# What is the concentration of all ions present in .300 mole of sodium phosphate in 600.0 mL of solution?

Aug 8, 2016

["Na"^(+)] = "1.50 mol L"^(-1)

["PO"_4^(3-)] = "0.500 mol L"^(-1)

#### Explanation:

This problem wants to test your understanding of what happens when you dissolve soluble ionic compounds in water.

Sodium phosphate, ${\text{Na"_3"PO}}_{4}$, is indeed soluble in aqueous solution, which means that its constituent ions, the sodium cation, ${\text{Na}}^{+}$, and the phosphate anion, ${\text{PO}}_{4}^{3 -}$, will dissociate completely.

You will have

${\text{Na"_ color(blue)(3) "PO"_ (4(aq)) -> color(blue)(3)"Na"_ ((aq))^(+) + "PO}}_{4 \left(a q\right)}^{3 -}$

Notice that every mole of sodium phosphate that dissolves in solution produces

• three moles of sodium cations, $\textcolor{b l u e}{3} \times {\text{Na}}^{+}$
• one mole of phosphate anions, $1 \times {\text{PO}}_{4}^{3 -}$

In your case, $0.300$ moles of sodium phosphate will produce

0.300 color(red)(cancel(color(black)("moles Na"_3"PO"_4))) * (color(blue)(3)color(white)(a)"moles Na"^(+))/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = "0.900 moles Na"^(+)

and

0.300 color(red)(cancel(color(black)("moles Na"_3"PO"_4))) * "1 mole PO"_4^(3-)/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = "0.300 moles PO"_4^(3-)

To find the concentration of the ions, use the volume of the solution, but not before converting it to liters. You will have

["Na"^(+)] = "0.900 moles"/(600.0 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)color(black)("1.50 mol L"^(-1))color(white)(a/a)|)))

["PO"_4^(3-)] = "0.300 moles"/(600.0 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.500 mol L"^(-1))color(white)(a/a)|)))

The answers are rounded to three sig figs.

SIDE NOTE It's worth mentioning that the phosphate anion acts as a base in aqueous solution. The anion accepts a proton from water to form hydrogen phosphate, ${\text{HPO}}_{4}^{2 -}$.

This means that the concentration of the phosphate anions will actually by slightly smaller than ${\text{0.500 mol L}}^{- 1}$, since some of the anions will be protonated.

However, the actual concentration of phosphate anions is well beyond the scope of your question.