Write the complete ionic equation for the reaction of magnesium phosphate with lead (II) nitrate. Include which ions will be aqueous and which compound will precipitate. (Valences: Mg=+2, PO4=-3, Pb=+2, NO3=-1)
Someone may be under the false impression that Mg3(PO4)2 is soluble but it isn't. Pb3(PO4)2 is insoluble, too, and which predominates depends upon the solubility product of each.
To write the complete ionic equation for the reaction of magnesium phosphate (Mg3(PO4)2) with lead (II) nitrate (Pb(NO3)2), we first need to identify the ions present in each compound and then determine the products of the reaction.
The formula for magnesium phosphate is Mg3(PO4)2. When it dissolves in water, it will dissociate into its respective ions: 3 Mg2+ ions and 2 (PO4)3- ions.
The formula for lead (II) nitrate is Pb(NO3)2. When it dissolves in water, it will dissociate into its respective ions: 1 Pb2+ ion and 2 NO3- ions.
Now, let's determine the products of the reaction based on the valence of each ion.
The valence of Mg2+ is +2, and the valence of (PO4)3- is -3. Therefore, for the reaction to occur, we need three Mg2+ ions to combine with two (PO4)3- ions to balance the charges. So, the product will be:
3 Mg2+(aq) + 2 (PO4)3-(aq)
The valence of Pb2+ is +2, and the valence of NO3- is -1. Therefore, we need two NO3- ions to combine with one Pb2+ ion to balance the charges. So, the product will be:
1 Pb2+(aq) + 2 NO3-(aq)
Now, let's write the balanced complete ionic equation, including the state of each ion:
2 Pb2+(aq) + 6 NO3-(aq) + 3 Mg2+(aq) + 6 (PO4)3-(aq) ⟶ 2 Pb(PO4)3(s) + 6 NO3-(aq) + 3 Mg2+(aq)
In this equation, the ions NO3- remain aqueous and do not undergo any change, while the compound Pb(PO4)3 will precipitate as a solid.
Therefore, the complete ionic equation for the reaction of magnesium phosphate with lead (II) nitrate is:
2 Pb2+(aq) + 6 NO3-(aq) + 3 Mg2+(aq) + 6 (PO4)3-(aq) ⟶ 2 Pb(PO4)3(s) + 6 NO3-(aq) + 3 Mg2+(aq)
Note: The state (s) represents a solid precipitate, (aq) represents aqueous ions in solution.