write below 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.

I understand how to balance the equation, but the example they give me is completely different from the question and I don't understand where I'm supposed to get my answer from. Can anyone help?

Mg3(PO4)2 + 3Pb(NO3)2 --> 3Mg(NO3)2 (s) + Pb3(PO4)2 (aq)

I think that's how you do it, just an ordinary double displacement reaction. To be honest, I just use my instinct to say the magnesium nitrate is solid and the other is aqueous... I can't remember, if only I have your text book with me.

Anyway, Good Luck!

Jake, you missed it on the products. Nitrates are soluble, but lead phosphate is not.

Mg3(PO4)2 + 3Pb(NO3)2 --> 3Mg(NO3)2 (aq) + Pb3(PO4)2 (s)

So the net ionic equation is...

2PO4-- + 3Pb++ >> Pb3(PO4)2 (s)

Here are a set of rules for solubility. Memorize them.

http://www.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html

Oh yeah, I forgot the NET IONIC thing :(. I took this inorganic class like 4 years ago.

To write the complete ionic equation for the reaction of magnesium phosphate with lead(II) nitrate, we first need to know the chemical formulas of these compounds:

- Magnesium phosphate has the formula Mg3(PO4)2.
- Lead(II) nitrate has the formula Pb(NO3)2.

Now let's break down each compound into its respective ions.

For magnesium phosphate:

- Magnesium ions (Mg^2+) come from the magnesium phosphate compound.
- Phosphate ions (PO4^3-) also come from the magnesium phosphate compound.

For lead(II) nitrate:

- Lead(II) ions (Pb^2+) come from the lead(II) nitrate compound.
- Nitrate ions (NO3^-) also come from the lead(II) nitrate compound.

Now, we can write the balanced chemical equation:

3Mg3(PO4)2(aq) + 2Pb(NO3)2(aq) → 6Mg^2+(aq) + 2PO4^3-(aq) + 2Pb^2+(aq) + 6NO3^-(aq)

To determine which ions are aqueous and which compound will precipitate, we need to consider the solubility rules.

- Magnesium compounds are generally soluble, so Mg^2+ and PO4^3- ions will stay in aqueous form.
- Lead compounds are generally insoluble, so Pb^2+ ions will form a precipitate.

The complete ionic equation, with the ions written in their aqueous forms, is:

6Mg^2+(aq) + 2PO4^3-(aq) + 2Pb^2+(aq) + 6NO3^-(aq) → 6Mg^2+(aq) + 2PO4^3-(aq) + 2Pb^2+(aq) + 6NO3^-(aq)

The net ionic equation, with the spectator ions (Mg^2+ and NO3^-) removed, simplifies to:

2PO4^3-(aq) + Pb^2+(aq) → Pb3(PO4)2(s)

Hence, lead(II) phosphate (Pb3(PO4)2) will be the precipitate in this reaction.