Titration

Lab Report Basic Experimental Work

Experiment 2: Volumetric Analysis (Titration)

Submitted By: Md. Rafiqul IslamM.Sc Utilities and Waste- Sustainable Processing

Tutor: Heiko Schwegmann

Date of Experiment: 02.11.10 Date of Submission: 09.11.10

Engler-Bunte-Institut Karlsruher Institut für Technologie(KIT)

Contents

Page no

1 Introduction 3

2 Acid- base titration 5

2.1 Chemicals 6

2.2 Instruments 6

2.3 Acid- base titration ( Titrant: 0.1 M HCl; Analyte: NaOH) 7

2.3.1 Procedure 7

2.3.2 Calculations 8

2.3.2 Results and Discussions 9

2.4 Acid- base titration ( Titrant: 0.1 M NaOH; Analyte: HCl) 10

2.4.1 Procedure 10

2.4.2 Calculations 11

2.4.3 Results and Discussions 12

3 Complexometric titration 13

3.1 Chemicals 13

3.2 Instruments 13

3.3. Procedure 14

3.4 Calculations 15

3.5 Results and Discussions 15

4 Conclusion 16

5 Questions and answers 17

6 References 20

Figures

Figure 1 Typical pH curves for the acid-base titration reactions. 17

Figure 2 (a) the structure of EDTA; (b) the structure of tetracarboxilate ion [EDTA]4- , forms from dissociation of EDTA.

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Basic Experimental Work Volumetric analysis

3

1. Introduction

Titration is used for measuring the unknown concentration of a known reactant with

the help of a standard reactant of known concentration. It is also used for finding the

endpoint of a reaction. Titration is also called volumetric analysis because volume

measurement has a vital role in this method. The standard solution is known as titrant

and the solution with unknown concentration is known as analyte. Almost in all

titration an indicator or a pH meter is used to detect the endpoint or equivalence point

of a reaction. But there are some titrations where the strong coloured reactants or

products act as an indicator. An indicator is a substance which changed its colour

when the reaction is completed and reached in endpoint. So we can easily understand

the completion of a reaction by visual observation with the help of an indicator. There

are different types of indicator and it has to be chosen for each type of titration

reaction. Usually each indicator worked in a specific pH range. Suppose an acid-base

indicator methyl orange is itself a weak acid or its conjugated base, both has different

colour.

Some titration requires a specific pH value to complete its reaction. So buffer solutions are added in the analyte to maintain the pH of the solution in the optimum range. Such as in this experiment we mixed NH3/NH4Cl -buffer solution of pH 10 with the analyte in complexometric titration. [1]

In this experiment we did different volumetric analysis or titration of different samples.

In acid-base titration we measured the concentration of acid and bases and in

complexometric titration we determined the hardness of drinking water.

2. Acid-Base Titration

In acid- base titration we determined the concentration of NaOH and HCl solution

separately by two different experiments. In the first acid-base titration NaOH is used as

analyte and 0.1 M HCl solution as titrant. On the other experiment HCl is used as

analyte and 0.1 M NaOH solution as titrant.

The reaction for both experiment is as follows:

HCl(aq) + NaOH(aq) NaCl + H2O(l)

In this experiment we conducted strong acid – strong base titrations. Therefore we

chose phenolphthalein as an indicator because it changes its colour in the pH range of

8.3-10. [2]

It is known that pH means the negative logarithm of [H]+ ion. For this reason, after

achieving the equilibrium condition in acid-base reaction, small amount of addition of

any reactants causes higher difference in pH value. On the other hand each indicator

has a specific pH range and if the pH changes above or below of this range indicator

changes its colour also. That’s mean immediate change in pH leads the immediate

colour change of the solution by the indicator.


