Question

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

Answer 1

`1.03*10^25 \ "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.

`"Na"_3"PO"_4 = "Na"^(+) + "Na"^(+) + "Na"^(+) + "PO"_4^(3-)`

We see that for every formula unit of `"Na"_3"PO"_4`, there are `3` ions of `"Na"^(+)`.

Step 1

We are given `"4.57 L"` of a `"1.25-M"` (molarity) solution of `"Na"_3"PO"_4`. From the formula

`"molarity"` `(c) = ("moles solute")/("1 L solution")`

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;

`"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"^(+)`