pAg in KI & AgNO3 Titration: Exploring 39mL & 44.30mL Volumes

Step 1 of 3

(a) When 39 mL of AgNO3 is added

Reaction for the titration

\(\mathrm{KI}(\mathrm{aq})+\mathrm{AgNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{AgI}(\mathrm{s})+\mathrm{KNO}_{3}(\mathrm{aq})\)

Calculating the concentration of unprecipitated \(\mathrm{I}^{-}\) 

\(\begin{aligned}
\text { Moles of } \mathrm{I}^{-} & =\text {Original moles of } \mathrm{I}^{-}-\text {Moles of } \mathrm{Ag}^{+} \text {added } \\
& =(0.025 \mathrm{~L})(0.08230 \mathrm{~mol} / \mathrm{L})-(0.039 \mathrm{~L})(0.05110 \mathrm{~mol} / \mathrm{L}) \\
& =0.0020575 \mathrm{~mol}-0.0019929 \mathrm{~mol} \\
& =0.0000646 \mathrm{~mol}
\end{aligned}\) 

The total volume of the solution

\(\begin{aligned}
V & =(0.025+0.039) \mathrm{L} \\
& =0.064 \mathrm{~L}
\end{aligned}\) 

The concentration of \(\mathrm{I}^{-}\) will be

\(\begin{aligned}
{\left[\mathrm{I}^{-}\right] } & =\frac{0.0000646 \mathrm{~mol}}{0.064 \mathrm{~L}} \\
& =0.001 \mathrm{~mol} / \mathrm{L}
\end{aligned}\) 

Ksp for AgI will be

\(\mathrm{Ksp}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right]\) 

Concentration of \(\mathrm{Ag}^{+}\) 

\(\begin{aligned}
{\left[\mathrm{Ag}^{+}\right] } & =\frac{\mathrm{Ksp}}{\left[\mathrm{I}^{-}\right]} \\
& =\frac{8.3 \times 10^{-17}}{0.01} \\
& =8.3 \times 10^{-15} \mathrm{M}
\end{aligned}\) 

The p function for \(\mathrm{Ag}^{+}\) will be

\(\begin{aligned}
\mathrm{pAg}^{+} & =-\log \left(\mathrm{Ag}^{+}\right) \\
& =-\log \left(8.3 \times 10^{-15}\right) \\
& =14.08
\end{aligned}\)