analytical chemistry ocw

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STK1094 Analytical Chemistry 1 Dayang Norafizan binti Awang Chee Faculty of Resource Science and Technology Universiti Malaysia Sarawak This OpenCourseWare@UNIMAS and its related course materials are licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

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Page 1: Analytical chemistry ocw

STK1094 Analytical Chemistry 1

Dayang Norafizan binti Awang Chee

Faculty of Resource Science and Technology

Universiti Malaysia Sarawak

This OpenCourseWare@UNIMAS and its related course materials are licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Page 2: Analytical chemistry ocw

Learning Objectives

At the end of this lesson, students should be able to:

Describe the difference between an β€œendpoint” and an

β€œequivalence point” in an acid-base titration

Identify the equivalence point in an acid-base titration

from the pH titration curve

Illustrate titration of a weak acid with a strong base.

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Defining Termsβ€’ Standard solution: A reagent of a known concentration which used in

the titrimetric analysis

β€’ Titration: This is performed by adding a standard solution from buretteor other liquid-dispensing device to a solution of the analyte until thepoint at which the reaction is believe to be completed

β€’ Equivalence point: Occurs in a titration at the point in which theamount of added titrant is chemically equivalent to the amount ofanalyte in a sample

β€’ Back-titration: This is a process in which an excess of the standardtitrant is added, and the amount of the excess is determined by backtitration with a second standard titrant.

In this instance, the equivalent point corresponds with the amount ofinitial titrant is chemically equivalent to the amount of analyte plus theamount of back titrant.

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Defining Termsβ€’ End point: The point in titration when a physical change occurs that is

associated with the condition of chemical equivalence

β€’ Indicators are used to give an observable physical change (end point)

or at near the equivalence point by adding them to the analyte. The

difference between end point and equivalence point should be very

small and this difference is referred to as titration error.

β€’ Titration error, Et

Et=Vep – Veq

Vep is the actual volume used to get to the end point.

Veq is the theoretical value of reagent required to reach the end point.

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Acid-Base Titrations

β€’ A quick and accurate method for determining acidic

or basic substances in many samples.

β€’ The titrant is typically a strong acid or base.

β€’ The sample species can be either a strong or weak

acid or base.

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Acid-Base Titration

Types of acid-base titrations:

1)strong acid – strong base titration

2)weak acid – strong base titration

3)strong acid – weak base titration

4) polyprotic acid – strong base titration

5) polybasic base – strong acid titration

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Strong Acid - Strong Base Titration

In strong acid – strong base titration, there are three

regions of the titration curve that represent different kinds

of calculations :

- before equivalence point

- at equivalence point

- after equivalence point

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Titration curve

Strong acid titrated with a strong base:

β€’ The net reaction is

H3O++ OH- β†’ 2H2O

β€’ Before the equivalence point, acid is present in excess

β€’ pH is determined by the concentration of excess HCl

[𝐻3𝑂+] =

π‘šπ‘šπ‘œπ‘™π‘’π‘  π‘Žπ‘π‘–π‘‘βˆ’π‘šπ‘šπ‘œπ‘™π‘’π‘  π‘π‘Žπ‘ π‘’

π‘‘π‘œπ‘‘π‘Žπ‘™ π‘£π‘œπ‘™π‘’π‘šπ‘’

8

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Strong acid titrated with a strong base

β€’ At equivalence point, moles of acid and moles of base are equal.

β€’ At equivalence point,

[H3O+] = [OH- ]

pKw = 14 = pH + pOH

pH = 7

β€’ So, the equivalence point for strong acid/base is always a pH=7

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Strong acid titrated with a strong baseOvertitration

Pass the equivalence point, we don’t have any acid remaining. All that we are doing

is diluting our titrant.

