13. A single extraction with 100 mL of CHCl3 extracts or removes 88.5% of the we

Question

13. A single extraction with 100 mL of CHCl3 extracts or removes 88.5% of the weak acid, HA, from 50 mL of aqueous solution. What is the partition coefficient, K, for HA at a pH 3.00? The Ka for HA is 3.5 x 10-5 C (a) 0.067 C (b) 3.98 (c) 3.85 7/30/2007 http://64.124.160.206/g/session dll?SESSION 0594219773010869&NAME; Marasinghe%2C+A 14. A single extraction with 100 mL of CHC, extracts or removes 88.5% of the weak acid, HA (Ka 3.5 x 10-5), from 50 mL of aqueous solution at a pH 3.00? If a student carries out the same extraction at a pH 7.00 and reports that fraction remaining in the aqueous phase is 0.01, is this result is reasonable? C (a) This result is reasonable. A is the principal species in solution, which is less soluble in the aqueous phase. Therefore, 11.5% should remain in aqueous phase. C (c) This result is not reasonable. A is the principal species in solution, which is more soluble in the aqueous phase. Therefore, >11.5% should remain in aqueous phase, not 1%.

Explanation / Answer

13. Partition function for the extraction of acid at the given pH would be,

(a) 0.067

14. This result in not resonable as,

(c)

15. number of extractions needed,

(c) 1

16. percent remained

%Extracted = 1 - [50/(50 + 4 x 25)]^4 x 100

% remained = 100 - %E

(c) 0.012 %

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