91Ó°ÊÓ

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

Molarity $\left(M\right)=\frac{Molesofsolute\left(ing\right)}{Volumeofsolution(inL)}$

Here Given,

25.00 mL of 0.08230 M KI

39.00mLof 0.05110 M AgNO₃

The titration reaction of ${\mathrm{AgNO}}_3\mathrm{vs}\mathrm{Kl}$ is as follows

${\mathrm{Ag}}^{+}+{\mathrm{I}}^{-}→\mathrm{Agl}\left(\mathrm{s}\right)$

There are more moles of iodide in solution than silver ions before the equivalency threshold.

Unprecipitated iodide's strength is computed as

Moles of l = original moles of ${\mathrm{l}}^{-}$moles of ${\mathrm{Ag}}^{+}$added

$$\begin{gathered} =\left(0.025\mathrm{L}\right)\left(0.08230\mathrm{mol}/\mathrm{L}\right)-\left(0.039\mathrm{L}\right)\left(0.05110\mathrm{mol}/\mathrm{L}\right) \\ =0.0000646\mathrm{mol} \end{gathered}$$

Total volume of both solution is$0.064\mathrm{L}\left(25.00\mathrm{mL}+39.00\mathrm{mL}\right)$

The concentration of iodide ion is

$\left[{\mathrm{l}}^{-}\right]=\frac{0.0000646\mathrm{mol}}{0.064\mathrm{L}}=0.001009375\mathrm{M}$

The concentration of silver ion which is in equilibrium with iodide ion is

$$\begin{gathered} \left[A{g}^{+}\right]=\frac{K_{sp}}{\left[l^{-}\right]}=\frac{8.3×{10}^{-17}}{0.001009375} \\ =8.223×{10}^{-14}A{g}^{+} \\ =-{\log }_{10}\left[A{g}^{+}\right] \\ =-{\log }_{10}\left(8.223×{10}^{-14}\right) \\ =13.08 \end{gathered}$$

Here Given,

25.00 mL of 0.08230 M KI

39.00mLof 0.05110 M AgNO₃

The titration reaction of ${\mathrm{AgNO}}_3\mathrm{vs}\mathrm{Kl}$is as follows

${\mathrm{Ag}}^{+}+{\mathrm{I}}^{-}→\mathrm{Agl}\left(\mathrm{s}\right)$

The moles of iodide in solution equal the silver ions at the equivalency point.

The silver ion and iodide ion concentrations are computed as

$\left[{\mathrm{Ag}}^{+}\right]=\left[{\mathrm{I}}^{-}\right]$

$$\begin{gathered} \left[{\mathrm{Ag}}^{+}\right]=\left[{\mathrm{I}}^{-}\right]=K_{sp} \\ \left(x\right)\left(x\right)=8.223×{10}^{-17} \\ x=\sqrt{8.3×{10}^{-17}}=9.1×{10}^{-9} \\ pA{g}^{+}=-{\log }_{10}\left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left(9.1×{10}^{-9}\right)=8.04 \end{gathered}$$.

Here Given,

25.00 mL of 0.08230 M

KI39.00mLof 0.05110 M AgNO₃

The titration reaction of ${\mathrm{AgNO}}_3\mathrm{vs}\mathrm{Kl}$ is as follows

${\mathrm{Ag}}^{+}+{\mathrm{I}}^{-}→\mathrm{Agl}\left(\mathrm{s}\right)$

The volume of silver ions at equivalence point is

$$\begin{gathered} =\left(0.025\mathrm{L}\right)\left(0.08230\mathrm{mol}/\mathrm{L}\right)={\mathrm{V}}_{\mathrm{e}}×\left(0.05110\right) \\ {\mathrm{V}}_{\mathrm{e}}=\frac{0.00205}{0.05110} \\ {\mathrm{V}}_{\mathrm{e}}=40.26\mathrm{mL} \end{gathered}$$

At the equivalency point, the majority of the iodide ion precipitates. More moles of silver ions are present in solution after the equivalence point.

After the equivalence point, the volume of silver ion is 44.30-40.12=4.18 mL.

Moles of Ag⁺

$$\begin{gathered} =\left(0.00418\mathrm{L}\right)\left(0.08230\mathrm{mol}/\mathrm{L}\right) \\ =0.000344{\mathrm{molAg}}^{+}\left[{\mathrm{Ag}}^{+}\right] \\ =\left(0.000344\mathrm{mol}\right)/\left(0.06930\mathrm{L}\right) \\ =0.00496{\mathrm{pAg}}^{+} \\ =-{\log }_{10}\left[{\mathrm{Ag}}^{+}\right] \\ =-\log \left(0.00496\right) \\ =2.53 \end{gathered}$$

Thus, the equivalence point is 2.53.