Question

The text claims that precipitation of ${\mathbf{l}}^{\mathbf{-}}$is not complete before ${\mathbf{Cl}}^{\mathbf{-}}$begins to precipitate in the titration in Figure 7-4. Calculate the concentration${\mathbf{Ag}}^{+}$of at the equivalence point in the titration of ${\mathbf{l}}^{\mathbf{-}}$alone. Show that this concentration ${\mathbf{Ag}}^{+}$of ${\mathbf{Cl}}^{-}$will precipitate.

Short answer

The concentration of silver ion at equivalence point in given titration is$9.1\times {10}^{-9}\mathrm{M}$.

Step-by-step solution

Concept used. 

• In titration, the equivalence point is the point at which the quantity of standard solution and analysis becomes equal and neutralisation occurs.
• The end point is when the colour changes at a specific point in the titration.

The concentration of Ag+  at the equivalence point in the titration of  alone.

The first equivalence point where silver iodide precipitates first because

$$\begin{gathered} {\mathrm{V}}_{\mathrm{c}}\times 0.0845\mathrm{M}=0.040\mathrm{L}\times 0.0502\mathrm{M} \\ {\mathrm{V}}_{\mathrm{c}}=\frac{0.040\mathrm{L}\times 0.0502\mathrm{M}}{0.0845\mathrm{M}} \\ =0.023676\mathrm{L} \\ =23.76\mathrm{mL} \\ \mathrm{At}\mathrm{the}\mathrm{equivalence}\mathrm{point},\mathrm{the}\mathrm{concentrations}\mathrm{of}\mathrm{silver}\mathrm{and}\mathrm{iodide}\mathrm{ions}\mathrm{are}\mathrm{the}\mathrm{same}. \\ \left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{I}}^{-}\right]={\mathrm{x}}^2 \\ ={\mathrm{K}}_{\mathrm{sp}} \\ \left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]={\mathrm{x}}^2 \\ =8.3\times {10}^{-17} \\ \left[{\mathrm{Ag}}^{+}\right]=9.1\times {10}^{-9}\mathrm{M} \\ \mathrm{Q}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right] \\ =2.8\times {10}^{-10} \\ \mathrm{The}\mathrm{Q}\mathrm{value}\mathrm{which}\mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{solubility}\mathrm{product}\mathrm{constant}\mathrm{of} \\ \mathrm{silver}\mathrm{chloride}.\mathrm{Thus}\mathrm{silver}\mathrm{ions}\mathrm{can}\mathrm{precipitate}. \end{gathered}$$