Question

What volume of 0.00300 m phosphoric acid is required to neutralize 45.00 mL of 0.00150 M calcium hydroxide? 3Ca(OH)₂(aq) + 2H₃PO₄(aq) → 1Ca₃(PO₄)2(aq) + 6H₂O(l) (M = mol/L) (1000 mL = 1L)

Answer 1

To neutralize 45.00 mL of 0.00150 M calcium hydroxide, 15 mL of 0.00300 M phosphoric acid is required. This is calculated using the molar ratio from the provided chemical equation and the definition of molarity.

Step-by-step explanation:

The question asks for the volume of 0.00300 M phosphoric acid required to neutralize 45.00 mL of 0.00150 M calcium hydroxide. The chemical reaction shows that it takes 2 moles of phosphoric acid to neutralize 3 moles of calcium hydroxide. This gives a mole ratio of 2/3 or 0.67.

First, we need to determine the moles of calcium hydroxide in the 45.00 mL solution. We can use the formula number of moles = molarity x volume (L). Therefore, for calcium hydroxide, it's moles = 0.00150 M x 45.00 mL / 1000 = 0.0000675 mol. Since the mole ratio of phosphoric acid to calcium hydroxide is 2/3 or 0.67, we need 0.000045 mol (0.67 x 0.0000675) of phosphoric acid to neutralize it. The volume of this phosphoric acid solution can be calculated by dividing number of moles by its molarity, yielding 15 mL (0.000045 mol / 0.00300 M x 1000).

Learn more about Acid-Base Neutralization