Jun 18, 2017

$\text{Molarity} = 0.189 \cdot m o l \cdot {L}^{-} 1$. As expected the concentration is almost halved.

#### Explanation:

$\text{Molarity"="moles of solute"/"volume of solution}$

And so......

$\text{moles of solute"="molarity"xx"volume of solution}$.

And if we assume, reasonably, that the volumes are additive, then we get the new concentration by the expression.....

$\left[{H}_{2} S {O}_{4} \left(a q\right)\right] = \frac{300.0 \cdot m L \times {10}^{-} 3 \cdot \frac{L}{m L} \times 0.377 \cdot m o l \cdot {L}^{-} 1}{0.600 \cdot L}$

$= 0.189 \cdot m o l \cdot {L}^{-} 1$ WITH RESPECT TO ${H}_{2} S {O}_{4}$.

Now of course we know that sulfuric acid ionizes in aqueous solution, but that is nothing that we need to concern ourselves with here.

What is the $p H$ of this solution?