How many moles of Na+ ions are present in 275.0 mL of 0.35 M Na3PO4 solution?

1 Answer
Peter M.
May 28, 2017

n(Na+)=3⋅n(Na3PO4)=0.29 mol

Explanation:

The definition of concentration is

C=nV

where concentration is is moles per litre (M), n is the number of moles (mol) and V is the volume in litres (L). Make sure to use these units to get the right answer.
I.e.

C=0.35 M
L=275.0⋅10−3 L

Now rearrange the concentration formula to solve for the number of moles of Na3PO4. I will round my answers to two significant figures, as that is the least amount given in the question.

n(Na3PO4)=C⋅V=0.35⋅275.0⋅10−3=9.6⋅10−2 mol

The number of moles is a measure of how many particles there are. The chemical formula shows that for every Na3PO4 salt particle, there are three Na+ ions, so we need to multiply the answer by three, giving

n(Na+)=3⋅n(Na3PO4)=3⋅9.6⋅10−2=0.29 mol

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