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Problem help!

  1. Feb 15, 2005 #1
    Problem:

    The distribution coefficient, k = (conc. in ligroin/ conc. in water), between ligroin and water for solute A is 7.5 . What weight of A would be removed from a solution of 10g of A in 100mL of water by a single extraction with 100 mL of ligroin? What weight of A would be removed by four successive extractions with 25mL portions of ligroin? How much ligroin would be required to remove 98.5% of A in a single extraction?

    I'm having so much trouble with this problem. I plugged in the numbers giving to find the first part. After calculations, I found out that depending on the way you divide and multiply to find the weight removed, I'm assuming that 750 grams of A is removed. However, I'm not sure, not even sure if I'm doing this right. ANY SUGGESTIONS PLEASE?!?!
     
  2. jcsd
  3. Feb 15, 2005 #2

    chem_tr

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    I remember that a formula called Jensen's formula is used to calculate this kind of problems. It involves logarithmic expressions. Please try to find it by yourself, as I lost the source. However, it will surely help you.

    Best wishes.
     
  4. Feb 15, 2005 #3

    Gokul43201

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    You have 10g of A in 100mL water + 100 mL ligroin.

    Let the amount of A in water be x. Then the amount in ligroin would be 7.5x (since the volume of ligroin is the same as that of water). The total amount, 8.5x is equal to 10g. This tells you what x is. And x is the required number since it the the weight of A left in water. Another important number is the fraction of A extracted, which in this case is 1/8.5

    Can you try the second part, where the volume of water is different from the volume of ligroin ?
     
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