Help with part c of this question. The pain reliever phenacetin is soluble in co

Question

Help with part c of this question.

The pain reliever phenacetin is soluble in cold water to the extent of 1.0 g/1310 mL and soluble in diethyl ether to the extent of 1.0 g/90 mL.

(a) Determine the approximate coefficient distribution for phenacetin in these solvent.
A) (1310 g/1310 mL)ether / (1.0 g/1310 mL)water = 1310

(b) If 50 mg of phenacetin were dissolved in 100 mL of water, how much ether would be required to extract 90% of the phenacetin in a single extraction?

B) 90% of the phenacetin = 45 mg

1315 = (0.045 g/x mL)ether / (0.005 g/100mL)water

        = 0.68 mL

(c) What percent of the phenacetin would be extracted from the aqueous solution in (b) by two 25-mL portions of ether?

Explanation / Answer

(a)
distribution coefficient in ether to water Kew = (1/90) ether /(1/1310) water = 14.56

(b) If 50 mg of phenacetin were dissolved in 100 mL of water, how much ether would be required

to extract 90% of the phenacetin in a single extraction?

total penacetin = 50mg
It is required to extract phenacetin in X ml of ether = 0.9*50 = 45mg
Remaining phenacetin in 100ml water = 50-45=5mg

Kew = (45/X) / (5/100) = 14.56
X = 45*100/(5*14.56) = 62ml

(c) What percent of the phenacetin would be extracted from the aqueous solution in (b) by two

25-mL portions of ether?

Extracting two times with 25mL of ether each time

First extraction:
Let W1 mg of penacetin extracted in 25ml ether,
Water will contain = (50 - W1)mg
Kew = 14.56 = (W1/25)/{(50-W1)/100}
penacetin extracted, W1 = 39.22mg

Second extraction:
Remaining total penacetin = 50-39.22 = 10.78mg

Let W2 mg of penacetin extracted in 25ml ether,
Water will contain = (10.78 - W2)mg
Kew = 14.56 = (W2/25)/{(10.78-W2)/100}
penacetin extracted, W2 = 8.46mg
Total penacetin extracted = W1+W2=39.22+8.46=47.7mg
percentage extracted = 47.7/50*100 = 95.4%

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