5

2.1 Chemicals:

Sodium hydroxide pellets , M= 40 g/mol, 0.1 M hydrochloric acid, HCl Concentrated hydrochloric acid, 32% ( wt percent), HCl Diluted NaOH Distilled water

Indicator (Phenolphthalein)

2.2 Instruments & Glassware:

Weighting machine ( Sartorius- BP310s; Max- 310 g, d= 0.01 g)

1 graduated pipette, 10 ml 1 Burette, 25 ml 3 Beaker, 200 ml Funnel Weighting boat Pasteur pipette Peleus ball

2.3 Acid- base titration ( Titrant: 0.1 M HCl; Analyte: NaOH)

2.3.1 Procedure:

At first, we took 24.9 ml of 0.1 M HCl in burette as a titrant. Before taking the acid solution in the burette we washed it with the sample to remove other constituents in it.

After that we prepare approximately 100 ml of 0.1 M NaOH solution. Because sodium hydroxide supplied as pellets we weighted 0.374 g of NaOH by the balance (Sartorius- BP310s; Max- 310 g, d= 0.01 g). Then we weighted 93.56 ml of distilled water (93.56 g, assume density= 1000 kg/m3). We added the NaOH pellets in the water (solvent) and shacked it for 2-3 minutes to get the desired solution.

By a 10 ml pipette we took NaOH solution in another beaker as an analyte.

In the analyte solution we put 2-3 drops of phenolphthalein indicator which changed the solution colour as pink.

We set all of the instruments in appropriate position and start titration by adding titrant drop wise in the analyte and observed the change of the analyte solution colour.


7

2.3.2 Calculation:

Preparation of 0.1 M NaOH: Molecular weight of NaOH = 40 g/mol Mass of sodium hydroxide, m = 0.374 g Volume of solvent (distilled water) = 93.5 ml Mole of sodium hydroxide, n = (0.374g)/(40g/mol) = 0.00935 molConcentration of sodium hydroxide, C = (0.00935 mol / 93.5ml)* (1L/1000ml) = 0.1mol/L = 0.1 M

Measurement of actual concentration of analyte (NaOH) Volume of titrant (HCl) = 9.75 ml Number of mole of titrant (0.1 M HCl), Nr = (0.1 mole/ L X 9.75 ml)* (1L/ 1000ml) = 9.75*10-4 mol Volume of analyte, Vr = 10 mlConcentration of analyte, Cr = Nr / Vr

= (9.75 *10-4 mol / 10ml)* (1000ml/L) = 0.0975 mol/L = 0.0975 M

Error = * 100 = * 100 = 2.5%

2.3.3 Results and discussion:

Test no.

Initialburettereading

(ml)

Final burettereading

(ml)

Titrate(HCl) Volume (ml)

Titrantaverage

Volume (ml)

NHCl (mol)

CNaoH

M or mol/L

1 0.1 9.7 9.69.75 0.975X10-3 0.0975

2 9.7 19.6 9.9

From the calculation we find that 9.75 ml of 0.1M HCl requires for neutralizing 10 ml of NaOH and the actual concentration of the analyte (NaOH) is 0.0975 M which is very close to 0.1M. Though we prepared 0.1 M NaOH solution and used as analyte in the titration. From the equation of reaction between HCl and NaOH we know that 1 mol of HCl requires 1 mol of NaOH to reach the equilibrium point. But the concentration of NaOH solution found different by the experiment than measured value. There may some error occurred during preparation of the NaOH solution or to find the exact end point of the reaction. The error is found 2.5%.


9

2.4 Acid- base titration ( Titrant: NaOH ; Analyte: HCl)

2.4.1 Procedure:

At first, we took 24.9 ml of 0.1 M NaOH in burette as a titrant. Before taking the NaOH solution in the burette we washed it with distilled water and sample respectively to remove other constituents in it.

After that we prepare approximately 100 ml of 0.1 M HCl solution from concentrated hydrochloric acid, 32% (wt percent). Because of high concentration of the stock solution of hydrochloric acid (9.79 M) and we required only 0.1 M HCl, we diluted the concentrated solution in 2 steps to get the desired value.

