2. Given a solution of 10 mg of traumatic acid dissolved in 100 mL of water, and

Question

2. Given a solution of 10 mg of traumatic acid dissolved in 100 mL of water, and using the partition coefficient given for traumatic acid in question 1, show that extracting the 100 mL of aqueous solution with two 10 mL portions of MtBE would recover more of the traumatic acid from the aqueous solution than one extraction using 20 mL of MtBE. From question 1, we are told the partition coefficient is 5.8.

Explanation / Answer

Let UNITs guide you; always USE THEM in your calculation to prevent errors Firstly, the eqn: D = (g/100ml organic)/g/100ml water) = 5.8 single extraction -----> 5.8 = (x g / 20ml)/((10-x)g / 100ml H2O) __________________5.8 *((10-x) /100 ml H2O) = (x g / 20 ml) __________________58/100 = x/20 + 5.8x/100 __________________58 = 5x + 5.8x = 10.8x __________________x = 58/10.8 g/20ml = 5.37 g/20ml double extraction ------> 5.8 = (y1 g /10ml)/((10-y1)g /100 ml H2O) ___________________5.8 *((10-y1) /100 ml H2O) = (y1 g / 10 ml) ___________________58/100 = y1/10 + 5.8y1/100 ___________________58 = 10y1 + 5.8y1 = 15.8y1 ___________________y1 = 58/15.8 g/10ml = 3.67 g/10ml ____________________5.8 = (y2 g/10ml) / ((10-3.67 -y2)g / 100ml H2O) ___________________5.8 *((6.33-y2) /100 ml H2O) = (y2 g / 10 ml) ___________________(5.8*6.33)/100 = y2/10 + 5.8y2/100 ___________________36.7 = 10y2 + 5.8y2 = 15.8y2 ___________________y2 = 367/15.8 g/10ml = 2.32 g/10ml y1 + y2 = ?? ( compare to single extraction value)

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