# 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)?

Feb 1, 2016

$6 \text{H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O} \left(l\right)$

OR

$\text{H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O} \left(l\right)$

#### Explanation:

$2 \text{H"_3"PO"_4(aq) + 3"Ca(OH)"_2(aq) -> "Ca"_3"(PO"_4")"_2(aq) + 6"H"_2"O}$

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

$6 \text{H"^(+)(aq) + 2"PO"_4^(3-)(aq) + 3"Ca"^(2+)(aq) + 6"OH"^(-)(aq) -> 3"Ca"^(2+)(aq) + 2"PO"_4^(3-)(aq) + 6"H"_2"O} \left(l\right)$

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

$6 \text{H"^(+)(aq) + cancel(2"PO"_4^(3-)(aq)) + cancel(3"Ca"^(2+)(aq)) + 6"OH"^(-)(aq) -> cancel(3"Ca"^(2+)(aq)) + cancel(2"PO"_4^(3-)(aq)) + 6"H"_2"O} \left(l\right)$

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

$6 \text{H"^(+)(aq) + 6"OH"^(-)(aq) -> 6"H"_2"O} \left(l\right)$

And by cancelling down coefficients we get:

$\text{H"^(+)(aq) + "OH"^(-)(aq) -> "H"_2"O} \left(l\right)$

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