In step 1 of dilution of hydrochloric acid, we weighted 100.3 g water in a beaker by the balance. Then we took 10 ml of concentrated HCl by a pipette and add in the water. So, we get the concentration of the solution 0.97 M by calculation.

In second step of dilution, we again weighted 128 g of distilled water in another beaker and after that we took 10 ml of HCl from 0.97 M HCl solution. We mixed the HCl in water and got 0.076 M HCl solution by calculation. This 0.076 M HCl used as analyte.

By a 10 ml pipette we took HCl solution in another beaker as an analyte.

In the analyte solution we put 2-3 drops of phenolphthalein indicator but the colour of the solution remained unchanged.

We set all of the instruments in appropriate position and start titration by adding titrant drop wise in the analyte and observed the change of the analyte solution colour.

2.4.2 Calculation:

Molarity of concentrated hydrochloric acid, 32% (wt percent), HCl: Molecular weight of HCl = 36.36 g/mol Concentration of the stock HCl solution = 32% (wt percent) Density of stock HCl solution = 1.16 g/cm3

= (1.16 g/cm3) * (1000 cm3/L) =1116 g/LMass of HCl in 1 L solution = (1116g/L)*0.32 =357.12 g/LMolarity of HCl, M = (357.12 g/L)/(36.46 g/mol) = 9.79 mol/L = 9.97 M

Dilution of concentrated HCl solution: Step 1(1:10 dilution) S1 V1= S2 V2

Molarity of stock solution, S1 = 9.79 MVolume of stock solution, V1 = 10 mlVolume of solvent (distilled water), V2 = 100.3 ml Molarity of diluted HCl solution, S2 = ? S2 = (9.79 M*10 ml)/ 100.3 ml = 0.97 M Step 2 (1:10 dilution) S1V1= S2 V2

Molarity of stock solution, S1 = 0.97 MVolume of stock solution, V1 = 10 mlVolume of solvent (distilled water), V2 = 128 mlMolarity of diluted HCl solution, S2 = ?


11

S2 = (0.97 M*10 ml)/ (128 ml) = 0.075 MSo, we prepared 0.075 M HCl which used as an analyte in the titration.

Measurement of actual concentration of analyte (HCl) Volume of titrant = 12.2 ml No. of mole of titrant (0.1 M NaOH), Nr = (0.1 mole/ L X 12.2 ml)* (1L/ 1000ml) = 1.22 *10-3 molVolume of analyte, Vr = 10 mlConcentration of analyte, Cr = Nr / Vr

= (1.22 *10-3 mol / 10ml)* (1000ml/L) = 0.122 mol/L = 0.122 M

Error = * 100 = * 100 = 38.5 %

2.4.3 Results and discussion:

Test no.

Initial burette reading

(ml)


(ml)

Titrant(NaOH) Volume (ml)

Titrantaverage

Volume (ml)

NNaOH(mol)

CHCl

mol/L

1 0.1 11.3 11.2 12.2 1.22*10-3 0.122

2 11.3 24.5 13.2

We prepared 0.075 M HCl which used as an analyte in the titration. But From the calculation we find that 12.2 ml of 0.1M NaOH requires for neutralizing 10 ml of HCl and the actual concentration of the analyte (NaOH) is 0.122 M which is not very close to the original concentration of HCl that we prepared. From the reaction of HCl and NaOH we know that 1 mol of HCl requires exactly 1 mol of NaOH to reach the equilibrium point. There may some error occurred during preparation of the HCl solution or to find the exact end point of the reaction. The error of this experiment is found 34.5%.

3. Complexometric Titration

There are many organic salts dissolve in natural water. It is also known as hard water and Ca+ & Mg+ ions are the main source of hardness. Hard water causes different problems like scaling, corrosion if it use as process water without treatment. So it is very important to determine the hardness of water at first before using in any process.

