How many grams of Na3PO4 will be needed to produce 375mL of a solution that has a concentration of Na+ ions of 1.40 M?

Answer 1

You can use regular stoichiometry to solve this—even if it looks like you can’t!

First, find the molar mass of Na₃PO₄: 163.94 g/mol (googled)

Then, on the left we start with the volume we want, and its units:

(375 mL)

We know that we’ll need this in Litre, because concentrations are in M, which is mol/L:

(375 mL)*(1 L /1000 mL)

Now that we’ve reached L, we need to convert it into concentration, so we use the second provided number:

(375 mL)*(1 L /1000 mL)*(1.4 mol Na⁺/ 1 L)

Then, we have to think about how many moles of Na+ ions are in 1 mole of Na3PO4... we can write this as:

3(Na⁺) + 1(PO₄^3-) so then we know that there are 3 moles of sodium, 1 mole of phosphate in 1 mole of sodium phosphate. Using this relationship, we add to our “stoichiometric” conversion:

(375 mL)*(1 L /1000 mL)*(1.4 mol Na⁺/ 1 L)*(1 mol Na₃PO₄/ 3 mol Na⁺)

Now, we still have to use the molar mass of Na3PO4, not only Na+, because when we’re weighing out our solid/powder, we haven’t changed the compound being dissolved (which is still Na3PO4), and we’ve already accounted for the Na+ ions by using the molar ratio (above), so finally, to get the # g of Na3PO4 needed, we have to convert from mol Na3PO4 to g Na3PO4 using the molar mass as a conversion factor:

(375 mL)*(1 L /1000 mL)*(1.4 mol Na+ 1 L)*(1 mol Na₃PO₄/ 3 mol Na⁺)*(163.94 g Na₃PO₄ /1 mol Na₃PO₄) = ...

Try it on your own, and see that the units cancel...if you still need help/just want the answer number, add a comment, and I’m happy to reply!