# caoh2+h3po4

In this reaction, phosphoric acid, ${\text{H"_3"PO}}_{4}$, a weak acid, will react with calcium hydroxide, #"Ca"("OH")_2#, a strong base, to lớn 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.

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

So, your starting equation looks lượt thích this

${\text{H"_3"PO"_text(4(aq]) + "Ca"("OH")_text(2(s]) -> "Ca"_3("PO"_4)_text(2(s]) darr + "H"_2"O}}_{\textrm{\left(l\right]}}$

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

${\text{H"_3"PO"_text(4(aq]) -> 3"H"_text((aq])^(+) + "PO}}_{\textrm{4 \left(a q\right]}}^{3 -}$

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

${\text{Ca"("OH")_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"OH}}_{\textrm{\left(a q\right]}}^{-}$

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

${\text{Ca"_3("PO"_4)_text(2(s]) -> 3"Ca"_text((aq])^(2+) + 2"PO}}_{\textrm{4 \left(a q\right]}}^{3 -}$

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This means that the unbalanced chemical equation can be written as

$3 {\text{H"_text((aq])^(+) + "PO"_text(4(aq])^(3-) + "Ca"_text((aq])^(2+) + 2"OH"_text((aq])^(-) -> 3"Ca"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) + "H"_2"O}}_{\textrm{\left(l\right]}}$

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

So, to lớn get $3$ calcium ions on the reactants' side, you'd have to lớn multiply the calcium hydroxide by $\textcolor{b l u e}{3}$. Likewise, to lớn get $2$ phosphate ions on the reactants' side, you'd have to lớn multiply the phosphoric acid by $\textcolor{red}{2}$.

#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

$6 {\text{H"_text((aq])^(+) + 2"PO"_text(4(aq])^(3-) + 3"Ca"_text((aq])^(2+) + 6"OH"_text((aq])^(-) -> 3"Ca"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) + "H"_2"O}}_{\textrm{\left(l\right]}}$

Now it all comes down to lớn 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 $\textcolor{p u r p \le}{6}$ to lớn get

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$6 {\text{H"_text((aq])^(+) + 2"PO"_text(4(aq])^(3-) + 3"Ca"_text((aq])^(2+) + 6"OH"_text((aq])^(-) -> 3"Ca"_text((aq])^(2+) + 2"PO"_text(4(aq])^(3-) + color(purple)(6)"H"_2"O}}_{\textrm{\left(l\right]}}$

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

$\textcolor{red}{2} {\text{H"_3"PO"_text(4(aq]) + color(blue)(3)"Ca"("OH")_text(2(s]) -> "Ca"_3("PO"_4)_text(2(s]) darr + color(purple)(6)"H"_2"O}}_{\textrm{\left(l\right]}}$