What are the products of this reaction? What is the complete balanced equation? #?H_3PO_4(aq) + ?Ca(OH)_2(aq) ->#

Although phosphoric acid and calcium hydroxide might be less familiar examples, they are an acid and a base, and we know that these react to yield a salt and water.

#? H_3PO_4 (aq) + ? Ca(OH)_2(aq) → ? #'salt'# + ? H_2O#

The salt formed in this case is calcium phosphate, and since we know the charges on the ions are #Ca^(2+)# and #PO_4^(3-)#m, calcium phosphate has the formula #Ca_3(PO_4)_2#.

A more complete equation, then, is:

#? H_3PO_4 (aq)# + ? #Ca(OH)_2(aq)# → ? #Ca_3(PO_4)_2(s)# + #? H_2O(l)#

Now we just need to replace all those question marks!

In the discussion that follows, I will treat the #PO_4^3-# ion as a unit, rather than as separate phosphorus and oxygen atoms. This means I will have to be careful when counting oxygen.

One mole of #Ca# on the left yields #3# mole of #Ca# on the right. Place a #3# in front of the #Ca(OH)_2#.

#? H_3PO_4 (aq)# + #3Ca(OH)_2(aq)# → ? #Ca_3(PO_4)_2# + #? H_2O#

Now there are #3# moles of #Ca# on both sides.

One mole of #PO_4# on the left yields #2# mole of #PO_4# on the right, so add a #2# in front of the phosphoric acid:

#2 H_3PO_4 (aq)# + #3Ca(OH)_2(aq)# → ? #Ca_3(PO_4)_2# + #? H_2O#

Now there are #2# moles of #PO_4# on both sides.

Ignore the oxygen in the #PO_4# groups and look at the rest of the oxygen. There are 6 #O# on the left and only one on the right (in the #H_2O#), so place a #6# in front of the #H_2O# so that there will be 6 on each side:

#2 H_3PO_4 (aq)# + #3Ca(OH)_2(aq)# → ? #Ca_3(PO_4)_2# + #6 H_2O#

Finally, check the H. There are now 12 on the left and 12 on the right, so that's OK.

That means we can replace the final '?' with 1, but we don't usually bother writing 1 as a coefficient in chemical equations, so the final chemical equation is:

#2 H_3PO_4 (aq)# + #3Ca(OH)_2(aq)# → #Ca_3(PO_4)_2(s)# + #6 H_2O(l)#