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8 April, 15:54

David prepared the potassium phosphate solution by adding 46.3g to 250 mL of water. Kim needs 0.10M potassium phosphate solution. How many stock solutions does she need if she wants to mix 150 ml of diluent?

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  1. 8 April, 17:42
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    17.2mL

    Explanation:

    Step 1:

    Determination of the molarity of potassium phosphate K3PO4 solution.

    This is illustrated below:

    Mass of K3PO4 = 46.3g

    Molar Mass of K3PO4 = (39x3) + 31 + (16x4) = 212g/mol

    Mole of K3PO4 = ... ?

    Mole = Mass / Molar Mass

    Mole of K3PO4 = 46.3/212 = 0.218 mole

    Volume of water = 250mL = 250/1000 = 0.25L

    Molarity = mole / Volume

    Molarity of K3PO4 = 0.218/0.25

    Molarity of K3PO4 = 0.872M

    Step 2:

    Determination of the volume of the stock solution of K3PO4 needed.

    Molarity of stock solution (M1) = 0.872M

    Volume of stock solution needed (V1) = ... ?

    Molarity of diluted solution (M2) = 0.1M

    Volume of diluted solution (V2) = 150mL

    The volume of the stock solution needed can be obtained as follow:

    M1V1 = M2V2

    0.872 x V1 = 0.1 x 150

    Divide both side by 0.872

    V1 = (0.1 x 150) / 0.872

    V1 = 17.2mL

    Therefore, the volume of the stock solution needed is 17.2mL
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