# Question #04a67

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

*Potassium sulfate*, **soluble** ionic compound that dissociates completely in aqueous solution to form potassium cations,

#"K"_2"SO"_text(4(aq]) -> 2"K"_text((aq])^(+) + "SO"_text(4(aq])^(2-)#

The problem wants you to express the **concentration** of the sulfate anions the same way the **pH** of a solution is calculated.

Here

#"pSO"_4^(2-) = - log(["SO"_4^(2-)])" "# , where

Notice the similarity with the definition of pH, which is

#"pH" = - log(["H"^(+)])#

So basically that little **negative log** of a concentration.

In order to find the concentration of sulfate anions, you need to find the concentration of the potassium sulfate. As you know, **molarity** tells you how many moles of solute you get per **liter** of solution.

Use the **molar mass** of potassium sulfate to find the number of moles of potassium sulfate you have in that sample - **do not** forget to convert the *milligrams* to *grams*!

#597 color(red)(cancel(color(black)("mg"))) * (1color(red)(cancel(color(black)("g"))))/(10^3color(red)(cancel(color(black)("mg")))) * ("1 mole K"_2"SO"_4)/(174.24color(red)(cancel(color(black)("g")))) = "0.003426 moles K"_2"SO"_4#

This means that the concentration of potassium sulfate will be - **do not** forget to convert the volume of the solution to *liters*!

#["K"_2"SO"_4] = "0.003426 moles"/(799 * 10^(-3)"L") = "0.004289 M"#

Now take a look at the chemical equation that describes the dissociation of the salt. Notice the **mole ratio** that exists between potassium sulfate and the sulfate anions.

This tells you that **mole** of potassium sulfate will produce **mole** of sulfate anions in solution. Therefore, you can say that

#["SO"_4^(2-)] = ["K"_2"SO"_4] = "0.004289 M"#

This means that you have

#"pSO"_4^(2-) = -log(0.004289) = color(green)(2.37)#

Finally, the problem wants you to determine the solution's **mass by volume percent concentration**. *mass of solute* divided by *100 mL of the solution*, and multiplied by

#color(blue)("% w/v" = "mass of solute"/"100 mL solution" xx 100)#

So, if

#100 color(red)(cancel(color(black)("mL"))) * (597 * 10^(-3)"g")/(799color(red)(cancel(color(black)("mL")))) = "0.07472 g"#

This means that the solution's

#"% w/v" = "0.07472 g"/(color(red)(cancel(color(black)(100)))"mL") * color(red)(cancel(color(black)(100))) = color(green)("0.0747%")#

This means that **every**