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

Answer:

#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"#