91Ó°ÊÓ

Figure 7.4 shows the titration of potassium halogenides with ${\mathrm{Ag}}^{+}{\mathrm{NO}}^3$. We notice the curve has two equivalence points when a mixture of two ions is titrated, the less soluble compounds will precipitate first. So, as AgI is significantly less soluble AgCl. AgI will precipitate first.

(i)Equivalence point is the middle of the vertical part of the titration curve. In equivalence point, the solution contains equivalent amonts of reactant. So, before the first equivalence point, the solution will contain more iodide and chloride ions than ${\mathrm{Ag}}^{+}$ions.

(ii)we’ve said before that in the equivalent point the solution contains chemically equal amounts of reactants. We’ve also mentionrd that first compound to precipitate will be AgI. So, in the first equivalence point, solution will contain chemically equal amouunts of ${\mathrm{I}}^{-}$and${\mathrm{Ag}}^{+}$.

(iii) After the first equivalence point iodide ions have precipitate whereas chloride ions are still in solution as AgI is significant less soluble than AgCl. By adding more ${\mathrm{AgNO}}_3$, Chloride ions begins to precipitate as AgI and chloride ions begin to precipitate as AgCl.

In the second equivalence point, the solution contains chemically equal amount of ${\mathrm{Ag}}^{+}$and role="math" localid="1655025030506" ${\mathrm{Cl}}^{-}$ions.

After the second equivalence point , the concentration of chloride ions keeps dropping until all chloride ions precipitate as AgCl.

When it comes to calculating $\left[{\mathrm{Ag}}^{+}\right]$, as the ions are not in the solution in equilibrium, the concentration of silver ions is close to zero throughout the titration. The reason is quite simple – as long as there are iodide and chloride ions in the solution, ${\mathrm{Ag}}^{+}$will precipitate as $\mathrm{Agl}$or $\mathrm{AglCl}$. However as it is not in the fact zero, just very low the concentration can be calculated using the following formula:

$\left[{\mathrm{Ag}}^{+}\right]=\mathrm{c}\left({\mathrm{Ag}}^{+}\right).\frac{\mathrm{V}{\left({\mathrm{Ag}}^{+}\right)}_{\mathrm{excess}}}{\mathrm{V}\left(\mathrm{solution}\right)}$