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The monochloroacetic acid (ClCH 2 COOH) preservative in 100.

The monochloroacetic acid (ClCH

2

COOH) preservative in 100.0 mL of a carbonated

beverage was extracted into diethyl ether and then returned to aqueous solution as

ClCH

3

COO

-

by extraction with 1M NaOH. This aqueous extract was acidified and

treated with 50.00 mL of 0.04521 M AgNO

3

. The reaction is


ClCH

2

COOH + Ag

+

+ H

2

O

HOCH

2

COOH + H

+

+ AgCl(s)


After filtration of the AgCl, titration of the filtrate and washings required 10.43 mL of an

NH

4

SCN solution. Titration of a blank taken through the entire process used 22.98 mL of

the NH

4

SCN. Calculate the weight (in milligrams) of ClCH

2

COOH in the sample

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The weight of ClCH 2... View the full answer

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