How many moles of Na^+ ions are present in 275.0 mL of 0.35 M Na_3PO_4 solution?

1 Answer
May 28, 2017

n(Na^+)=3*n(Na_3PO_4)=0.29" mol"

Explanation:

The definition of concentration is

C=n/V

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 Na_3PO_4. I will round my answers to two significant figures, as that is the least amount given in the question.

n(Na_3PO_4)=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 Na_3PO_4 salt particle, there are three Na^+ ions, so we need to multiply the answer by three, giving

n(Na^+)=3*n(Na_3PO_4)=3*9.6*10^-2=0.29" mol"