Question

Fajans’ method is used to determine Chloride in salty water. A 10.00 mL of the solution is treated with 15.00 mL of a standard 0.1100 M AgNO3 solution. The blank titration is done, which need 0.50 mL of standard solution.

1. Show an example of indicator to determine end point and find out the PH requirement for the Fajans’ method.
2. Find out the concentration of chloride in sample solution.

Answer 1

Fajan's method of chloride analysis utilizes an adsorption indicator called dichlorofluorescein. During the titration of chloride ions containing sample with standard solution of silver nitrate, precipitation of silver chloride takes place. As the precipitation reaction approaches its end-point, the excess of silver ions in the solution gets adsorbed on to the AgCl surface. There, the indicator that exist as dianion, adsorbs alongwith and marks the end-point as the appearance of pinkish color. The indicator is a weakly basic molecule and thus works best in the alkaline pH range as 5 to 9. At lower pH than 5, H+ ions would react with the free indicator (a dianion) and will hinder its adsorption on the AgCl surface (near the equivalence point). And at higher pH (pH>9), OH- ions will interfere in the precipitation reaction and may form Ag(OH) precipitate. Thus pH range should be 5-9.

Titration reaction:

Cl- + Ag+ $\rightarrow$ AgCl(s)

1 mole of Ag+ reacts with 1 mole of Cl-

Cl- concentration in blank solution:

number of millimoles of Ag+ in 0.5 mL of 0.1100 M AgNO3 standard solution = Molarity * volume

millimoles of Ag+ = 0.1100*0.5 = 0.055 millimole

from the reaction millimoles of Ag+ = millimoles of Cl-

$\therefore$ millimoles of Cl- = 0.055 millimole

1 mole of Cl- = 35.5 g/mol

0.055 millimole of Cl- = 35.5* 0.055* 10-3 g

Cl- = 0.0019 g

Sample titration

number of millimoles of Ag+ in 15 mL of 0.1100 M AgNO3 standard solution = Molarity * volume

millimoles of Ag+ = 0.1100*15 = 1.65 millimole

from the reaction, millimoles of Ag+ = millimoles of Cl-

$\therefore$ millimoles of Cl- = 1.65 millimole

1 mole of Cl- = 35.5 g/mol

1.65 millimole of Cl- = 35.5* 1.65* 10-3 g

Cl- = 0.0586 g

Amount of chloride ion present in salty water = chloride amount in sample - chloride amount in blanck

= 0.0586 - 0.0019 = 0.0566 g

therefore Amount of chloride ion present in salty water = 0.0566 g