# Write the net ionic equation for the following molecular equation. (Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)...

Write the net ionic equation for the following molecular equation.

(Use the solubility rules provided in the OWL Preparation Page to determine the solubility of compounds.)

3Pb(NO3)2(aq) + 2K3PO4(aq) -> Pb3(PO4)2(s)6KNO3(aq)

sharikendrick | Certified Educator

The balanced equation is:

`~3Pb(NO_3)_2` (aq) + `~2K_3PO_4` (aq) -> `~Pb_3(PO_4)_2` (s) + `~6KNO_3` (aq)

Solubility

We can determine the solubility of the compounds in this reaction two ways:

Use the state symbols:

• The state symbol "aq" indicates that the substance is soluble in water. Therefore, `~3Pb(NO_3)_2` , `~2K_3PO_4` , and `~6KNO_3` are soluble in water.
• The state symbol "s" indicates that the substance is a solid that is not soluble in water. Therefore, `~Pb_3(PO_4)_2`  is not soluble in water.

Use the solubility rules:

• "Nitrates: soluble ionic compounds" - `~3Pb(NO_3)_2 ` and `~6KNO_3` are soluble in water.
• "Phosphates: insoluble except w/ sodium, potassium, or ammonium" - `~2K_3PO_4 ` is soluble in water, `~Pb_3(PO_4)_2` is insoluble in water.

Net Ionic Equation

First, write the total ionic equation by separating all soluble substances into ions:

`~3Pb^2^+` + `~6NO_3^- ` + `~6K^+` + `~2PO_4^3^-` -> `~Pb_3(PO_4)_2` + `~6K^+` + `~6NO_3^-`

Now, write the net ionic equation by removing all of the ions that are exactly the same on both sides of the equation:

`3Pb^2^+` + `~2PO_4^3^-` -> `~Pb_3(PO_4)_2`