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...
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`
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