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Chapter 12: Q8P-i (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

(i) For 60 mL the value of pMn2+is 10.82.

Step by step solution

01

Introduction

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

Titration鈥夆赌�Reaction:鈥�Mn2++EDTA鈬�MnY2鈭�Kf=1013.89At鈥夆赌夆��pH鈥夆赌夆��8鈥夆赌�伪Y4鈭�=4.2脳10鈭�3鈥�Table鈥�12鈭�1

02

Determine equilibrium constant

Kf'=伪Y4鈭�脳Kf=4.2脳10鈭�3脳1013.89=3.3脳1011

Equivalence point=50 mL

03

Determine the value of pMn2+

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

There is 10 mL of Excess EDTA

EDTA=鈥�1025+60脳0.01M=1.176脳10鈭�3MMnY2鈭�=鈥�2525+60脳0.02M=5.88脳10鈭�3M

Kf'=MnY2鈭�Mn2+EDTA3.3脳1011=5.88脳10鈭�3Mn2+1.176脳10鈭�3Mn2+=1.5脳10鈭�11M

Therefore, the value of pCu2+

pMn2+=鈭�logMn2+=鈭�log1.5脳10鈭�11=10.82

04

Graph

The titration curve combining all the data from SID 135385-12-8P-a to SID 135385-12-8P-i was obtained

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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

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饾啅ML尾1=[ML][M][L]M+2L饾啅ML2尾2=[ML2][M][L]2

Let 伪M be the fraction of metal in the form M, 伪ML be the fraction in the form ML, and 伪ML2be the fraction in the form ML2. Following the derivation in Section 12-5, you could show that these fractions are given by

role="math" localid="1667801924683" 伪M1=11+尾1[L]+尾2[L]2伪ML=尾1[L]1+尾1[L]+尾2[L]2伪ML2=尾2[L]21+尾1[L]+尾2[L]2

The concentrations of ML and ML2are

[ML]=伪MLCMVMVM+VL[ML2]=伪ML2CMVMVM+VL

because CMVMVM+VLis the total concentration of all metal in the solution. The mass balance for ligand is

[L]+[ML]+2[ML2]=CMVMVM+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

蠒=CLVLVM+VM=伪ML+2伪ML2+LCM1-LCL

A 1.000-mL sample of unknown containing Co2+ and Ni2+ was treated with 25.00 mL of 0.038 72 M EDTA. Back titration with 0.021 27 M Zn2+ at pH 5 required 23.54 mL to reach the xylenol orange end point. A 2.000-mL sample of unknown was passed through an ion-exchange column that retards Co2+ more than Ni2+. The Ni2+ that passed through the column was treated with 25.00 mL of 0.038 72 M EDTA and required 25.63 mL of 0.021 27 M Zn2+ for back titration. The Co2+ emerged from the column later. It, too, was treated with 25.00 mL of 0.038 72 M EDTA. How many milliliters of 0.021 27 M Zn2+ will be required for back titration?

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

Calculate pCu2+ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu2+ with 0.00100 M EDTA at pH 11.00 in a solution with [NH3] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL
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