# Question 04a67

Feb 4, 2016

Here's what I got.

#### Explanation:

Potassium sulfate, ${\text{K"_2"SO}}_{4}$, is a soluble ionic compound that dissociates completely in aqueous solution to form potassium cations, ${\text{K}}^{+}$, and sulfate anions, ${\text{SO}}_{4}^{2 -}$

${\text{K"_2"SO"_text(4(aq]) -> 2"K"_text((aq])^(+) + "SO}}_{\textrm{4 \left(a q\right]}}^{2 -}$

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

Here ${\text{pSO}}_{4}^{2 -}$ is equivalent to having

$\text{pSO"_4^(2-) = - log(["SO"_4^(2-)])" }$, where

$\left[{\text{SO}}_{4}^{2 -}\right]$ - the concentration of the sulfate anions in solution

Notice the similarity with the definition of pH, which is

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

So basically that little $\text{p}$ is used to symbolize the 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 $1 : 1$ mole ratio that exists between potassium sulfate and the sulfate anions.

This tells you that $1$ mole of potassium sulfate will produce $1$ 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

${\text{pSO}}_{4}^{2 -} = - \log \left(0.004289\right) = \textcolor{g r e e n}{2.37}$

Finally, the problem wants you to determine the solution's mass by volume percent concentration. $\text{% w/v}$, which is defined as the mass of solute divided by 100 mL of the solution, and multiplied by $100$.

$\textcolor{b l u e}{\text{% w/v" = "mass of solute"/"100 mL solution} \times 100}$

So, if $\text{799 mL}$ of solution (you can assume that the volume does not change upon the addition of the solute) contains $597 \cdot {10}^{- 3} \text{g}$ of solute, you can say that $\text{100 mL}$ of solution will contain

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 $\text{% w/v}$ will be equal to

"% 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 $\text{100 mL}$ of this solution will contain $\text{0.0747 g}$ of potassium sulfate.