# 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.

##### 1 Answer

#### 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

#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

#"K"_ 2"SO"_ (3(aq)) -> 2"K"_ ((aq))^(+) + "SO"_ (3(aq))^(2-)#

Since **every mole** of potassium sulfite dissolved in the solution will produce **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_"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**.