# What's V2 in general/how do you calculate it step by step?

This is the question: If I add 25 mL of water to 125 mL of a 0.15 M NaOH solution, what will the molarity of the diluted solution be?

Well volumes are generally additive, and of course the concentration will be diluted.

#### Explanation:

By one of the definitions, $\text{concentration"="Moles of solute"/"Volume of solution} .$

And thus $\text{moles of solute"="concentration"xx"volume of solution}$

And so.....the new concentration will be given by the quotient....

$\frac{125 \times {10}^{-} 3 \cdot \cancel{L} \times 0.15 \cdot m o l \cdot \cancel{{L}^{-} 1}}{125 \times {10}^{-} 3 \cdot L + 25 \times {10}^{-} 3 \cdot L}$

$= 0.125 \cdot m o l \cdot {L}^{-} 1$, i.e. concentration has reduced slightly.

This goes back to the old equality, ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, when a given concentration is diluted, the new concentration is readily accessible.

Here we solved for ${C}_{2} = \frac{{C}_{1} {V}_{1}}{V} _ 2$.........

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