Question

Construct a graph of ${\mathbf{pAg}}^{\mathbf{+}}$versus milliliters of ${\mathbf{Ag}}^{\mathbf{+}}$for the titration of $\mathbf{40}\mathbf{.}\mathbf{00}\mathbf{mL}$of solution containing $\mathbf{0}\mathbf{.}\mathbf{05000}{\mathbf{MBr}}^{\mathbf{-}}$and$\mathbf{0}\mathbf{.}\mathbf{05000}{\mathbf{MCI}}^{\mathbf{-}}$. The titrant is$\mathbf{0}\mathbf{.}\mathbf{08454}\mathbf{MAgNO}$. Calculate ${\mathbf{pAg}}^4$at the following volumes:

localid="1654845944495" $\mathbf{2}\mathbf{.}\mathbf{000}\mathbf{,}\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{22}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{23}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{24}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{30}\mathbf{.}\mathbf{00}\mathbf{,}\mathbf{40}\mathbf{.}\mathbf{00}\mathbf{mL}\mathbf{,}$second equivalence point,$\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{mL}$.

Short answer

The graph:

Step-by-step solution

Concept used.

The ratio of moles of solute (in grams) to the volume of the solution is known as molarity (in litres). The formula for calculating a solution's molarity is:

$\mathrm{Molarity}\left(\mathrm{M}\right)=\frac{\mathrm{MOLES}\mathrm{of}\mathrm{solute}\left(\mathrm{ing}\right)}{\mathrm{Volume}\mathrm{ofsolution}\left(\mathrm{in}\mathrm{L}.\right)}$

Derive Equation 10-11for potassium hydrogen phthalate.

Given,

$$\begin{gathered} 0.5000{\mathrm{MBr}}^{-} \\ 0.5000{\mathrm{MCI}}^{-} \\ 0.8454\mathrm{MAgNO}3 \end{gathered}$$

Moles of silver ion equals moles of bromide ions is the first equivalence point when silver bromide precipitates.

$$\begin{gathered} {\mathrm{V}}_{\mathrm{e}}\times 0.8454\mathrm{M}=0.040\mathrm{L}\times 0.0500\mathrm{M} \\ {\mathrm{V}}_{\mathrm{e}}=\frac{0.040\mathrm{L}\times 0.0500\mathrm{M}}{0.8454\mathrm{M}} \\ =0.023366\mathrm{L} \\ =23.66\mathrm{mL} \end{gathered}$$

Silver bromide is partially precipitated upto$23.66\mathrm{mL}$and more bromide ions are still in solution. When $2\mathrm{mL}$of silver ion is added, the concentration of silver ion is

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left|\mathrm{Br}\right|} \\ =\frac{5.0\times {10}^{-13}}{\left({\displaystyle \frac{2.86\mathrm{mL}-200\mathrm{mL}}{22.66\mathrm{mL}.}}\right)\left(0.05000\mathrm{M}\right)\left({\displaystyle \frac{4000\mathrm{mL}}{4200\mathrm{mL}.}}\right)} \\ =1.15\times {10}^{-11}\mathrm{M} \\ {\mathrm{pAgg}}^{+}=-\log \left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left[1.15\times {10}^{-11}\right] \\ =10.94 \end{gathered}$$

When $10\mathrm{mL}$of silver ion is added, the concentration of silver ion is

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left|{\mathrm{Br}}^{-1}\right|} \\ =\frac{5.0\times {10}^{-13}}{\left({\displaystyle \frac{2.66\mathrm{mL}-10.0\mathrm{mL}}{2.66\mathrm{mL}}}\right)\left(0.05000\mathrm{M}\right)\left({\displaystyle \frac{40.00\mathrm{mL}}{50.00\mathrm{mL}.}}\right)} \\ =2.88\times {10}^{-20}\mathrm{M} \\ {\mathrm{pAg}}^{+}=-\log \left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left[2.88\times {10}^{-20}\right] \\ =19.66 \end{gathered}$$

When $22\mathrm{mL}$of silver ion is added, the concentration of silver ion is

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left|\mathrm{Br}\right|} \\ =\frac{5.0\times {10}^{-13}}{\left({\displaystyle \frac{2.66\mathrm{mL}-2.0\mathrm{mL}.}{23.66\mathrm{mL}}}\right)\left(0.05000\mathrm{M}\right)\left({\displaystyle \frac{40.00\mathrm{mL}}{62.20\mathrm{mL}}}\right)} \\ =2.19\times {10}^{-10}\mathrm{M} \\ {\mathrm{pAg}}^{+}=-\log \left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left[2.19\times {10}^{-10}\right] \\ =9.66 \end{gathered}$$

