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

1 Answer
Aug 8, 2016

Answer:

#["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, #"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.