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

Mar 11, 2017

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, ${\text{Na}}^{+}$, and phosphate anions, ${\text{PO}}_{4}^{3 -}$.

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

Notice that every formula unit of trisodium phosphate dissociates in aqueous solution to produce $\textcolor{red}{3}$ sodium cations and $1$ phosphate anion.

This implies that every mole of this salt will produce $\textcolor{red}{3}$ 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"