Question

In a Mohr titration of chloride with silver nitrate, an error is made while preparing the indicator. Instead of 0.011 M chromate indicator in the titration flask at the endpoint, there is only 0.0011 M. If the flask contains 100 mL at the end point, what is the error in the titration in milliliters of 0.100 M silver nitrate?

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Answer 1

Ans: There will not be any error in the above titration. It is explained below:

Indicators are used to show the end point by color change as soon as any excess titratnt is added. In Mohr titration chromate ions are used to show the end point by forming the brick red colored silver chromate.

The Ag+ first reacts with all the available Cl- ions as per the below reaction:

Ag+ (aq.) + Cl– (aq.) --> AgCl (s) (white ppt.)

After all the Cl- ions are reacted and formed the white ppt., the excess Ag+ ions when added to the titration flask (as AgNO3 solution) react with the indicator chromate as per the below reaction and immediately forms the brick red silver chromate precipitate. From the below balanced reaction it is seen that 2 moles of silver ions reacts with the 1 Mole of chromate ions to give the silver chromate precipitate (ppt).

2Ag + (aq.) + CrO42– (aq.) --> Ag2CrO4(s) (brick red color)

The concentration of indicator used is 0.0011 M. The volume present in the titration flask is 100 mL at the end point.

Therefore, even if one mL of excess silver nitrate is used after the end point, which will again be diluted to final 101 mL (100 mL + 1mL) and the efective molarity of silver nitrate in the final volume would be

= 0.100 / 101 = 0.001 M

This will need = 0.001/2 = 0.0005 M chromate solution which is already in present in the flask to show the brick red precipitae.

Therefore, we will not get any error in the titration in milliliters of 0.100 M silver nitrate