Question

Ba₃(PO₄)₂ (s) <---> 3Ba²⁺ (aq) + 2PO₄^3- (aq)

My question is, for Ba₃(PO₄)₂, how do you obtain 3Ba²⁺ + 2PO₄^3- ? Where are the "3" in front of the Ba and the "2" in front of the PO coming from? I know it has something to do with net ionic charges having balance out, but I am having a difficult time with this.

Also, for (PO₄)_{2 ...} is the subscript 2 having to be applied to both the P and the O₄? As in, am I supposed to take the charge on P, which is -3, and multiply that by 2? And am I supposed to take the charge on oxygen, which is -2, and multiply that by 8? (Taking the subscript 2 and multiplying it by the 4 to get 8).

Please explain as clearly as possible, I want to be sure that I can understand this.

Thank you!

Answer 1

Ba₃(PO₄)₂ (s) <---> 3Ba²⁺ (aq) + 2PO₄^3- (aq)

Writing the formulafor an ionic compound: we follow the following procedure.

Actually Ba₃(PO₄)2 is formed like this:

the ionic symbol of Barium is Ba+2    because it is a metal and it is in IIA group and it loses easily 2 electrons.

(PO4)3- is a compound ion, it exists as as one unit.

If you consider (PO4), the oxidation state of P-atom is +5 and the O-atom is -2.

Here 4 oxygen atoms are there, so the total oxidation number is ------    -8.

Now if we add +5 and -8 , the total is -3.

Hence the symbol of Phosphate ion is (PO₄)^-3 .

Now if we write the formula:

1. Ba⁺²                (PO₄)^-3

2. WRITE THE VALENCIES: WITHOUT THE CHARGES:

Ba²                  (PO₄)³

3. Cris-cross the valencies:

(Ba)₃ (PO₄)₂

Hence the formula is : Ba₃ (PO₄)₂

This formula represents 3 Ba ions and 2 phosphate ions in it.

So when it is dissolved in water , it forms 3 Ba ions and 2 phosphate ions .

Hope this is clear!