In this experiment complexometric titration was used to determine the hardness of the sample water. In complexometric titration Ca+ & Mg+ ions in water react with ethylenediamintetraacitic acid (EDTA) and Eriochrome Black-T is used as an indicator. EDTA is itself a weak acid and at pH 10 it dissociates to form maximum amount of tetracarboxylate ion [EDTA]4-. The complex ion has six bonding site and bonded with Ca+ & Mg+ ions and form a strong 1:1 complex. For this reason we added a buffer solution of pH 10 with titrant (EDTA) in this experiment. [3]

3.1 Chemicals:

EDTA(0.1 M) Drinking water Distilled water Buffer (pH=10) Indicator (Eriochrome Black-T)

3.2 Instruments:

Weighting machine ( Sartorius- BP310s; Max- 310 g, d= 0.01 g)

2 graduated pipette, 10 ml 1 Burette, 25 ml 2 Beaker, 200 ml Funnel Weighting boat Pasteur pipette


13

Peleus ball

3.3 Procedure

At first we took 50 ml of tap water or drinking water in a beaker which used as the analyte in this titration.

We took 5 ml buffer NH3/NH4Cl solution of pH 10 by a pipette and mixed it with the tap water or analyte. This buffer solution used to maintain a certain pH range for the titration reaction.

We also added 2-3 drops of indicator (Eriochrome Black-T) in

the analyte to visualize the endpoint of reaction. After adding indicator it changed the analyte solution colour to red wine.

After that we took 25 ml of 0.01 M EDTA in a burette and it

was used as titrant. Before taking the titrant in the burette we washed it with the sample EDTA solution to remove all other constituents in it.

We set all of the instruments in appropriate position and start titration by adding titrant drop wise in the analyte and observed the change of the analyte solution colour. The titration stopped when we observed the change of colour from red wine to purple.

3.4 Calculation:

Measurement of Hardness in drinking water

Concentration of EDTA solution = 0.01 MVolume of EDTA solution = 19.5 ml Volume of sample water = 50 ml

No. of moles of Ca+ & Mg+ ions, mol = no. of moles of EDTA required for for titration, mol = (vol. of required EDTA solution)* (Conc. of EDTA solution) = (19.5 ml)* (0.01mol/L)* (1L/1000ml) = 1.95*10-4 mol = 0.195 mmol

Total Ca+ & Mg+ ions concentration in sample, mmol(LH2O)-1

= (0.195 mmol/50ml)*(1000ml/1L) = 3.9 mmol (LH2O)-1

= (3.9 mmol/ 1L of H2O)*( 1mol/1000 mmol) *( 1000L/1m3) = 3.9 mol/m3


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3.5 Results and discussion:

Test no.

Initialburette

Reading (ml)


(ml)

Titrant (EDTA)Volume

(ml)

Concentration of EDTA

solution (M)

No. of moles of Ca+ & Mg+

ions,mmol

Total Ca+ & Mg+ ions

concentrationmol/m3

1 0.1 19.6 19.5 0.01 0.195 3.9

From the calculation we found that the hardness of the sample water is 3.9 mol/m3. By definition this water is very hard water because it contains huge amount of Ca+ & Mg+

ions (390 mg/L as CaCO3 > 181 mg/L as CaCO3). [4]

4. Conclusion

By this experiment we learn about two basic type of titration acid-base and

complexometric titration. From acid-base titration we determine the concentration of

the analyte and hardness of the lab tap water by complexometric titration. We found

some difference in actual or measured value then the original value and mentioned

this difference in values as error of the specific titration. This error might occur for

different reasons. The original endpoint of a reaction and the indication of endpoint by

the indicator by changing its colour are found different for all case. This is because of

the logarithmic nature of pH it may change significantly with the small addition of titrant

in the analyte solution just before the endpoint of a reaction. This is also lead the

indicator to change its colour. But there is a little difference between the change in

indicator colour and the actual equivalence point of the titration reaction. This is knoen

as indicator error and it is undetermined.


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5. Questions and answers

a) What is an indicator? How does an indicator work? Why is necessary to add only few drops of indicator in a titration?