When $23\mathrm{mL}$of silver ion is added, the concentration of silver ion is

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left[{\mathrm{Br}}^{\_}\right]} \\ =\frac{5.0\times {10}^{-13}}{\left({\displaystyle \frac{2,.66\mathrm{mL}-2.0.0\mathrm{mL}}{2.66\mathrm{mL}}}\right)\left(0.05000\mathrm{M}\right)\left({\displaystyle \frac{40.00\mathrm{mL}}{60.00\mathrm{mL}}}\right)} \\ =5.623\times {10}^{-10}\mathrm{M} \\ {\mathrm{pAg}}^{+}=-\log \left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left[5.623\times {10}^{-10}\right] \\ =9.25 \end{gathered}$$

Silver chloride starts to precipitate beyond first equivalence point.

When $24\mathrm{mL}$of silver ion is added, the concentration of silver ion is

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left[{\mathrm{CI}}^{-1}\right]} \\ =\frac{1.8\times {10}^{-10}}{\left({\displaystyle \frac{4732\mathrm{mL}-24.0\mathrm{LL}}{23.66\mathrm{LL}}}\right)\left(0.05000\mathrm{M}\right)\left({\displaystyle \frac{40.00\mathrm{mL}}{64.00\mathrm{mL}}}\right)} \\ =5.8\times {10}^{-9}\mathrm{M} \\ {\mathrm{pAg}}^{+}=-\log \left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left[5.8\times {10}^{-9}\mathrm{M}\right] \\ =8.23 \end{gathered}$$

When $30\mathrm{mL}$of silver ion is added, the concentration of silver ion is

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left|{\mathrm{CI}}^{-}\right|} \\ =\frac{1.8\times {10}^{-10}}{\left({\displaystyle \frac{8732\mathrm{mL}-3.0.0mL}{23.86mL}}\right)\left(0.05000\mathrm{M}\right)\left({\displaystyle \frac{40.00\mathrm{mL}}{70.06\mathrm{mL}}}\right)} \\ =8.51\times {10}^{-9}\mathrm{M} \\ \mathrm{pAg}=-\log \left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left[5.8\times {10}^{-9}\right] \\ =8.07 \end{gathered}$$

When $40\mathrm{mL}$of silver ion is added, the concentration of silver ion is

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{sp}}}{\left|{\mathrm{CI}}^{-}\right|} \\ =\frac{1.8\times {10}^{-10}}{\left({\displaystyle \frac{4732\mathrm{mL}-0.0.0mL}{21.66mL}}\right)\left(0.05000\mathrm{M}\right)\left({\displaystyle \frac{40.00\mathrm{mL}}{80.00\mathrm{mL}}}\right)} \\ =2.34\times {10}^{-8}\mathrm{M} \\ \mathrm{pAg}=-\log \left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left[2.34\times {10}^{-8}\right] \\ =7.63 \end{gathered}$$

The concentration of silver ion and chloride ion are equal at equivalence point.

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{CI}}^{-}\right]={\mathrm{x}}^2={\mathrm{K}}_{\mathrm{sp}}\mathrm{for}\mathrm{AgCI} \\ \left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{CI}}^{-}\right]={\mathrm{x}}^2=1.8\times {10}^{-10} \\ \left[{\mathrm{Ag}}^{+}\right]=134\times {10}^{-5} \\ {\mathrm{pAg}}^{+}=4.87 \end{gathered}$$

When $50\mathrm{mL}$of silver ion is added, there is only silver ions present in excess.

Volume of silver ions $=\left(60.00-47.32\right)=2.68\mathrm{mL}\mathrm{of}{\mathrm{Ag}}^{+}$

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\left(\frac{2.68mL}{90.00mL}\right)\left(0.08454M\right) \\ =2.5\times {10}^{-3}M \\ {\mathrm{pAg}}^{+}=-\log \left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left[2.5\times {10}^{-3}\right] \\ =2.60 \end{gathered}$$

The graph of $pA{g}^{+}Vs{V}_{A{g}^{+}}$is plotted using the above calculated $pA{g}^{+}$