10}^{-} 3 \cdot \cancel{L} \times 1.00 \cdot m o l \cdot {L}^{-} 1}{2.50 \cdot \cancel{L}} = 2.00 \cdot m o l \cdot {L}^{-} 1$. At least our answer is dimensionally consistent.

Note that the given concentration is with respect to $\text{sulfuric acid}$. What would it be with respect to $S {O}_{4}^{2 -}$, and to ${H}_{3} {O}^{+}$?