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Posted by on Tuesday, October 11, 2011 at 12:00am.

You have a solution that contains 5.0 grams of oxalic acid dissolved in 150 mL of water. This solution is placed in a separatory funnel, and the 100 mL of diethyl ether is added to the separatory funnel and an extraction is performed. Use the following constants to calculate how much oxalic acid is in solvent after the extraction.
Solubility of oxalic acid in water= 9.5 g/100 mL
Solubility of oxalic acid in ether= 16.9 g/100 mL

  • Chemistry - , Tuesday, October 11, 2011 at 12:59am

    Ko/aq = 16.9/9.5 = ?
    How much is in WHICH solvent after the extraction.
    Let x = oxalic acid in water
    then 5.0-x = oxalic acid in ether.

    K = [(5.0-x)/100]/(x/150)
    Solve for x.

  • Chemistry - , Monday, October 28, 2013 at 6:34am

    1234

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