What is the concentration of sodium ions in a solution obtained by dissolving #"3.25 g"# of sodium phosphate, #"Na"_3"PO"_4#, in enough water to make #"250.0 mL"# of solution?

The idea here is that you need to use the fact that all the moles of sodium phosphate that you dissolve to make this solution will dissociate to produce sodium cations to calculate the concentration of the sodium cations.

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

Use the molar mass of sodium phosphate to calculate the number of moles of salt used to make this solution.

#3.25 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_3"PO"_4)/(163.9color(red)(cancel(color(black)("g")))) = "0.01983 moles Na"_3"PO"_4#

Now, notice that every #1# mole of sodium phosphate that you dissolve in water dissociates to produce #3# moles of sodium cations in aqueous solution.

This means that the number of moles of sodium cations present in this solution will be

#0.01983 color(red)(cancel(color(black)("moles Na"_3"PO"_4))) * "3 moles Na"^(+)/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = "0.05949 moles Na"^(+)#

Finally, to figure out the molarity of the sodium cations, you need to figure out the number of moles of sodium cations present in #10^3"mL" = "1 L"# of this solution.

To do that, you can use the fact that this sample has a volume of #"250.0 mL"#.

#10^3 color(red)(cancel(color(black)("mL solution"))) * "0.05949 moles Na"^(+)/(250.0color(red)(cancel(color(black)("mL solution")))) = "0.23796 moles Na"^(+)#

You can thus say that the molarity of the solution is equal to

#color(darkgreen)(ul(color(black)("molarity = 0.238 mol L"^(-1))))#

The answer is rounded to three sig figs, the number of sig figs you have for the mass of sodium phosphate.