Answer:
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×[tex]\frac{2molH_{3}PO_{4}}{3molCa(OH)_{2}}[/tex] = 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×[tex]\frac{1L}{0,0300mol}[/tex] = 8,33x10⁻³L ≡ 8,33mL of 0,0300M phosphoric acid
I hope it helps!
