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9 March, 01:49

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

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  1. 9 March, 02:11
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    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.
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