Question

An organic acid was isolated and purified by recrys- tallization of its barium salt. To determine the equiva- lent weight of the acid, a 0.393-g sample of the salt was dissolved in about 100 mL of water. The solution was passed through a strong-acid ion-ex- change resin, and the column was then washed with water; the eluate and washings were titrated with 18.1 mL of 0.1006 M NaOH to a phenolphthalein end point. (a) Calculate the equivalent weight of the organic 24-12. acid.

Answer 1

Answer :

To determine equivalent weight of organic acid :

NaOH base used : 18.1 ml of 0.1006 M ( for NaOH base Molarity and normality are equal because it is monoacidic base)

So, Normality of NaOH = 0.1006 N , volume of NaOH = 18.1 * 10^-3 L

No. of gram equivalent of NaOH used in titration = Normality * Volume(L)

=> No. of gram equivalent of base = 0.1006 * 18.1 * 10^-3 = 1.86 * 10^-3

In the titration , no. of gram equivalent of acid should equal to no. of gram equivalent of base

when salt of organic acid dissolved in water , organic acid is produced due to salt Hydrolysis . Assuming complete salt Hydrolysis so that

No. Of gram equivalent of barium salt of acid = no. of gram equivalent of acid = no. of gram equivalent of base

No. of gram equivalent of barium salt of acid = mass taken / equivalent weight = 1.86 * 10^-3

Mass of barium salt of acid taken = 0.393 g

Equivalent weight of barium salt of acid = mass taken / no. of gram equivalent

Equivalent weight = 0.393 / (1.8 * 10^-3 ) = 0.2113 * 10³ = 211.3 g

So, equivalent weight of barium salt of acid = 211.3 g

Now, in acid barium should be replaced by H

Equivalent weight of Ba = 137.3 / 2 = 68.65 g

Equivalent weight of acid = equivalent weight of salt - equivalent weight of barium + equivalent weight of H

Equivalent weight of acid = 211.3 - 68.65 + 1 = 143.65 g

So, equivalent weight of organic acid = 143.65 g