# Question 7b635

Nov 19, 2015

$\text{K"^(+): "0.144 moles}$
$\text{SO"_4^(2-): "0.072 moles}$

#### Explanation:

Potassium sulfate, ${\text{K"_2"SO}}_{4}$, is a soluble ionic compound, which means that it will dissociate completely in aqueous solution to give potassium cations, ${\text{K}}^{+}$, and sulfate anions, ${\text{SO}}_{4}^{2 -}$.

The important thing to notice here is that you need two potassium cations to balanced the negative charge of the sulfate anion. This means that when potassium sulfate dissolves in water, you will have

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

This tells you that every mole of potassium sulfate will produce 2 moles of potassium cations and one mole of sulfate anions.

Use the solution's molarity and volume to determine how many moles of potassium sulfate you start with

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{{K}_{2} S {O}_{4}} = {\text{0.24 M" * 300. * 10^(-3)"L" = "0.0720 moles K"_2"SO}}_{4}$

This means that your solution will contain

0.0720color(red)(cancel(color(black)("moles K"_2"SO"_4))) * ("2 moles K"^(+))/(1color(red)(cancel(color(black)("mole K"_2"SO"_4)))) = color(green)("0.144 moles K"^(+))

and

0.0720color(red)(cancel(color(black)("moles K"_2"SO"_4))) * ("1 mole SO"_4^(2-))/(1color(red)(cancel(color(black)("mole K"_2"SO"_4)))) = color(green)("0.072 moles SO"_4^(2-))#