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VOLUMETRIC ANALYSIS Volumetric analysis is an analysis in which the amount of the unknown is calculated from a known volume of added solution.

VOLUMETRIC ANALYSIS Volumetric analysis is an analysis in which the amount of the unknown is calculated from a known volume of added solution

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VOLUMETRIC ANALYSIS

Volumetric analysis is an analysis in which the amount of the unknown is calculated

from a known volume of added solution.

TITRATIONS

In a titration, the titrant (solution with a known concentration of a reagent) area added to analyte until their reaction is complete.

(a) In a titration, the titrant must react with the analyte to completion, with a reproducible stoichiometry, with an adequate reaction rate, and it must be stable over the time required for the analysis.

(b) The volume of the titrant and the weight or volume of the sample must be measured to the desired precision and accuracy.

(c) Pretreatment of the sample must be carried out to remove interference if the titration reaction is not specific to eliminate matrix effects.

(d) There must be a technique to determine the equivalence point. The equivalent point is a point in the titration when the mole of the titrant added is equivalent to the moles of analyte.

In titration methods, the analytical results depend on knowing the amount of one of the reactants used.

Direct titration:titrant is added to the analyte until the reaction is complete.

Back titration

Indirect titration

titration can be divided into for types.

•Acid-base or neutralization

•Precipitation

•Complex formation

•Oxidation-reduction (redox)

ACID-BASE TITRATIONS

•Titration of a strong acid and strong base

•Titration of a weak acid and strong base

•Titration of a weak base and strong acid

•Titration of a weak acid and weak base

ACID-BASE INDICATORSIn an acid-base titration, addition of titrant near the equivalence point causes the solution pH to change drastically. This pH change is detectable with indicators that change color as a function of pH. Indicators are weak acids that change color when they gain or lose their acidic proton(s).

COMMON INDICATORS

Indicator Color pH RangeAcidic End

PointBasic

Bromo-cresol green

yellow green blue 4.0-5.6

Methyl red red yellow yellow 4.4-6.2

Bromo-thymol blue

yellow green blue 6.2-7.6

Phenol-phthalein

colorless Light pink

red 8.0-10

The point at which the indicator changes color

called the end point.

HOW DOES AN INDICATOR WORK?•Weak acids are titrated in the presence of indicators which change under slightly alkaline/base conditions.

•Weak bases should be titrated in the presence of indicators which change under slightly acidic conditions.

HIn(aq) + H2O(l) In- (aq) + H3O+(aq) Acid color A Base color B

The acid and its conjugate base have different colors.

•At low pH: [H3O+] is high and so the equilibrium position lies to the left. The equilibrium solution has the color A.

•At high pH: [H3O+] is low and so the equilibrium position thus lies to the right and the equilibrium solution has color B.

PROPERTIES OF AQUEOUS ACID-BASE INDICATORS AT 25C

Indicator pH range pKa Acid Form Base Form

Methyl violet 0.0-1.6 0.8 Yellow Blue

Methyl violet 1.2-2.8 1.6 Red Yellow

Methyl yellow 2.9-4.0 3.3 Red Yellow

Methyl orange 3.1-4.4 4.2 Red Yellow

Bromocresol green 3.8-5.4 4.7 Yellow Blue

Methyl red 4.2-6.2 5.0 Red Yellow

Chlorophenol red 4.8-6.4 6.0 Yellow Red

Bromothymol blue 6.0-7.6 7.1 Yellow Blue

Phenol red 6.4-8.0 7.4 yellow Red

Cresol purple 7.4-9.0 8.3 yellow Purple

Thymol blue 8.0-9.6 8.9 Yellow Blue

phenolphthalein 8.0-9.8 9.7 Colorless Red

thymolphthalein 9.3-10.5 9.9 Colorless Blue

Alizarin yellow R 10.1-12.0 11.0 Yellow Red

Indigo carmine 11.4-13.0 12.2 Blue Yellow

ACID-BASE TITRATIONS CURVESThe titration curve for an acid-base titration is a plot of the solution pH, normally on the vertical axis, against the volume of titrant added.

TITRATION OF A STRONG ACID ANDSTRONG BASE

We can identify three different regions in this titration experiment:

Before the Equivalence Point: The pH is determined by the concentration of unneutralized strong acid.

At the Equivalence Point: The pH, 7, is determined by the dissociation of water.

After the Equivalence Point: The pH is determined by the concentration of excess strong base that we are adding.

