# What volume of 0.10 M NaOH can be prepared from 250 mL of 0.30 M NaOH?

Jul 29, 2016

$\text{750 mL}$

#### Explanation:

The thing to notice here is that the initial solution is $3$ times more concentrated than the target solution, since

$\left(0.30 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{M"))))/(0.10color(red)(cancel(color(black)("M}}}}\right) = 3$

This tells you that the volume of the target solution must be $3$ times greater than the volume of the initial solution.

That is the case because when you dilute a solution, you decrease its concentration by increasing its volume while keeping the number of moles of solute constant.

This implies that when the concentration of a solution decreases by a factor, which is usually called dilution factor, $\text{DF}$, upon dilution, the volume of the diluted solution increases by the same factor $\text{DF}$.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{DF" = V_"final"/V_"initial" = c_"initial"/c_"final} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You thus have

V_"final" = 3 xx "250 mL" = color(green)(|bar(ul(color(white)(a/a)color(black)("750 mL")color(white)(a/a)|)))

The answer is rounded to two sig figs.

So, you can decrease the concentration of $\text{250 mL}$ of sodium hydroxide solution from $\text{0.30 M}$ to $\text{0.10 M}$ by adding enough water to get its final volume to $\text{750 mL}$.