H3po4 + ca(oh)2 = ca3(po4)2 + h2o

     

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 lượt thích 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 lớn 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 vì chưng the same for the insoluble salt.


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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 lớn multiplying the ionic compound as a whole.

So, khổng lồ get #3# calcium ions on the reactants" side, you"d have khổng lồ 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)# lớn 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>)#