# Question #d9b58

Jan 28, 2018

See below:

#### Explanation:

${H}_{3} P {O}_{4} \left(a q\right) + C a {\left(O H\right)}_{2} \left(a q\right) \rightarrow C {a}_{3} {\left(P {O}_{4}\right)}_{2} + {H}_{2} O \left(l\right)$

Balance the non-hydrogen atoms to get:

$2 {H}_{3} P {O}_{4} \left(a q\right) + 3 C a {\left(O H\right)}_{2} \left(a q\right) \rightarrow C {a}_{3} {\left(P {O}_{4}\right)}_{2} + {H}_{2} O \left(l\right)$

To balance the oxygen atoms, we need to add 5 water molecules to the right hand side of the equation:

$2 {H}_{3} P {O}_{4} \left(a q\right) + 3 C a {\left(O H\right)}_{2} \left(a q\right) \rightarrow C {a}_{3} {\left(P {O}_{4}\right)}_{2} + {H}_{2} O \left(l\right) + 5 {H}_{2} O$
You'll see now that the hydrogen atoms have automatically balanced our. So, you don't need to add any hydrogen ions here.

Thus, our final equation is:

$2 {H}_{3} P {O}_{4} \left(a q\right) + 3 C a {\left(O H\right)}_{2} \left(a q\right) \rightarrow C {a}_{3} {\left(P {O}_{4}\right)}_{2} + 6 {H}_{2} O$