Question

What is wrong with this procedure? According to Table 7-1, carbonate can be measured by a Volhard titration. Removal of the precipitate is required. To analyse an unknown solution of, ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$, l acidified the solution with freshly boiled and cooled ${\mathbf{HNO}}_{\mathbf{3}}$to give ,0.5 M ${\mathbf{HNO}}_{\mathbf{3}}$. Then I added excess standard, but no ${\mathbf{Ag}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$precipitate formed. What happened?

Short answer

The silver nitrate precipitate does not form because silver salts dissolve in acidic solutions ${{\mathrm{CO}}_3}^{-2},{\mathrm{C}}_2{{\mathrm{O}}_4}^{-2}$and ${{\mathrm{ASO}}_4}^{3-}$easily dissolves.

Step-by-step solution

Define Volhard method

The titration of silver with ${\mathrm{NH}}_4\mathrm{SCN}$using ferric alum as an indicator is an example of a titration in which a coloured substance is formed in the solution. During the titration, $\mathrm{AgSCN}$is generated, while excess ${\mathrm{NH}}_4\mathrm{SCN}$combines with Fe(III) to form deep red ${\left[\mathrm{FeSCN}\right]}^{2+}$The amount of thiocyanate required to produce a noticeable colour is quite minimal. As a result, the end point error is minimal, but the solution should be vigorously shaken at the end point because silver ions are absorbed on the precipitate and then desorbed.

Explain the reason behind why silver nitrate precipitate is not formed.

The reason why the silver nitrate precipitate does not form.

Volhard titration is performed in an acidic solution$0.2\mathrm{M}(\mathrm{nitric}\mathrm{acid}).$

It eliminates some of the interference from other titrations. Certain silver salts, such as, dissolve in acidic solutions.

${{\mathrm{CO}}_3}^{-2},{\mathrm{C}}_2{{\mathrm{O}}_4}^{2-}$and${{\mathrm{ASO}}_4}^{-3}$ easily dissolves.

Thus, silver nitrate precipitate does not form.