Balance the chemical equation given below, and determine the number of milliliters of 0.0300m phosphoric acid required to neutralize 25.00ml of 0.0150m calcium hydroxide.

__ca(oh)2(aq) + __h3po4(aq) --> __ca3(po4)2(s) + __h2o(l)

3Ca(OH)₂(aq) + 2H₃PO₄(aq) → 1Ca₃(PO₄)₂(s) + 6H₂O(l)

8,33mL of 0,0300M phosphoric acid

Explanation:

For the reaction:

_Ca(OH)₂(aq) + _H₃PO₄(aq) → _Ca₃(PO₄)₂(s) + _H₂O(l)

You can see there are three moles of Ca formed in Ca₃(PO₄)₂(s), thus, you can write in Ca(OH)₂ and Ca₃(PO₄)₂(s),:

3Ca(OH)₂(aq) + _H₃PO₄(aq) → 1Ca₃(PO₄)₂(s) + _H₂O(l)

As 2 moles of PO₄²⁻ are produced:

3Ca(OH)₂(aq) + 2H₃PO₄(aq) → 1Ca₃(PO₄)₂(s) + _H₂O(l)

Now, you have 6 oxygens in Ca(OH)₂, that means:

3Ca(OH)₂(aq) + 2H₃PO₄(aq) → 1Ca₃(PO₄)₂(s) + 6H₂O(l)

Now, the reaction is balanced.

25,00mL of 0,0150M Ca(OH)₂ are:

0,02500L×0,0150mol/L = 3,750x10⁻⁴ moles of Ca(OH)₂.

Using the balanced equation (3 moles of Ca(OH)₂ reacts with 2 moles of H₃PO₄), moles of H₃PO₄ you need to neutralize the calcium hydroxide solution are:

3,750x10⁻⁴ moles of Ca(OH)₂× = 2,50x10⁻⁴ moles of H₃PO₄.

The volume of 0,0300M phosphoric acid you require to supply these moles is:

2,50x10⁻⁴ moles of H₃PO₄× = 8,33x10⁻³L ≡ 8,33mL of 0,0300M phosphoric acid

I hope it helps!