# How do you determine the molarity of 2.25 mole of sulfuric acid, H_2SO_4, dissolved in 725 mL of solution?

Apr 24, 2017

With relative ease...........

#### Explanation:

Now, by definition, $\text{molarity"="moles of solute (mol)"/"volume of solution (L)}$.

Now the wording of your problem gave us that $2.25 \cdot m o l$ sulfuric acid were dissolved in water to give us $725 \cdot m L$ of solution. Had it said that $2.25 \cdot m o l$ sulfuric acid were dissolved in $725 \cdot m L$ of water, we would be uncertain as to the volume of the sulfuric acid solution. Do you appreciate the distinction I make here?

And thus, we simply have to address the quotient:

$\frac{2.25 \cdot m o l}{725 \cdot m L \times {10}^{-} 3 L \cdot m {L}^{-} 1} = 3.10 \cdot m o l \cdot {L}^{-} 1$ WITH RESPECT TO SULFURIC ACID.

Again, I am trying to be cautious in my wording, and chemists often speak of a $\text{formal concentration}$ to indicate that of course in solution, the solute might speciate, and certainly sulfuric acid does, to ${H}_{3} {O}^{+}$, and $H S {O}_{4}^{-}$, and $S {O}_{4}^{2 -}$. As a practical chemist, you would treat this solution as $\left[S {O}_{4}^{2 -}\right] = 3.10 \cdot m o l \cdot {L}^{-} 1$, and $\left[{H}_{3} {O}^{+}\right] = 6.20 \cdot m o l \cdot {L}^{-} 1$. Capisce?