Asked by sweetpea
what is the total ionic equation for
AlPO4(aq)+3NaOH(aq)=Al(OH)3(s)+ Na3PO4, I am not sure I am right
I have
Al+PO4(aq)+Na3+OH3() --> Al+OH3+Na3+PO4
but then everything crosses itself out and I don't have a net.
what am I doing wrong?
AlPO4(aq)+3NaOH(aq)=Al(OH)3(s)+ Na3PO4, I am not sure I am right
I have
Al+PO4(aq)+Na3+OH3() --> Al+OH3+Na3+PO4
but then everything crosses itself out and I don't have a net.
what am I doing wrong?
Answers
Answered by
DrBob222
It makes it more complicated but we really should include the charges. Your main problem is that Al(OH)3 is a solid and does NOT break out ions. If you included the phases it might have kept you from making that error.
Al^3+(aq) + PO4^3-(aq) + 3Na+(aq) + 3OH^-(aq) ==>Al(OH)3(s) + 3Na^+(aq) + PO4^3-(aq)
Now when you start crossing out you can't eliminate Al^3+ nor 3OH^- so the net ionic equation is
Al^3+(aq) + 3OH^-(aq) ==> Al(OH)3(s)
Have you memorized the simple solubility rules. That would have told you Al(OH)3 is insoluble. Here is a simplified set of solubility rules.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
Al^3+(aq) + PO4^3-(aq) + 3Na+(aq) + 3OH^-(aq) ==>Al(OH)3(s) + 3Na^+(aq) + PO4^3-(aq)
Now when you start crossing out you can't eliminate Al^3+ nor 3OH^- so the net ionic equation is
Al^3+(aq) + 3OH^-(aq) ==> Al(OH)3(s)
Have you memorized the simple solubility rules. That would have told you Al(OH)3 is insoluble. Here is a simplified set of solubility rules.
http://www.files.chem.vt.edu/RVGS/ACT/notes/solubility_rules.html
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