# What is the concentration (M) of sodium ions in 4.57 L of a 1.25 M Na_3PO_4 solution?

Mar 1, 2016

$1.03 \cdot {10}^{25} \setminus {\text{ions Na}}^{+}$

#### Explanation:

When asked to find the concentration of sodium ions in a given concentration, you need to first find the ratio of ions per molecule.

${\text{Na"_3"PO"_4 = "Na"^(+) + "Na"^(+) + "Na"^(+) + "PO}}_{4}^{3 -}$

We see that for every formula unit of ${\text{Na"_3"PO}}_{4}$, there are $3$ ions of ${\text{Na}}^{+}$.

Step 1

We are given $\text{4.57 L}$ of a $\text{1.25-M}$ (molarity) solution of ${\text{Na"_3"PO}}_{4}$. From the formula

$\text{molarity}$ $\left(c\right) = \left(\text{moles solute")/("1 L solution}\right)$

we can derive that

"1.25 M Na"_3"PO"_4 = ("1.25 mol")/("1 L solution")

Step 2

With the two pieces of information, we can solve straight across;

${\text{4.57 L solution" xx ("1.25 mol Na"_3"PO"_4)/("1 L solution") xx (6.022*10^23 \ "for. units Na"_3"PO"_4)/("1 mol Na"_3"PO"_4) xx ("3 ions Na"^(+))/("1 mol Na"_3"O"_4) = 1.03*10^25 \ "ions Na}}^{+}$