Answer:
[tex]2Al^{3+} (aq) +2PO_{4}^{3-} (aq) = 2AlPO_{4} (s)[/tex]
Explanation:
1. First, you want to break up the aqueous solutions into their respective ions (remember to write the charge and number of each element). Note that you can only do this to compounds marked as aqueous, so leave the solid be.
[tex]2Al^{3+} + 3SO_{4}^{2-} + 2K_{3}^{1+} + 2PO_{4}^{3-} = 2AlPO_{4} (s) + 3K_{2}^{1+} +3SO_{4}^{2-}[/tex]
(Technically, both Potassium ions would have a coefficient of 6. I am leaving them as they were originally just for ease)
2. Now, you want to get rid of compounds on each side that are the same. These are called spectator ions, and they are exactly as the name says: they simply "spectate" the reaction, meaning they can be removed from the complete ionic equation.
Both the sulfate and potassium ions would be considered spectator ions and would be canceled out. Once removed the equation would look like:
[tex]2Al^{3+} (aq) +2PO_{4}^{3-} (aq) = 2AlPO_{4} (s)[/tex]
3. The complete ionic equation would be:
[tex]2Al^{3+} (aq) +2PO_{4}^{3-} (aq) = 2AlPO_{4} (s)[/tex]
The complete ionic equation of the given reaction is 2Al³⁺ (aq) + 2PO³⁻₄ (aq) = 2AlPO₄ (S).
When a reaction is occurring in a solution and the representation of ions is required, ionic equations are developed. An ionic equation shows the substance present in the equation, whereas a net ionic equation shows the change that happens during the reaction.
The sulfate and the potassium ions are the spectator ions, so these ions are canceled out and the rest of the reaction is written as the ionic equation.
Thus, the complete ionic equation of the given reaction is 2Al³⁺ (aq) + 2PO³⁻₄ (aq) = 2AlPO₄ (S).
Learn more about ionic equations, here:
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