FOUR DISTINCT REGIONS • i) Solution of only strong acid (solution of H3O+) • ii) Excess moles of strong acid + limiting moles of strong base

(solution of H3O+) • iii) Equivalence Point with equal moles of strong acid + strong

base (solution of H2O) • iv) Excess moles of strong base + limiting moles of strong acid

(solution of OH-)

i & ii

iii

iv

Example:Sample: Receiving flask containing 50.00 mL of a 0.100 M HBr(aq) solution.Titrant: Stepwise addition of a 0.200 M KOH(aq) solution from a buret using the following volumes:

Titrant Added (mL)

Region Classification

pH of Resulting Solution

0.00 1

10.00 2

25.00 3

30.00 4

a)Addition of 0.00 mL of Strong Base (pH of initial solution)•The solution contains only 0.100 M strong acid. •Strong acids completely ionize in aqueous solution.

HBr(aq) + H2O(l) H3O+(aq) + Br(aq)

Therefore, [H3O+] = 0.100 MpH = - log [0.100]pH = 1.000

b) Addition of 10.00 mL of Strong Base (Excess of strong acid present)

First solve for the final concentrations at the end of the strong acid/strong base reaction: HBr(aq) + KOH(aq) H2O(l) + KBr(aq)Initial mmol of H3O+ = 50.00 mL x 0.1 mmol mL-1 = 5. 00 mmol NaOH added = 10.00 mL x 0.20 mmol mL-1 = 2.00 mmol HBr consumed = 2.00 mmol HBr left = 3.00Total Volume = (50.00 + 10.00) mL = 60 mLTherefore, [H3O+]final = 3.00 mmol / 60mL = 0.0500 M

pH = - log 0.0500 pH = 1.30

c) Addition of 25.00 ml of Strong Base (the equivalence point)

First solve for the final concentrations at the end of the strong acid/strong base reaction:H3O+(aq) + OH-(aq) 2 H2O(l)Initial mmol of H3O+ = 50.00 mL x 0.1 mmol mL-1 = 5. 00 mmol NaOH added = 25.00 mL x 0.20 mmol mL-1 = 5.00 mmol H3O+ consumed = 5.00 mmol HBr left = 0.00Total Volume = (50.00 + 25.00) mL = 75.00 mLAt equivalent point, [H3O+] = [ OH-]Kw = [H3O+][ OH-] = 1 10-14, pada 25 oC

[H3O+] = Kw = 1 x 10-14 = 1.0 10-7 MpH = - log 1.0 10-7 = 7.00

d) Addition of 30.00 mL of Strong Base(excess strong base added)First solve for the final concentrations at the end of the strong acid/strong base reaction:

H3O+(aq) + OH-(aq) 2 H2O(l)Initial mmol of H3O+ = 50.00 mL x 0.1 mmol mL-1 = 5. 00 mmol NaOH added = 30.00 mL x 0.20 mmol mL-1 = 6.00 mmol H3O+ consumed = 5.00 (none left) mmol NaOH left = 1.00Total Volume = (50.00 + 30.00) mL = 80.00 mLTherefore, [OH-]final = 1.00 mmoles / 80.00 mL = 1.25 x 10-2 M pOH = -log 1.25 x 10-2 = 1.90pH = 14 – 1.90 = 12.10.

Basic solution in buret (titrant)

Acidic solution in Erlenmeyer flask

-------

Buret clamp

Magnetic stirring bar

Magnetic stirrer

Titrant Added (mL)

Region Classification

pH of Resulting Solution

0.00 1 1.00

10.00 2 1.30

25.00 3 7.00

30.00 4 12.10

TITRATION OF A WEAK ACID WITH A STRONG BASE

1. The pH of the Starting SolutionAt the start of the titration the solution contains only the weak acid. The pH is calculated from the concentration and its Ka.

2. Before the Equivalence Point. After titrant been added, the solution consists of buffer. The pH is determined by the Ka, [conjugate base] and [acid].

3. At the Equivalence PointThe solution contains only the conjugate base (salt) and the pH is calculated based on the hydrolysis of the salt.

4. After the Equivalence PointThe excess of strong base titrant represses the basic character that the pH is governed largely by the concentration of the excess titrant.

Example:

Let us calculate the pH values along the course of a particular titration, the titration of exactly 20 mL of 0.100 M acetic acid with 0.100 M sodium hydroxide.

The stoichiometric reaction,

CH3COOH + Na+ + OH- CH3COO- + Na+ + H2O.

