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

May 28, 2017

$n \left(N {a}^{+}\right) = 3 \cdot n \left(N {a}_{3} P {O}_{4}\right) = 0.29 \text{ mol}$

#### Explanation:

The definition of concentration is

$C = \frac{n}{V}$

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.

$C = 0.35 \text{ M}$
$L = 275.0 \cdot {10}^{-} 3 \text{ L}$

Now rearrange the concentration formula to solve for the number of moles of $N {a}_{3} P {O}_{4}$. I will round my answers to two significant figures, as that is the least amount given in the question.

$n \left(N {a}_{3} P {O}_{4}\right) = C \cdot V = 0.35 \cdot 275.0 \cdot {10}^{-} 3 = 9.6 \cdot {10}^{-} 2 \text{ mol}$

The number of moles is a measure of how many particles there are. The chemical formula shows that for every $N {a}_{3} P {O}_{4}$ salt particle, there are three $N {a}^{+}$ ions, so we need to multiply the answer by three, giving

$n \left(N {a}^{+}\right) = 3 \cdot n \left(N {a}_{3} P {O}_{4}\right) = 3 \cdot 9.6 \cdot {10}^{-} 2 = 0.29 \text{ mol}$