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

Since you"re taking care of the reaction in between a weak acid and a solid base, you can say that this is a neutralization reaction.

Judging from the assets of the reaction, you"re taking care of a complete neutralization.

So, your starting equation looks prefer 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 technique here will be to balance this equation by making use of ions. Because that a complete neutralization reaction, you have the right to 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 an extremely soluble in aqueous solution. However, its solubility rises significantly in the existence of an acid, so you have the right to say that

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

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

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

This way that the unbalanced chemistry equation deserve to 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. Save in mind the multiplying ions is tantamount to multiplying the ionic compound as a whole.

So, to obtain #3# calcium ions on the reactants" side, you"d have to multiply the calcium hydroxide through #color(blue)(3)#.


You are watching: Calcium hydroxide and phosphoric acid react to form calcium phosphate and water


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Likewise, to gain #2# phosphate ions on the reactants" side, you"d have to multiply the phosphoric acid through #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 provide 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 every comes down to balancing the hydrogen and oxygen atoms. Since you now have actually #12# hydrogen atoms and also #6# oxygen atoms on the reactants" side, multiply thew water molecule by #color(purple)(6)# come 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 because that this neutralization reaction will certainly 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>)#