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 13.21P
Interpretation Introduction

Interpretation:

Value of pMn2+ for 0 mL, 20.0 mL, 40.0 mL, 49.0 mL, 49.9 mL, 50.0 mL, 50.1 mL, 55.0 mL and 60.0 mL of EDTA has to be determined. Also, titration curve for the same 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
Check Mark

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 7.41×1013 for Kf and 4.2×103 for αY4 in equation (1).

  Kf'=(4.2×103)(7.41×1013)=31.1×1010

At 0 mL of EDTA solution,

Expression to calculate pMn2+ is as follows:

  pMn2+=log[Mn2+]        (2)

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

  pMn2+=log(0.0200 M)=1.6981.70

When 0 mL of EDTA is added to Mn2+ solution, pMn2+ is 1.70.

At 20.0 mL of EDTA solution,

Expression to calculate total volume of solution is as follows:

  Total volume of solution=(Volume of Mn2++Volume of EDTA)        (3)

Substitute 25.0 mL for volume of Mn2+ 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 Mn2+ is as follows:

  Excess [Mn2+]=(([Mn2+])(Volume of Mn2+)([EDTA])(Volume of EDTA)Total volume of solution)        (4)

Substitute 25.0 mL for volume of Mn2+, 0.0200 M for [Mn2+], 20.0 mL for volume of EDTA, 0.0100 M for [EDTA] and 45 mL for total volume of solution in equation (4).

  Excess [Mn2+]=((0.0200 M)(25.0 mL)(0.0100 M)(20.0 mL)45 mL)=0.00667 M

Substitute 0.00667 M for [Mn2+] in equation (2).

  pMn2+=log(0.00667 M)=2.18

At 40.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Mn2+ 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 Mn2+, 0.0200 M for [Mn2+], 40.0 mL for volume of EDTA, 0.0100 M for [EDTA] and 65 mL for total volume of solution in equation (4).

  Excess [Mn2+]=((0.0200 M)(25.0 mL)(0.0100 M)(40.0 mL)65 mL)=0.00154 M

Substitute 0.00154 M for [Mn2+] in equation (2).

  pMn2+=log(0.00154 M)=2.81

At 49.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Mn2+ 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 Mn2+, 0.0200 M for [Mn2+], 49.0 mL for volume of EDTA, 0.0100 M for [EDTA] and 74 mL for total volume of solution in equation (4).

  Excess [Mn2+]=((0.0200 M)(25.0 mL)(0.0100 M)(49.0 mL)74 mL)=0.000135 M

Substitute 0.000135 M for [Mn2+] in equation (2).

  pMn2+=log(0.000135 M)=3.87

At 49.9 mL of EDTA solution,

Substitute 25.0 mL for volume of Mn2+ and 49.9 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+49.9) mL=74.9 mL

Substitute 25.0 mL for volume of Mn2+, 0.0200 M for [Mn2+], 49.9 mL for volume of EDTA, 0.0100 M for [EDTA] and 74.9 mL for total volume of solution in equation (4).

  Excess [Mn2+]=((0.0200 M)(25.0 mL)(0.0100 M)(49.9 mL)74.9 mL)=0.0000013 M

Substitute 0.0000013 M for [Mn2+] in equation (2).

  pMn2+=log(0.0000013 M)=5.89

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 0.0200 M for molarity and 25.0 mL for volume of solution in equation (6) to calculate millimoles of Mn2+.

  Millimoles of Mn2+=(0.0200 M)(25.0 mL)=0.5 mmol

Substitute 0.0100 M for molarity and 50.0 mL for volume of solution in equation (6) to calculate millimoles of EDTA.

  Millimoles of EDTA=(0.0100 M)(50.0 mL)=0.5 mmol

Substitute 25.0 mL for volume of Mn2+ 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:

  Mn2++Y2MnY2

Concentration of MnY2 at equivalence point is calculated as follows:

  [MnY2]=0.5 mmol75 mL=0.0067 M

Consider change in concentrations of ionic species to be negligible and amount of Mn2+ and EDTA to be x.

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

  Kf'=[MnY2][Mn2+][EDTA]        (7)

Substitute 0.0067 M for [MnY2], x for [Mn2+], x for [EDTA] and 31.1×1010 for Kf' in equation (7).

  31.1×1010=0.0067 M(x)(x)

Solve for x,

  x=1.27×107 M

Substitute 1.27×107 M for [Mn2+] in equation (2).

  pMn2+=log(1.27×107 M)=6.89

At 50.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Mn2+ and 50.1 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+50.1) mL=75.1 mL

Chemical reaction occurs as follows:

  Mn2++Y2MnY2

Concentration of MnY2 at equivalence point is calculated as follows:

  [MnY2]=0.5 mmol75.1 mL=0.00066 M

Concentration of excess EDTA is calculated as follows:

Excess [EDTA]=(0.0100 M)(0.1 mL)75.1 mL=1.33×105 M

Rearrange equation (7) for [Mn2+].

  [Mn2+]=[MnY2]Kf'[EDTA]        (8)

Substitute 0.00066 M for [MnY2], 1.33×105 M for [EDTA] and 31.1×1010 for Kf' in equation (8).

  [Mn2+]=0.00066 M(31.1×1010)(1.33×105 M)=1.60×1010 M

Substitute 1.60×1010 M for [Mn2+] in equation (2).

  pMn2+=log(1.60×1010 M)=9.80

At 55.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Mn2+ 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:

  Mn2++Y2MnY2

Concentration of MnY2 at equivalence point is calculated as follows:

  [MnY2]=0.5 mmol80 mL=6.25×103 M

Concentration of excess EDTA is calculated as follows:

  Excess [EDTA]=(0.0100 M)(0.1 mL)80 mL=6.25×104 M

Substitute 6.25×103 M for [MnY2], 6.25×104 M for [EDTA] and 31.1×1010 for Kf' in equation (8).

  [Mn2+]=6.25×103 M(31.1×1010)(6.25×104 M)=3.215×1011 M

Substitute 3.215×1011 M for [Mn2+] in equation (2).

  pMn2+=log(3.215×1011 M)=10.49

At 60.0 mL of EDTA solution,

Substitute 25.0 mL for volume of Mn2+ and 60.0 mL for volume of EDTA in equation (3).

  Total volume of solution=(25.0+60.1) mL=85 mL

Chemical reaction occurs as follows:

  Mn2++Y2MnY2

Concentration of MnY2 at equivalence point is calculated as follows:

  [MnY2]=0.5 mmol85 mL=5.88×103 M

Concentration of excess EDTA is calculated as follows:

  Excess [EDTA]=(0.0100 M)(0.1 mL)85 mL=1.18×103 M

Substitute 5.88×103 M for [MnY2], 1.18×103 M for [EDTA] and 31.1×1010 for Kf' in equation (8).

  [Mn2+]=5.88×103 M(31.1×1010)(1.18×103 M)=1.602×1011 M

Substitute 1.602×1011 M for [Mn2+] in equation (2).

  pMn2+=log(1.602×1011 M)=10.80

Titration curve for the same is drawn as follows:

Exploring Chemical Analysis, Chapter 13, Problem 13.21P

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