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28 July, 08:42

How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solution using 100 mL of ethyl acetate per extraction if the partition coefficient is 8

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  1. 28 July, 10:54
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    4.94g of material

    Explanation:

    Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

    Kp = 8 = Concentration in Ethyl acetate / Concentration in water

    100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

    8 = X / 100mL / (5g-X) / 100mL

    Where X is the amount of material in grams that comes to the organic phase.

    8 = X / 100mL / (5g-X) / 100mL

    8 = 100X / (500-100X)

    4000 - 800X = 100X

    4000 = 900X

    4.44g = X

    Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.

    And will remain 5g-4.44g = 0.56g.

    In the second extraction:

    8 = X / 100mL / (0.56g-X) / 100mL

    8 = 100X / (56-100X)

    448 - 800X = 100X

    448 = 900X

    0.50g = X

    In the second extraction, you will extract 0.50g of material

    Thus, after the two extraction you will lost:

    4.44g + 0.50g = 4.94g of material
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