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

1 Answer
Mar 1, 2016

Answer:

#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"^(+)#