Question

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

Answer 1

`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"`