Exploring Chemical Analysis
Exploring Chemical Analysis
5th Edition
ISBN: 9781429275033
Author: Daniel C. Harris
Publisher: Macmillan Higher Education
Question
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Chapter 13, Problem 13.24P
Interpretation Introduction

Interpretation:

Value of pCu2+ for 0 mL, 20.0 mL, 40.0 mL, 49.0 mL, 50.0 mL, 51.0 mL, and 55.0 mL of EDTA has to be determined. Also, graph of pCu2+ versus volume of titrant has to be drawn.

Concept Introduction:

EDTA or ethylenediaminetetraacetic acid is used for titration of majority of metals with help of formation of strong metal complexes in the ratio of 1:1. It is utilized to bind metals, most commonly used in industrial processes. It is also used for prevention of food oxidation and in environmental chemistry. It can be used either directly or indirectly for analysis of most of the elements present in periodic table. EDTA titrations can be performed in many ways. Some of these are mentioned below.

1. Direct titration

2. Back titration

3. Displacement titration

4. Masking

Expert Solution & Answer
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Explanation of Solution

Expression for Kf' is as follows:

  Kf'=(αY4)(Kf)        (1)

Here,

Kf' is conditional rate constant.

Kf is rate constant.

αY4 is fraction of free EDTA in form of Y4.

Substitute 6.31×1018 for Kf and 3.7×107 for αY4 in equation (1).

  Kf'=(6.31×1018)(3.7×107)=23347×108

At 0 mL of EDTA solution,

Expression to calculate pCu2+ is as follows:

  pCu2+=log[Cu2+]        (2)

Substitute 0.0200 M for [Mn2+] in equation (2).

  pCu2+=log(0.0800 M)=1.09

When 0 mL of EDTA is added to Cu2+ solution, pCu2+ is 1.09.

At 20.0 mL of EDTA solution,

Expression to calculate total volume of solution is as follows:

  Total volume of solution=(Volume of Cu2++Volume of EDTA)        (3)

Substitute 25.0 mL for volume of Cu2+ and 20.0 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+20.0) mL=45 mL

Since EDTA forms strong metal complexes in the ratio of 1:1, moles of EDTA will be equivalent to moles of metal reacted.

Expression to calculate excess concentration of Cu2+ is as follows:

  Excess [Cu2+]=(([Cu2+])(Volume of Cu2+)([EDTA])(Volume of EDTA)Total volume of solution)        (4)

Substitute 25.0 mL for volume of Cu2+, 0.0800 M for [Mn2+], 20.0 mL for volume of EDTA, 0.0400 M for [EDTA] and 45 mL for total volume of solution in equation (4).

  Excess [Cu2+]=((0.0800 M)(25.0 mL)(0.0400 M)(20.0 mL)45 mL)=0.02667 M

Substitute 0.02667 M for [Cu2+] in equation (2).

  pCu2+=log(0.02667 M)=1.57

At 40.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Cu2+ and 40.0 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+40.0) mL=65 mL

Substitute 25.0 mL for volume of Cu2+, 0.0800 M for [Cu2+], 40.0 mL for volume of EDTA, 0.0400 M for [EDTA] and 65 mL for total volume of solution in equation (4).

  Excess [Cu2+]=((0.0800 M)(25.0 mL)(0.0400 M)(40.0 mL)65 mL)=0.00615 M

Substitute 0.00615 M for [Cu2+] in equation (2).

  pCu2+=log(0.00615 M)=2.21

At 49.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Cu2+ and 49.0 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+49.0) mL=74 mL

Substitute 25.0 mL for volume of Cu2+, 0.0800 M for [Cu2+], 49.0 mL for volume of EDTA, 0.0400 M for [EDTA] and 74 mL for total volume of solution in equation (4).

  Excess [Cu2+]=((0.0800 M)(25.0 mL)(0.0400 M)(49.0 mL)74 mL)=0.000541 M

Substitute 0.000541 M for [Cu2+] in equation (2).

  pCu2+=log(0.000541 M)=3.27

At 50.0 mL of EDTA solution,

Formula to calculate molarity of solution is as follows:

  Molarity of solution(M)=Moles of soluteVolume (L) of solution        (5)

Rearrange equation (5) for moles of solute.

  Moles of solute=[(Molarity of solution)(Volume of solution)]        (6)

Substitute 0.0800 M for molarity and 25.0 mL for volume of solution in equation (6) to calculate millimoles of Cu2+.

  Millimoles of Cu2+=(0.0800 M)(25.0 mL)=2 mmol

Substitute 0.0400 M for molarity and 50.0 mL for volume of solution in equation (6) to calculate millimoles of EDTA.

  Millimoles of EDTA=(0.0400 M)(50.0 mL)=2 mmol

Substitute 25.0 mL for volume of Cu2+ and 50.0 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+50.0) mL=75 mL

Chemical reaction occurs as follows:

  Cu2++Y2CuY2

Concentration of CuY2 at equivalence point is calculated as follows:

  [CuY2]=2 mmol75 mL=0.0267 M

Consider change in concentrations of ionic species to be negligible and amount of Cu2+ and EDTA to be x.

Expression to calculate Kf' at equivalence point is as follows:

  Kf'=[CuY2][Cu2+][EDTA]        (7)

Substitute 0.0267 M for [CuY2], x for [Cu2+], x for [EDTA] and 23347×108 for Kf' in equation (7).

  23347×108=0.0267(x)(x)

Solve for x,

  x=1.0694×107 M

Substitute 1.0694×107 M for [Cu2+] in equation (2).

  pCu2+=log(1.0694×107 M)=6.97

At 51.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Cu2+ and 51.0 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+51.0) mL=76 mL

Chemical reaction occurs as follows:

  Cu2++Y2CuY2

Concentration of CuY2 at equivalence point is calculated as follows:

  [CuY2]=2 mmol76 mL=0.0263 M

Concentration of excess EDTA is calculated as follows:

  Excess [EDTA]=(0.0400 M)(1.0 mL)76 mL=0.000526 M

Rearrange equation (7) for [Cu2+].

  [Cu2+]=[CuY2]Kf'[EDTA]        (8)

Substitute 0.0263 M for [CuY2], 0.000526 M for [EDTA] and 23347×108 for Kf' in equation (8).

  [Cu2+]=0.0263 M(23347×108)(0.000526 M)=2.142×1011 M

Substitute 2.142×1011 M for [Cu2+] in equation (2).

  pCu2+=log(2.142×1011 M)=10.67

At 55.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Cu2+ and 55.0 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+55.1) mL=80 mL

Chemical reaction occurs as follows:

  Cu2++Y2CuY2

Concentration of CuY2 at equivalence point is calculated as follows:

  [CuY2]=2 mmol80 mL=0.025 M

Concentration of excess EDTA is calculated as follows:

  Excess [EDTA]=(0.0400 M)(5.0 mL)80 mL=0.0025 M

Substitute 0.025 M for [CuY2], 0.0025 M for [EDTA] and 23347×108 for Kf' in equation (8).

  [Cu2+]=0.025 M(23347×108)(0.0025 M)=4.283×1012 M

Substitute 4.283×1012 M for [Cu2+] in equation (2).

  pCu2+=log(4.283×1012 M)=11.37

Graph of pCu2+ versus volume of EDTA or titrant added is as follows:

Exploring Chemical Analysis, Chapter 13, Problem 13.24P

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