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

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

Sodium phosphate, #"Na"_3"PO"_4#, is indeed soluble in aqueous solution, which means that its constituent ions, the sodium cation, #"Na"^(+)#, and the phosphate anion, #"PO"_4^(3-)#, will dissociate completely.

You will have

#"Na"_ color(blue)(3) "PO"_ (4(aq)) -> color(blue)(3)"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)#

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

• three moles of sodium cations, #color(blue)(3) xx "Na"^(+)#
• one mole of phosphate anions, #1 xx "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, #"HPO"_4^(2-)#.

This means that the concentration of the phosphate anions will actually by slightly smaller than #"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.