# 60.0 mL of 0.45 M K_3PO_4 is mixed with 240.0 mL of 0.20 M K_2SO_3. What is the final concentration of all three ions in the solution?

Feb 3, 2016

["K"^(+)] = "0.59 M"" " ["PO"_4^(3-)] = "0.090 M"" " ["SO"_3^(2-)] = "0.16 M"

#### Explanation:

The key to this problem lies with the dissociation of the two salts, both soluble in aqueous solution, to form positively charged ions called cations and negatively charged ions called anions.

Tripotassium phosphate, ${\text{K"_3"PO}}_{4}$, will dissociate in aqueous solution to form

${\text{K"_3"PO"_text(4(aq]) -> color(red)(3)"K"_text((aq])^(+) + "PO}}_{\textrm{4 \left(a q\right]}}^{3 -}$

Potassium sulfite, ${\text{K"_2"SO}}_{3}$, will dissociate in aqueous solution to form

${\text{K"_2"SO"_text(3(aq]) -> color(blue)(2)"K"_text((aq])^(+) + "SO}}_{\textrm{3 \left(a q\right]}}^{2 -}$

This means that the three ions present in the target solution will be

• the potassium cation, ${\text{K}}^{+}$
• the phosphate anion, ${\text{PO}}_{4}^{3 -}$
• the sulfite anion, ${\text{SO}}_{3}^{2 -}$

In order to be able to determine the concentration of each ion in the final solution, you need to know two things

• the number of moles of each ion present in solution
• the total volume of the solution

As you know, molarity is defined as the number of moles of solute per liter of solution. Use the volume and molarities of the two solutions to determine how many moles of each salt you're mixing

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

${n}_{{K}_{3} P {O}_{4}} = {\text{0.45 M" * 60.0 * 10^(-3)"L" = "0.0270 moles K"_3"PO}}_{4}$

and

${n}_{{K}_{2} S {O}_{3}} = {\text{0.20 M" * 240.0 * 10^(-3)"L" = "0.0480 moles K"_2"SO}}_{3}$

Now look at the two balanced chemical equations and determine how many moles of each ion are produced in solution when the salts dissolve.

Tripotassium phosphate will produce $\textcolor{red}{3}$ moles of potassium cations and $1$ mole of phosphate anions, so you can say that

${n}_{{K}^{+}} = \textcolor{red}{3} \times {\text{0.0270 moles" = "0.0810 moles K}}^{+}$

${n}_{P {O}_{4}^{3 -}} = 1 \times {\text{0.0270 moles" = "0.0270 moles PO}}_{4}^{3 -}$

Potassium sulfite will produce $\textcolor{b l u e}{2}$ moles of potassium cations and $1$ mole of sulfite anions, so you can say that

${n}_{{K}^{+}} = \textcolor{b l u e}{2} \times {\text{0.0480 moles" = "0.0960 moles K}}^{+}$

${n}_{S {O}_{3}^{2 -}} = 1 \times {\text{0.0480 moles" = "0.0480 moles SO}}_{3}^{2 -}$

As you can see, you have potassium cations being delivered to the final solution by both salts, which means that you'll have

n_(K^(+)"total") = "0.0810 moles" + "0.0960 moles" = "0.177 moles K"^(+)

The total volume of the final solution will be

${n}_{\text{total}} = {V}_{{K}_{3} P {O}_{4}} + {V}_{{K}_{2} S {O}_{3}}$

${n}_{\text{total" = "60.0 mL" + "240.0 mL" = "300.0 mL}}$

Therefore, the concentration of each ion in the final solution will be

$\left[\text{K"^(+)] = "0.177 moles"/(300.0 * 10^(-3)"L") = color(green)("0.59 M}\right)$

$\left[\text{PO"_4^(3-)] = "0.0270 moles"/(300.0 * 10^(-3)"L") = color(green)("0.090 M}\right)$

$\left[\text{SO"_3^(2-)] = "0.0480 moles"/(300.0 * 10^(-3)"L") = color(green)("0.16 M}\right)$

The answers are rounded to two sig figs.