# How would you balance the following chemical equation: mercury (II) hydroxide + phosphoric acid --> mercury (II) phosphate + water?

Dec 15, 2015

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

#### Explanation:

$\text{Hg(OH)"_2 + "H"_3"PO"_4 -> "Hg"_3("PO"_4)_2 + "H"_2"O}$

Begin by fixing the obvious. We need 3 $\text{Hg}$ on the left side and we only have 1, so add a coefficient of 3 to ${\text{Hg(OH)}}_{2}$

$\text{3 Hg(OH)"_2 + "H"_3"PO"_4 -> "Hg"_3("PO"_4)_2 + "H"_2"O}$

This solves our mercury problem, now on to phosphate. There is one phosphate ion on the left and two on right, so add a coefficient of two to ${\text{H"_3"PO}}_{4}$

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

The only thing left to the balance is the water. On the left side, we have a total of 12 $\text{H}$ and 6 $\text{O}$ (not counting the phosphate). This gives us exactly 6 $\text{H"_2"O}$, so adding a coefficient of 6 will give us our final answer:

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

Always double check your answer to make sure all the elements balance. In this case, they do, so we are done.