Question

How would you balance: `H_3PO_(4(aq)) + Ca(OH)_(2(s)) -> Ca_3(PO_4)_(2(s)) + H_2O_((l))`?

Answer 1

Here's one method.

Explanation:

In this reaction, phosphoric acid, `"H"_3"PO"_4`, a weak acid, will react with calcium hydroxide, `"Ca"("OH")_2`, a strong base, to produce calcium phosphate, `"Ca"_3("PO"_4)_2`, an insoluble salt, and water.

Since you're dealing with the reaction between a weak acid and a strong base, you can say that this is a neutralization reaction.

Judging from the products of the reaction, you're dealing with a complete neutralization.

So, your starting equation looks like this

`"H"_3"PO"_text(4(aq]) + "Ca"("OH")_text(2(s]) -> "Ca"_3("PO"_4)_text(2(s]) darr + "H"_2"O"_text((l])`

A useful approach here will be to balance this equation by using ions. For a complete neutralization reaction, you can say that you'll have

`"H"_3"PO"_text(4(aq]) -> 3"H"_text((aq])^(+) + "PO"_text(4(aq])^(3-)`

Now, calcium hydroxide is not very soluble in aqueous solution. However, its solubility increases significantly in the presence of an acid, so you can say that

`"Ca"("OH")_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"OH"_text((aq])^(-)`

Now, for the purpose of balancing the equation, you can do the same for the insoluble salt. Keep in mind that it is not correct to represent an insoluble salt as ions in the balanced chemical equation.

`"Ca"_3("PO"_4)_text(2(s]) -> 3"Ca"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-)`

This means that the unbalanced chemical equation can be written as

`3"H"_text((aq])^(+) + "PO"_text(4(aq])^(3-) + "Ca"_text((aq])^(2+) + 2"OH"_text((aq])^(-) -> 3"Ca"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) + "H"_2"O"_text((l])`

Now all you have to do is balance out the ions. Keep in mind that multiplying ions is equivalent to multiplying the ionic compound as a whole.

So, to get `3` calcium ions on the reactants' side, you'd have to multiply the calcium hydroxide by `color(blue)(3)`. Likewise, to get `2` phosphate ions on the reactants' side, you'd have to multiply the phosphoric acid by `color(red)(2)`.

`color(red)(2) xx overbrace((3"H"_text((aq])^(+) + "PO"_text(4(aq])^(3-)))^(color(red)("phosphoric acid")) + color(blue)(3) xx overbrace(("Ca"_text((aq])^(2+) + 2"OH"_text((aq])^(-)))^(color(blue)("calcium hydroxide")) -> 3"Ca"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) + "H"_2"O"_text((l])`

This will give you

`6"H"_text((aq])^(+) + 2"PO"_text(4(aq])^(3-) + 3"Ca"_text((aq])^(2+) + 6"OH"_text((aq])^(-) -> 3"Ca"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) + "H"_2"O"_text((l])`

Now it all comes down to balancing the hydrogen and oxygen atoms. Since you now have `12` hydrogen atoms and `6` oxygen atoms on the reactants' side, multiply thew water molecule by `color(purple)(6)` to get

`6"H"_text((aq])^(+) + 2"PO"_text(4(aq])^(3-) + 3"Ca"_text((aq])^(2+) + 6"OH"_text((aq])^(-) -> 3"Ca"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) + color(purple)(6)"H"_2"O"_text((l])`

Finally, the balanced chemical equation for this neutralization reaction will be

`color(red)(2)"H"_3"PO"_text(4(aq]) + color(blue)(3)"Ca"("OH")_text(2(s]) -> "Ca"_3("PO"_4)_text(2(s]) darr + color(purple)(6)"H"_2"O"_text((l])`