In precipitation titrations of halides by the ion pair$\mathbf{AgX}\left(\mathrm{aq}\right)\left(\mathrm{CI},\mathrm{Br},I\right)$is in equilibrium with the precipitate. Use Appendix J to find the concentrations of $\mathbf{AgCI}\left(\mathrm{aq}\right)\mathbf{,}\mathbf{AgBr}\left(\mathrm{aq}\right)$and $\mathbf{Agl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}$during the precipitations.

Precipitation titration is a type of titration that involves precipitation at the end of the process. The precipitation method is used to titrate the majority of metallic halides. The argentometric titration method is also known as argentometric titration. In precipitation titration, there are three ways for determining the end point.

Find the logarithm of the formation.

$$\begin{gathered} M\mathit{\left(}\mathrm{aq}\right)+L\mathit{\left(}aq\right)\leftrightarrow ML\mathit{\left(}aq\mathit{\right)} \\ A{g}^{\mathit{+}}\left(aq\right)\mathit{+}C{l}^{\mathit{-}}\mathit{\to }AgCl\mathit{\left(}aq\mathit{\right)}\mathit{;}\log \mathit{\left(}{K}_{\mathit{F}}\mathit{\right)}\mathit{=}3.31 \\ {\mathrm{Ag}}^{\mathit{+}}\mathit{\left(}\mathit{a}\mathit{q}\mathit{\right)}\mathit{+}B{r}^{\mathit{-}}\mathit{\to }\mathrm{AgBr}\mathit{\left(}\mathrm{aq}\mathit{\right)}\mathit{;}\log \mathit{\left(}{\mathrm{K}}_{\mathrm{F}}\mathit{\right)}\mathit{=}\mathit{4}\mathit{.}\mathit{46} \\ A{g}^{\mathit{+}}\mathit{\left(}\mathit{a}\mathit{q}\mathit{\right)}\mathit{+}{l}^{\mathit{-}}\mathit{\to }Agl\mathit{\left(}aq\mathit{\right)}\mathit{;}log\mathit{\left(}{K}_{\mathit{F}}\mathit{\right)}\mathit{=}\mathit{6}\mathit{.}\mathit{6} \\ \mathrm{Now}\mathrm{formation}\mathrm{constant}\mathrm{is}\mathrm{given}\mathrm{as}{K}_{\mathit{F}}\mathit{=}\frac{\left[ML\right]}{\left[M\right]\left[L\right]} \\ \mathit{log}\mathit{\left(}{k}_{\mathit{F}}\mathit{\right)}\mathit{=}\mathit{log}\left[ML\right]\mathit{-}\mathit{log}\left(\left[M\right]\right)\mathit{-}\mathit{log}\left(\left[L\right]\right) \end{gathered}$$

$$\begin{gathered} \left[ML\right]=\left[M\right] \\ =\left[L\right]K_F=\frac{1}{\left[ML\right]}\log \left(K_F\right) \\ =-\log \left[ML\right]\left(ML\right)={10}^{-\log (k_{f)}} \\ \mathrm{calculate}\mathrm{the}\mathrm{concentrations}\mathrm{of}\mathrm{AgCl},\mathrm{AgBr}\mathrm{and}\mathrm{Agl}.\mathrm{AgCl} \\ \mathit{(}AgCl\mathit{,}aq\mathit{)}\mathit{=}{10}^{\mathit{-}\mathit{l}\mathit{o}\mathit{g}\mathit{(}{\mathit{k}}_{\mathit{f}\mathit{)}}}c\mathit{(}AgCl\mathit{,}aq\mathit{)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}{10}^{\mathit{-}\mathit{l}\mathit{o}\mathit{g}\mathit{(}\mathit{3}\mathit{.}\mathit{31}\mathit{)}}c\mathit{(}AgCl\mathit{,}aq\mathit{)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=4.9.{10}^{-4}\mathrm{mol}/\mathrm{L} \\ \mathrm{AgBr} \\ \mathit{(}AgBr\mathit{,}aq\mathit{)}\mathit{=}{10}^{\mathit{-}\log \mathit{(}{\mathrm{k}}_{\mathrm{f}\mathit{)}}}\mathrm{c}(AgBr\mathit{,}aq) \\ ={10}^{-\log (4.6)}\mathrm{c}\mathit{(}AgBr\mathit{,}aq\mathit{)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=2.5.{10}^{-5}\mathrm{mol}/\mathrm{L} \\ \mathrm{Agl}: \\ \mathit{(}Agl\mathit{,}aq\mathit{)}\mathit{=}{10}^{\mathit{-}\log \mathit{(}{\mathrm{k}}_{\mathrm{f}\mathit{)}}}\mathrm{c}\mathit{(}\mathrm{Agl}\mathit{,}\mathrm{aq}\mathit{)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=2.5.{10}^{-7}\mathrm{mol}/\mathrm{L} \\ \mathrm{c}\mathit{(}\mathrm{AgBr}\mathit{,}\mathrm{aq}\mathit{)}\mathit{=}4.9.{10}^{-4}\mathrm{mol}/\mathrm{L} \\ \mathrm{Thus}\mathrm{the}\mathrm{formation}\mathrm{constant}\mathrm{for}\mathrm{the}\mathrm{following}\mathrm{are}c\mathit{(}AgBr\mathit{,}aq\mathit{)}=2.5.{10}^{-5}\mathrm{mol}/\mathrm{L} \\ \mathrm{c}\mathit{(}\mathrm{AgBr}\mathit{,}\mathrm{aq}\mathit{)}\mathit{=}2.5.{10}^{-7}\mathrm{mol}/\mathrm{L} \end{gathered}$$