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Chapter 12: Q8P-e (page 283)

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

(e) For 49.9 mL the value ofpMn2+is 4.87.

Step by step solution

01

Introduction

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

Titration  Reaction: Mn2++EDTA⇌MnY2−Kf=1013.89At   pH   8  αY4−=4.2×10−3 Table 12−1

02

Determine equilibrium constant

Kf'=αY4−×Kf=4.2×10−3×1013.89=3.3×1011

03

Determine the value of pMn2+

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

=Fraction remaining × Initial concentration × Dilution factor

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

Mn2+=50−49.9500.02 M2525+49.9=1.34×10−5 M

Therefore, the value of pMn2+

pMn2+=−logMn2+=−log1.34×10−5=4.87

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Most popular questions from this chapter

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

State the purpose of an auxiliary complexing agent and give an example of its use.

Spreadsheet equation for formation of the complexes ML and ML2.Consider the titration of metal M (initial concentration = CM, initial volume = VM) with ligand L (concentration = CL, volume added = VL), which can form 1:1 and 2 : 1 complexes:

M+L⇋MLβ1=[ML][M][L]M+2L⇋ML2β2=[ML2][M][L]2

Let αM be the fraction of metal in the form M, αML be the fraction in the form ML, and αML2 be the fraction in the form ML2. Following the derivation in Section 12-5, you could show that these fractions are given by

αM=11+β1[L]+β2[L]2αML=(β1[L])1+β1[L]+β2[L]2αML2=β2[L]21+β1[L]+β2[L]2

The concentrations of ML and ML2 are

[ML]=αMLCMVMVM+VL[ML]=αML2CMVMVM+VL

becauseCMVMVM+VL is the total concentration of all metal in the solution. The mass balance for ligand is

[L]+[ML]+2[ML2]=CLVLVM+VL

By substituting expressions for [ML] and [ML2] into the mass balance, show that the master equation for a titration of metal by ligand is

ϕ=CLVLCMVM=αML+2αML2+LCM1-LCL

Find [Ca2+] in 0.10 M CaY2- at pH 8.00

Calculate pCu2+ (to the 0.01 decimal place) at each of the following points in the titration of 50.0 mL of 0.040 0 M EDTA with 0.080 0 M Cu (NO3)2 at pH 5.00: 0.1, 5.0, 10.0, 15.0, 20.0, 24.0, 25.0, 26.0, and 30.0 mL. Make a graph of pCu2+ versus volume of titrant.
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