# How much water would I need to add to 500 mL of a 2.4 M KCl solution to make a 1.0 M solution?

Feb 16, 2016

$\text{700 mL}$

#### Explanation:

Let's assume that you're not familiar with the formula for dilution calculations.

The underlying concept of a dilution is that you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.

Now, a solution's molarity tells you the number of moles of solute present in one liter of that solution.

$\textcolor{b l u e}{c = {n}_{\text{solute"/V_"solution}}}$

In a dilution, you know that the number of moles of solute must remain constant. This means that you can use the molarity and volume of the initial solution to figure out how many moles it contains

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \times V}$

${n}_{K C l} = 2.4 \text{moles"/color(red)(cancel(color(black)("L"))) * 500 * 10^(-3)color(red)(cancel(color(black)("L"))) = "1.2 moles KCl}$

This is exactly how many moles of potassium chloride must be present in the target solution, which means that you have

$\textcolor{b l u e}{c = \frac{n}{V} \implies V = \frac{n}{c}}$

${V}_{\text{target" = (1.2color(red)(cancel(color(black)("moles"))))/(1.0color(red)(cancel(color(black)("moles")))/"L") = "1.2 L}}$

So, you need the volume of the target solution to be equal to

${V}_{\text{target" = "1.2 L" = 1.2 * 10^3"mL" = "1200 mL}}$

This means that you must add

V_"added" = "1200 mL" - "500 mL" = color(green)("700 mL")

to your initial solution to get its molarity down from $\text{2.4 M}$ to $\text{1.0 M}$.

This is exactly how the formula for dilution calculations works

$\textcolor{b l u e}{{\overbrace{{c}_{1} \times {V}_{1}}}^{\textcolor{b r o w n}{\text{moles of solute in initial solution")) = overbrace(c_2 xx V_2)^(color(brown)("moles of solute in diluted solution}}}}$

Here

${c}_{1}$, ${V}_{1}$ - the concentration and volume of the initial solution
${c}_{2}$, ${V}_{2}$ - the concentration and volume of the target solution