91Ó°ÊÓ

Equations and data obtained in order to proceed for calculation are as follows

$\begin{array}{c}Titration\text{ â¶Ä‰}\mathrm{Re}action:\text{ }M{n}^{2+}+EDTA⇌Mn{Y}^{2−}\\ K_f={10}^{13.89}\\ At\text{ â¶Ä‰â€‰}pH\text{ â¶Ä‰â€‰}8\text{ â¶Ä‰}{\mathrm{Î\pm }}_{Y^{4−}}=4.2×{10}^{−3}\text{ }\left(Table\text{ }12−1\right)\end{array}$

$\begin{array}{c}{K}_f^{\text{'}}={\mathrm{Î\pm }}_{Y^{4−}}×K_f\\ =4.2×{10}^{−3}×{10}^{13.89}\\ =3.3×{10}^{11}\end{array}$

Equivalence point=50 mL

The concentration of the remaining productcan be calculated using the following equation

=Fraction remaining × Initial concentration × Dilution factor

Concentration of MnY^2-

^{$\begin{array}{c}\left[M{n}^{2+}\right]=\left(0.02\text{ }M\right)\left(\frac{25}{25+50}\right)\\ =0.00667\text{ }M\end{array}$}

Therefore, the value of pMn²⁺

At equivalence point (50mL) the following can be written

$\begin{array}{l}M{n}^{2+}+EDTA⇌Mn{Y}^{2−}\\ \text{ â¶Ä‰â€‰â¶Ä‰}x\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰}x\text{ â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰â¶Ä‰â€‰}0.00667−x\end{array}$

$\begin{array}{c}{K}_f^{\text{'}}=\frac{0.00667−x}{x^2}\\ \frac{0.00667−x}{x^2}=3.3×{10}^{11}\\ x=1.4×{10}^{−7}M\\ M{n}^{2+}=1.4×{10}^{−7}M\end{array}$

Therefore, the value of pMn²⁺

^{$\begin{array}{c}pM{n}^{2+}=−\log \left(M{n}^{2+}\right)\\ =−\log \left(1.4×{10}^{−7}\right)\\ =6.85\end{array}$}