# Calculate the molarity (in mol/L) of "K"^(+) ions present in an aqueous solution prepared by adding "23.4 mL" of 9.32*10^(-1)"M" aqueous "K"_2"SO"_3 to "185 mL" of water? Report your answer to three significant figures.

Sep 21, 2017

${\text{0.209 mol L}}^{- 1}$

#### Explanation:

The first thing that you need to do here is to figure out how many moles of potassium sulfite are present in the initial sample.

To do that, use the volume and the molarity of the solution--keep in mind that $\text{1 L" = 10^3color(white)(.)"mL}$

23.4 color(red)(cancel(color(black)("mL solution"))) * overbrace((9.32 * 10^(-1)color(white)(.)"moles K"_2"SO"_3)/(10^3color(red)(cancel(color(black)("mL solution")))))^(color(blue)(=9.32 * 10^(-1)color(white)(.)"M")) = "0.02181 moles K"_2"SO"_3

Now, you know that potassium sulfite is soluble in aqueous solution, which implies that it dissociates completely when dissolved in water

${\text{K"_ 2"SO"_ (3(aq)) -> 2"K"_ ((aq))^(+) + "SO}}_{3 \left(a q\right)}^{2 -}$

Since every mole of potassium sulfite dissolved in the solution will produce $2$ moles of potassium cations, you can say that your initial solution contains

0.02181 color(red)(cancel(color(black)("moles K"_2"SO"_3))) * "2 moles K"^(+)/(1color(red)(cancel(color(black)("mole K"_2"SO"_3)))) = "0.04362 moles K"^(+)

Next, calculate the total volume of the resulting solution, i.e. the volume of the diluted solution

${V}_{\text{total" = "23.4 mL + 185 mL = 208.4 mL}}$

This means that the molarity of the potassium cations in the diluted solution will be--don't forget to convert the volume of the solution to liters!

["K"^(+)] = "0.04362 moles"/(208.4 * 10^(-3)color(white)(.)"L") = color(darkgreen)(ul(color(black)("0.209 mol L"^(-1))))

The answer is rounded to three sig figs.