A weak organic acid (pKa = 4.82) is being studied as a potential broadleaf herbi
ID: 1005222 • Letter: A
Question
A weak organic acid (pKa = 4.82) is being studied as a potential broadleaf herbicide. One aspect of the study includes testing for its presence in water samples taken from a run-off of a field. The acid has a distribution ratio of 62 between water and chloroform at a pH of 3.42. a. Using 30.0 mL of chloroform, what fraction of the compound will be extracted from a 50.0 mL field sample at pH = 3.42? b. What fraction of the acid in the 50.0 mL sample would be extracted, again at pH 3.42, using three 10.0 mL portions of chloroform? c. If the pH of the solution was changed to 6.00, what fraction would be extracted using 30.0 mL of chloroform with a 50.0 mL field sample? d. Qualitatively, predict what effect a lower pH would have on the extraction.
Explanation / Answer
(a): Given Kd = 62
pH = 3.42
=> [H3O+] = 10-3.42 = 3.802x10-4 M
Ka of acid = 10-4.82 = 1.514x10-5
D = Kd x [H3O+] / ([H3O+] + Ka)
=> D = 62x3.802x10-4 M / (3.802x10-4 M + 1.514x10-5) = 59.626
Number of extractions, n = 1
Vaq = 50.0 mL , Vorg = 30.0 mL
Fraction of the acid present in organic phase after one extraction, (q)org
= DxVorg / (DxVorg + Vaq)
= 59.626x30.0 mL / (59.626x30.0 + 50.0)
= 0.973 (answer)
(b):
Fraction of the acid present in organic phase after one extraction, (q)org
= DxVorg / (DxVorg + Vaq)
= 59.626x10.0 mL / (59.626x10.0 + 50.0)
= 0.923 (answer)
(c): given pH = 6.00
=> [H3O+] = 10-6.00 = 1.0x10-6 M
Now
D = Kd x [H3O+] / ([H3O+] + Ka)
=> D = 62x1.0x10-6 M / (1.0x10-6 M + 1.514x10-5) = 3.841
Fraction of the acid present in organic phase after one extraction, (q)org
= DxVorg / (DxVorg + Vaq)
= 3.841x30.0 mL / (3.841x30.0 + 50.0)
= 0.697 (answer)
(d): The extraction to chloroform(organic medium) is higher at lower pH. As the pH increases the fraction that is extracted decreases.