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

Expert verified

(f) For 50 mL the value of pMn2+is 6.85.

Step by step solution

01

Introduction

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

Titration  Reaction:Mn2++EDTAMnY2Kf=1013.89At   pH   8  αY4=4.2×103Table121

02

Determine equilibrium constant

Kf'=αY4×Kf=4.2×103×1013.89=3.3×1011

Equivalence point=50 mL

03

Determine the value of pMn2+

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

=Fraction remaining × Initial concentration × Dilution factor

Concentration of MnY2-

Mn2+=0.02M2525+50=0.00667M

Therefore, the value of pMn2+

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

Mn2++EDTAMnY2    x             x               0.00667x

Kf'=0.00667xx20.00667xx2=3.3×1011x=1.4×107MMn2+=1.4×107M

Therefore, the value of pMn2+

pMn2+=logMn2+=log1.4×107=6.85

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Most popular questions from this chapter

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