Question

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

Short answer

(c) For 40 mL the value ofpMn²⁺is 2.81.

Step-by-step solution

Introduction

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

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

Determine equilibrium constant

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

Equivalence point=50 mL

Determine the value of pMn2+

The concentration of the remaining product can be calculated using the following equation

=Fraction remaining × Initial concentration × Dilution factor

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

$\begin{array}{c}\left[M{n}^{2+}\right]=\left(\frac{50-40}{50}\right)\left(0.02M\right)\left(\frac{25}{25+40}\right)\\ =1.54\times {10}^{-3}M\end{array}$

Therefore, the value of pMn²⁺

^{$\begin{array}{c}pM{n}^{2+}=-\log \left(M{n}^{2+}\right)\\ =-\log \left(1.54\times {10}^{-3}\right)\\ =2.81\end{array}$}