# To 225 mL of a 0.80M solution of KI, a student adds enough water to make 1.0 L of a more dilute KI solution. What is the molarity of the new solution?

Mar 21, 2016

${\text{0.18 mol L}}^{- 1}$

#### Explanation:

Right from the start, you know that the molarity of solution decreased upon the addition of water, which is what diluting a solution implies.

The underlying principle behind 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.

So, you can use the molarity and volume of the initial solution to find the number of moles of potassium iodide, $\text{KI}$, present in the solution

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Remember, molarity uses liters of solution, so don't forget to convert the volume from milliliters to liters

${n}_{K I} = \text{0.80 mol" color(red)(cancel(color(black)("L"^(-1)))) * 225 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.180 moles KI}$

This is exactly how many moles of solute must be present in the diluted solution, so you can say that

${c}_{\text{diluted" = n_(KI)/V_"solution}}$

c_"diluted" = "0.180 moles"/"1.0 L" = color(green)(|bar(ul(color(white)(a/a)"0.18 mol L"^(-1)color(white)(a/a)|)))

This is exactly what the formula for dilution calculations allows you to do

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

Here

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

Plug in your values to get

${c}_{2} = {V}_{/} 1 {V}_{2} \cdot {c}_{1}$

c_2 = (225 * 10^(-3)color(red)(cancel(color(black)("L"))))/(1.0color(red)(cancel(color(black)("L")))) * "0.80 mol L"^(-1) = color(green)(|bar(ul(color(white)(a/a)"0.18 mol L"^(-1)color(white)(a/a)|)))

The answer is rounded to two sig figs.