How do you determine the molar concentrations of #Na^+# and #PO_4^(3-)# in a 1.50 M #Na_3PO_4# solution?

1 Answer
Mar 11, 2017

Answer:

Here's how you can do that.

Explanation:

The trick here is to look at the chemical formula of the compound.

Trisodium phosphate is a soluble ionic compound that dissociates completely in aqueous solution to produce sodium cations, #"Na"^(+)#, and phosphate anions, #"PO"_4^(3-)#.

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

Notice that every formula unit of trisodium phosphate dissociates in aqueous solution to produce #color(red)(3)# sodium cations and #1# phosphate anion.

This implies that every mole of this salt will produce #color(red)(3)# moles of sodium cations and #1# mole of phosphate anions in aqueous solution.

Therefore, the concentrations of the two ions in solution will be

#["Na"^(+)] = color(red)(3) xx ["Na"_3"PO"_4]" "# and #" "["PO"_4^(3-)] = 1 xx ["Na"_3"PO"_4]#

In your case, this will produce

#["Na"^(+)] = color(red)(3) xx "1.50 M" = "4.50 M"#

#["PO"_4^(3-)] = 1 xx "1.50 M" = "1.50 M"#