[π‘Άπ‘―βˆ’] = π’Žπ’Žπ’π’π’†π’” 𝒆𝒙𝒄𝒆𝒔𝒔

𝒕𝒐𝒕𝒂𝒍 π’—π’π’π’–π’Žπ’†

pH = 14-pOH

Eg. Construct a titration curve for the titration of 100 mL 0.1 M HCl with 0.1 M NaOH

1) Volume of NaOH needed to reach eq. point

Moles HCl = moles NaOH

VNaOH = 100.0 mL

2) Before addition of NaOH

pH = - log [0.1] = 1

3) After addition of 10mL NaOH

[𝐻3𝑂+] =

100π‘šπ‘™ 0.10𝑀 βˆ’(10π‘šπ‘™)(0.10𝑀)

100π‘šπ‘™+10π‘šπ‘™

= 0.082 M, pH= 1.09

10

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mLtitrant

Total mL [H3O+] pH

0 100 0.10 1.00

10 110 0.082 1.09

20 120 0.067 1.17

30 130 0.054 1.28

40 140 0.043 1.37

50 150 0.033 1.48

60 160 0.025 1.60

70 170 0.018 1.74

80 180 0.011 1.96

90 190 0.0053 2.28

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Titration curve

4) At equivalence point

β€’ Equivalence point, moles of HCl= moles of NaOH

β€’ Since neither is in excess, pH is determined by Kw

Kw = 1.00 x 10-14 =

[H3O+][OH-] = [H3O

+]2

[H3O+] = 1.00 x 10-7

pH= 7

β€’ Note that for the first 90 mL of titration, pH = 2.28

β€’ At eq. point, the pH value jump of 4.72 pH unit

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Titration curve

5) Overtitration

β€’ Account for the dilution of titrant

β€’ 10 mL overtitration

[OH-] = moles excess NaOH = MVNaOH- MVHCl

Vtotal

= (0.1M)(110mL)-(0.1M)(100mL)

210 mL

= 0.0048M

pOH = 2.32

pH = 14-2.32 = 11.68

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mL titrant

Total volume

[OH-] pH

110 210 0.0048 11.68

120 220 0.0091 11.96

130 230 0.013 12.11

140 240 0.017 12.23

150 250 0.020 12.30

160 260 0.023 12.36

170 270 0.026 12.41

180 280 0.029 12.46

190 290 0.031 12.49

200 300 0.033 12.52

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Titration curve

Titration of a strong base with a

strong acid:

β€’ If we plot pOH rather than pH,

the result still look identical.

β€’ Typically, we still plot pH verses

mL titrant, so the curve is

inverted.

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Titration of weak acids & weak bases with

strong titrant

β€’ Must concerned with conjugate acid/base pairs & their

equilibrium

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Titration of weak acids & weak bases with

strong titrantBefore titration:

β€’ If the sample is weak acid, then use

β€’ 𝐾𝐴= 𝐻3𝑂

+ [π΄βˆ’]

[𝐻𝐴]

β€’ [H3O+]=[A-]

β€’ Calculate the pH value

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β€’ If the sample is weak base, then use

β€’ 𝐾𝐡= π‘‚π»βˆ’ [π»π΄βˆ’]

[π΄βˆ’]

β€’ [OH-]=[HA]

β€’ Calculate the pH value = 14 - pOH

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Titration of weak acids & weak bases

with strong titrant

Before equivalence point:

β€’ Equilibrium expression used is the Henderson-Hasselbalch equation

β€’ Starting with an acid

β€’ pH= 𝑝𝐾𝐴 + log[π΄βˆ’]

[𝐻𝐴]

β€’ Starting with base

β€’ pH= 14 βˆ’ (𝑝𝐾𝐡 + log[𝐻𝐴]

[π΄βˆ’])

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Titration of weak acids & weak bases

with strong titrant

At equivalence point

β€’ All sample is converted to its conjugate form

β€’ If the sample was an acid-solve the pH using KB

relationship

β€’ If the sample was a base-solve the pH using KA

relationship

𝐾𝐡+ 𝐾𝐴= 14

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Page 20: Analytical chemistry ocw

Titration of weak acids & weak bases with

strong titrant

Overtitration:

β€’ Identical to strong acid/strong base example.

