Our teacher gaves us the answers so that we could figure out the process. I am t
ID: 859395 • Letter: O
Question
Our teacher gaves us the answers so that we could figure out the process. I am terrible with figuring out the Kd of somethings.... can someone help me with how to get to the answers from the info given?
Five grams of compound A is dissolved in 90 mL of water. The distribution coefficient for compound A between hexanes and water is 5:
a) How much of compound A will be in the hexanes if you extract it from water one time with 90 mL of hexanes. What is the percent recovery? [answer: 4.2 g, 84%]
b) how much of compound A will be in the hexanes if you extract it from the water with threee sequential extractions using 30 mL of hexanes each time, and the combine with hexane extractions? What is the percent recovery? [answer: 4.74 g, 95%]
Explanation / Answer
X / 90 mL hexane
Kc = 5.0 = ????????????????--------------- where x is amount extracted with hexane
(5 ? X) / 90 mL water
so 25 - 5x = x
x = 4.16 g or 4.16 X 100 / 5 = 83.33%
2. first extraction
X / 30 mL hexane
Kc = 5.0 = ????????????????--------------- where x is amount extracted with hexane ( in first extraction)
(5 ? X) / 90 mL water
25 - 5x = 3x
x = 3.125 g ( first extraction)
unextracted amount is 1.875
second extraction1
X / 30 mL hexane
Kc = 5.0 = ????????????????--------------- where x is amount extracted with hexane in second extraction
(1.875 ? X) / 90 mL water
9.375 - 5x = 3x
x = 1.171 g ( in second extraction)
unextracted amount is 0.730
Third extraction
X / 30 mL hexane
Kc = 5.0 = ????????????????--------------- where x is amount extracted with hexane in third extraction
(0.730 ? X) / 90 mL water
3.65 - 5x = 3x
x = 0.456 g
Total amount extracted in three steps = 4.752g
% recovery = 4.752 / 5 = 95.04 %
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