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

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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?

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Answer 1 · 2017-08-17T16:26:51

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

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

And thus $moles of solute = concentration \times volume of solution$ $"moles of solute"="concentration"xx"volume of solution"$

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

$(125xx10^-3*cancelLxx0.15*mol*cancel(L^-1))/(125xx10^-3*L+ 25xx10^-3*L)$

$=0.125*mol*L^-1$ , i.e. concentration has reduced slightly.

This goes back to the old equality, $C_1V_1=C_2V_2$ , when a given concentration is diluted, the new concentration is readily accessible.

Here we solved for $C_2=(C_1V_1)/V_2$ .........