Consider the titration of 25.0 mL of 0.020 0 M MnSO₄ with 0.010 0 M EDTA in a solution buffered to pH 8.00. Calculate pMn²⁺ at the following volumes of added EDTA and sketch the titration curve:

(a) 0 mL (b) 20.0 mL (c) 40.0 mL (d) 49.0 mL (e) 49.9 mL (f) 50.0 mL (g) 50.1 mL

(h) 55.0 mL (i) 60.0 mL

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

$\begin{array}{c}Titration\mathrm{Re}action:M{n}^{2+}+EDTA\rightleftharpoons Mn{Y}^{2-}\\ K_f={10}^{13.89}\\ AtpH8{\alpha }_{Y^{4-}}=4.2\times {10}^{-3}\left(Table12-1\right)\end{array}$

$\begin{array}{c}{K}_f^{\text{'}}={\alpha }_{Y^{4-}}\times K_f\\ =4.2\times {10}^{-3}\times {10}^{13.89}\\ =3.3\times {10}^{11}\end{array}$

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

=Fraction remaining × Initial concentration × Dilution factor

If 0 mL solution is added then metal (Mn) concentration will be C_Mn²⁺ =0.02 M

$\begin{array}{c}\left[M{n}^{2+}\right]=C_{M{n}^{2+}}\\ =0.02M\\ \end{array}$

Therefore, the value of pMn²⁺

^{$\begin{array}{c}pM{n}^{2+}=-\log \left(M{n}^{2+}\right)\\ =-\log \left(0.02\right)\\ =1.7\end{array}$}

If 20 mL solution is added then the reaction will be 20/50 completed as the equivalence point is at 50 mL. Then the metal (Mn) concentration will be

Therefore, the value of pMn²⁺

If 40 mL solution is added then the reaction will be 40/50 completed as the equivalence point is at 50 mL. Then the metal (Mn) concentration will be

Therefore, the value of pMn²⁺

If 49 mL solution is added then the reaction will be 49/50 completed as the equivalence point is at 50 mL. Then the metal (Mn) concentration will be

Therefore, the value of pMn²⁺

If 49 .9mL solution is added then the reaction will be 49.9/50 completed as the equivalence point is at 50 mL. Then the metal (Mn) concentration will be

Therefore, the value of pMn²⁺

Concentration of MnY^2-

Therefore, the value of pMn²⁺

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

Therefore, the value of pMn²⁺

Past equivalence point (50.1mL) we need to calculate the metal concentration

There is 0.1 mL of Excess EDTA

Past equivalence point (55mL) we need to calculate the metal concentration

There is 5 mL of Excess EDTA

Therefore, the value of pCu²⁺

Past equivalence point (60mL) we need to calculate the metal concentration

There is 10 mL of Excess EDTA

Therefore, the value of pCu²⁺