β€’ Need to account for the amount of excess titrant & how much it has been diluted.

β€’ Eg. 100 mL solution of 0.1 M benzoic acid is titrated with 0.1 M NaOH. Construct a titration curve.

For benzoic acid

Ka=6.31 x 10-5

pKa=4.20

1) Volume of NaOH needed to reach eq. point

Moles C5H6COOH = moles NaOH

VNaOH = 100.0 mL

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Page 21: Analytical chemistry ocw

Titration of weak acids & weak bases with

strong titrant

2) Before titration:

β€’ 𝐾𝐴= 𝐻3𝑂

+ [π΄βˆ’]

[𝐻𝐴]

β€’ [𝐻3𝑂+]= [π΄βˆ’]

β€’ Assume [A-] is negligible compared to [HA]

𝐾𝐴= 6.31 Γ— 10βˆ’5= π‘₯2

0.10

= 6.31 Γ— 10βˆ’5 (0.10)

= 0.025 M

pH= 2.60

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Titration of weak acids & weak bases with

strong titrant3) After addition of 10mL NaOH

Henderson-Hasselbalch equationpH = pKa + log [C5H6COO-]

[C5H6COOH]

[C5H6COOH] = moles unreacted C5H6COOH = MVC5H6COOH- MVNaOH

Vtotal Vtotal

= (0.1M)(100mL)-(0.1M)(10mL)

110 mL

= 0.082M

[C5H6COO-] = moles NaOH added = MVNaOH

Vtotal Vtotal

= (0.1M)(10mL)

110 mL

= 0.009M

pH = 4.2 + log (0.009/0.082) = 3.24

β€’ Calculate other point by repeating this process

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Titration of weak acids & weak bases with

strong titrant

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mLtitrant

pH

0 2.60

10 3.24

20 3.60

30 3.83

40 4.02

50 4.20

60 4.38

70 4.57

80 4.80

90 5.15

mL titrant

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Titration of weak acids & weak bases with

strong titrant

4) At equivalence point

100mL titrant:

β€’ All acid has been converted to its conjugate base – benzoate

β€’ Use KB relationship.

β€’ 𝐾𝐡= [π‘‚π»βˆ’][𝐻𝐴]

[π΄βˆ’]

β€’ 𝐾𝐡= πΎπ‘Š/𝐾𝐴= 1.58 Γ— 10βˆ’10

n benzoic acid = n NaOH

β€’ Predominate ion in solution is A-, which is a weak base

[A-] = moles acid/ total volume = 0.05M

β€’ We have diluted the sample & the total volume at this point is 200 mL.

β€’ We can assume that [benzoic acid] is negligible compared to [benzoate].

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Titration of weak acids & weak bases with

strong titrant

4) At equivalence point

C5H6COO- (aq) + H2O (l) OH- (aq) + C5H6COOH

𝐾𝑏= 1.58 Γ— 10βˆ’10 =π‘₯2

0.050

π‘₯ = (1.58 Γ— 10βˆ’10)(0.050)

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mL titrant

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Titration of weak acids or weak bases with

strong titrant

5) Overtitration

β€’ Need to account for the dilution of titrant.

β€’ Eg: 10 mL excess.

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Titration of weak acids or weak bases with

strong titrant

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mL titrant

Total volume

[OH-] pH

110 210 0.0048 11.68

120 220 0.0091 11.96

130 230 0.013 12.11

140 240 0.017 12.23

150 250 0.020 12.30

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Β© 2009, Prentice-Hall, Inc.

Titrations of Polyprotic Acids

When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.

Titration curve for the reaction of

50.0 mL of 0.10 M H3PO3 with 0.10 M NaOH

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Titrations of Polybasic Base

8-29

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Self-Reflection

β€’ What is the difference between end-point and equivalence point?

β€’ How to build the titration curve for strong acid/strong base with weak acid/weak base and vice versa?