What is the balanced net ionic equation for H_3PO_4(aq) + Ca(OH)_2(aq) -> Ca_3(PO_4)_2(aq) + H_2O(l)H3PO4(aq)+Ca(OH)2(aq)→Ca3(PO4)2(aq)+H2O(l)?

1 Answer
Owen Bell
Feb 1, 2016

6"H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O"(l)6H+(aq)+6OH−(aq)→6H2O(l)

OR

"H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O"(l)H+(aq)+OH−(aq)→H2O(l)

Explanation:

Firstly, balance your equation:

2"H"_3"PO"_4(aq) + 3"Ca(OH)"_2(aq) -> "Ca"_3"(PO"_4")"_2(aq) + 6"H"_2"O"2H3PO4(aq)+3Ca(OH)2(aq)→Ca3(PO4)2(aq)+6H2O

Then, split up your aqueous substances into their respective ions:

6"H"^(+)(aq) + 2"PO"_4^(3-)(aq) + 3"Ca"^(2+)(aq) + 6"OH"^(-)(aq) -> 3"Ca"^(2+)(aq) + 2"PO"_4^(3-)(aq) + 6"H"_2"O"(l)6H+(aq)+2PO3−4(aq)+3Ca2+(aq)+6OH−(aq)→3Ca2+(aq)+2PO3−4(aq)+6H2O(l)

Then we cancel ions that appear on both sides of the equation: these are spectator ions that do not participate in the reaction.

6"H"^(+)(aq) + cancel(2"PO"_4^(3-)(aq)) + cancel(3"Ca"^(2+)(aq)) + 6"OH"^(-)(aq) -> cancel(3"Ca"^(2+)(aq)) + cancel(2"PO"_4^(3-)(aq)) + 6"H"_2"O"(l)

And write out the new net ionic equation that excludes these ions:

6"H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O"(l)

And by cancelling down coefficients we get:

"H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O"(l)

Which is the balanced net ionic equation for any neutralisation reaction : the type of reaction we were dealing with in the first place.

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