What is the molarity of a Na3PO4 solution if adding AgNO3 to 50ml of it produces 2.05g Ag3PO4?
Answer:
Na3PO4 + 3AgNO3 = Ag3PO4 + 3NaNO3
net ionic is
PO4(3-) + 3Ag+ = Ag3PO4
(2.05g Ag3PO4)(1 mol Ag3PO4/ 418.58g Ag3PO4)
=
.00490 moles Ag3PO4
(.00490 moles Ag3PO4)( 1 mol Na3PO4/ 1 mol Ag3PO4)
(Get this from the reaction^^^)
=
.00490 moles Na3PO4
(.00490 moles Na3PO4)/(.050L) = .098M Na3PO4
Hope that helps
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