An aqueous of H3PO4 was prepared by dissolving 8.85 g in enough water to make 350.0 mL solution. Also, an aqueous solution of Ca (OH) 2 was prepared by dissolving 15.76 g of Ca (OH) 2 in enough water to make 550 mL solution. Calculate the volume of H3PO4 required to neutralize 25.0 mL of Ca (OH) 2

25.0 mL

Explanation:

1. Gather the information in one place.

MM: 98.00 74.09

2H3PO4 + 3Ca (OH) 2 → Ca3 (PO4) 2 + 6H2O

m/g: 8.85 15.76

V/mL: 350.0 550

2. Moles of H3PO4

n = 8.85 g * (1 mol/98.00 g) = 0.09031 mol H3PO4

3. Moles of Ca (OH) 2

n = 15.76 g * (1 mol/74.09 g) = 0.2126 mol Ca (OH) 2

4. Moles of Ca (OH) 2 in 25.0 mL Solution

n = 0.2126 mol * (25.0 mL/550 mL) = 0.009 663 mol Ca (OH) 2

5. Moles of H3PO4 needed

From the balanced equation, the molar ratio is 2 mol H3PO4: 3 mol Ca (OH) 2

n = 0.009 663 mol Ca (OH) 2 * (2 mol H3PO4/3 mol Ca (OH) 2)

= 0.006 442 mol H3PO4

6. Volume of H3PO4

V = 0.006 442 mol * (350.0 mL/0.09031 mol) = 25.0 mL H3PO4

It will take 25.0 mL of the H3PO4 solution to neutralize 25.0 mL of the Ca (OH) 2 solution.