Titrant Added (mL)

Region Classificatio

n

pH of Resulting Solution

0.00 1 2.88

10.00 2 4.75

20.00 3 8.73

30.00 4 12.30

1. The pH of the Starting SolutionAt the start of the titration the solution contains only the weak acid CH3COOH, for which

CH3COOH + H2O H3O+ + CH3COO- (0.10 – x) x x

Assume x << 0.10 M, so (0.100 – x) 0.100 M,

]COOHCH[]COOCH][OH[

3

33

Ka =

]COOHCH[]OH[

3

23

100.01076.1 5

= =1.76 x 10-5

[H3O+]2 = 1.76 x 10-5 x 0.100

[H3O+] = 1.32 x 10-3

pH = - log 1.32 x 10-3 = 2.88

2. Before the Equivalence Point (Addition of 10.00 mL NaOH)

Certain portion of the acetic acid has been titrated by the NaOH solution and producing CH3COONa. The solution in the flask become a buffer solution. CH3COOH + NaOH CH3COONa + H2O

Initial mmol of CH3COOH = 20.00 mL x 0.1 M = 2. 00mmol NaOH added = 10.00 mL x 0.100 M = 1.00 mmol CH3COOH consumed = 1.00 mmol CH3COOH left = 1.00mmol CH3COONa produced = 1.00

Total Volume = (20.00 + 10.00) mL = 30.00 mL

4.75

mL) mmol/30.00 (1.00

mL) mmol/30.00 (1.00 log 10 1.76 log -

COOHCH

COONaCH log K log -

5-

3

3a

pH

3. At the Equivalence Point (Addition of 20.00 mL NaOH)

Initial mmol of CH3COOH = 20.00 mL x 0.1 M = 2. 00mmol NaOH added = 20.00 mL x 0.100 M = 2.00 mmol CH3COOH consumed = 2.00 mmol CH3COOH left = 0.00mmol NaOH left = 0.00

At the equivalence point, all the acetic acid solution and NaOH solution react completely. Only a solution of the weak base CH3COONa, sodium acetate is present.

CH3COOH + NaOH CH3COONa + H2O

mmol of CH3COONa produced = 2. 00

Total Volume = (20.00 + 20.00) mL = 40.00 mL

Since, only CH3COONa present in the solution, CH3COONa CH3COO- + Na+The weak base CH3COO- which hydrolyzes, giving CH3COOH and OH- in a stoichiometric 1:1 ratio, following the reaction equilibrium

CH3COO- + H2O CH3COOH + OH-

73.8 27.5 00.14 pH

27.5 10 5.33 log - pOH

M 10 33.5 OH x

mL 00.40/mmol 00.2

x

10 76.1

10 00.1

COOCH

OHCOOHCH

K

K K

6-

6

2

5

14

3

3

a

wh

4. Beyond/after the Equivalence Point (Addition of 30.00 mL NaOH) Initial mmol of CH3COOH = 20.00 mL x 0.1 M = 2. 00mmol NaOH added = 30.00 mL x 0.100 M = 3.00 mmol CH3COOH consumed = 2.00 mmol CH3COOH left = 0.00mmol NaOH left = 1.00

The resulting solution is then a mixture of the strong base NaOH and the weak base CH3COONa. The strong base controls the pH,

30.12 1.70 00.14 pH

70.1 0.02 log- pOH

M 02.0 mL 00.50

mmol 00.1 OH

TITRATION CURVE A WEAK ACID-STRONG BASE

COMPARISON OF TITRATION CURVE FOR

STRONG ACID-STRONG BASE

AND

WEAK ACID-STRONG BASE

Example. A total of 25 mL of ammonium hydroxide, NH4OH, solution which is approximately 0.2000 M is to be titrated with 0.2000 M HCl. (Kb NH4OH = 3.39 x 10–5).

Titrant Added (mL)

Region Classificatio

n

pH of Resulting Solution

0.00 1 11.42

10.00 2 9.71

25.00 3 5.27

30.00 4 1.74

Anda diberi 50 mL larutan HCl, pH larutan tersebut ialah 1.0. Berapa isipadu NaOH 0.25 M diperlukan untuk merubah pH larutan kepada 7.0.

Anda diberi 50 mL larutan HCl, pH larutan tersebut adalah 1.0. Berapa isipadu NaOH 0.20 M yang telah ditambah jika pH larutan ialah 6.50.

Anda diberi 50 mL larutan HCl, pH larutan tersebut adalah 1.0. Berapa isipadu NaOH 0.25 M yang telah ditambah jika pH larutan adalah 11.15.

(Jwp: 20.4 mL)… buktikan