Question

Two chemicals are present in the same sample, where compound A is the analyte of interest and B is a possible interfering agent. It is known that A has a distribution ratio of 26.7 at pH = 7.0 when extracted from water with toluene. Compound B has a distribution coefficient ratio of 0.24 under the same conditions.

A) What percent of each compound will be extracted from a 100.0 mL sample of water when using 20.0 mL of toluene in a single-step extraction?

B) What percent of A and B will be extracted after two, three and four extractions if each extraction step uses a fresh 20.0 mL portion of toluene?

C) Describe how the recovery and purity of compound A will change as the

number of extractions increases in this sample.

Answer 1

A): Given the volume of water, Vw = 100.0 mL

Volume of Toluene, Vt = 20.0 mL

For compound A, partition coefficient, K = 26.7

For compound A, partition coefficient, K = 0.24

Since this is a single step extraction, n = 1

(B): For 2 extractions, n = 2

Similarly after 3 extractions we can calculate in a similar way

n = 3.

Hence percentage of A extracted = 99.2547 %

Percentage of B extracted = 47.933 %

After 4 extractions we can calculate in a similar way

n = 4.

Hence percentage of A extracted = 99.2544 %

Percentage of B extracted = 46.91 %

(C): From tha above calculated datas it is clear that, when we increase the number of extractions, the percentage extraction of compound A remains almost same. However the extraction of unwanted B decreases considerably from 49.83 % to 46.91 % .