Ans: An indicator is a substance which changed its colour when the reaction is completed and reached in endpoint. If the indicator expressed by Hln and its conjugated base by ln- then we can write the

following equation:

Hln(aq) + H2O(l) H3O+ (aq) + ln- (aq) Red Yellow

According to the LeChatelier's Principle,

= or, =

So, the ratio of ln-(aq)/ Hln(aq) depends on pH and changed the colour of solution

when its pH change due to the reaction. [5]

It is necessary to add few drops of indicator in titration because indicators also react

with the reactants and form intermediate substances. But it changes its colour or

solution colour when the reaction reach at equivalence point. If we add more indicator

it take more time to dissociate their structure or changing colour so it confuse the

determination of the endpoint

b) Which indicator is appropriate for titration of 0.1 M acetic acids with a 0.2M NaOH solution? Why?

Ans: The choice of an indicator is determined by the pH of the solution at the

equivalence point.[5] For example phenolphthalein is the perfect indicator for weak

acid (0.1 M acetic acid) and strong base (0.2 M NaOH) titration. Because of acetic acid

and sodium hydroxide produce sodium acetate and H2O by the reaction. In equilibrium

condition the pH of sodium acetate solution in the range of 8.2-10 where

phenolphthalein can work perfectly and changes its colour. [1]

c) Draw and explain a typical titration curve of HCl with NaOH. How would the

curve change if H2SO4 would be used for the titration instead of HCl? Why?

Ans: In figure 1 the first graph shows a pH curve of a strong acid-strong base titration

where HCl used as analyte and NaOH as titrant and both of reactants has same

concentration. At the equivalence point the pH is 7, but has increased sharply from 3.5

to 11 just before this point. As indicator shows different colour in acidic and basic

media so any suitable indicator like phenolphthalein or methyl orange can easily use to

understand the endpoint of the reaction. [6]

Figure 1: Typical pH curves for the acid-base titration reactions.

H2SO4 is a diprotic acid and it has two ionizable hydrogen atoms. So in reaction with

NaOH the first H+ of H2SO4 has been completely neutralized.

H2SO4 + H2O H3O+ + H2SO-

H2SO- + H2O H3O+ + HSO-


19

Sulfuric acid is a strong acid in its first ionization step and a weak acid in the second.

Ionization is complete in the first step. The second dissociation constant of H2SO4 is

quite large and the H3O+ from the second ionization step contributes quite substantially

to the total concentration of hydronium ion in solution. Therefore, when we observe the

titration curve of sulfuric acid, we cannot distinguish the first equivalence point, since

the pH of the solution is still very low at this point and the neutralization proceeds to

the second equivalence point. [7]

d) Why was the NH3/NH4Cl-buffer added to the drinking water during determination of hardness? Explain shortly how a buffer acts.

Ans:

EDTA is itself a weak acid and at pH 10 it dissociates to form maximum amount of tetracarboxylate ion [EDTA]4-. The complex ion has six bonding site in which four for caroxylate ion and other two sites for N atoms. These six sites are bonding with Ca+ & Mg+ ions and form a strong 1:1 complex. For this reason we added a NH3/NH4Cl-buffer solution which pH is 10 with the analyte in this experiment.

(a)

(b)

Figure 2: (a) the structure of EDTA; (b) the structure of tetracarboxilate ion [EDTA]4- , forms from dissociation of EDTA.

A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its

conjugate base or a weak base and its conjugate acid. It has to be added to maintain a

certain pH range for any reaction. In the other word it resists the change of pH in any

solution by adding or removing of H3O+ or OH- by itself.

6. References

1. Supplied lab manual2. http://en.wikipedia.org/wiki/Titration3. Determination of Hardness by EDTA titration – prepared by University of

California , Irvine.4. http://en.wikipedia.org/wiki/Hard_water5. http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm 6. http://www.tutorvista.com/content/chemistry/chemistry-iii/ionic-

equilibrium/titrations-acid-base.php7. Determining ka2 of sulfuric acid by titration - Sophia Nussbaum, PhD, University

of British Columbia; Dale Hammond, PhD, Brigham Young University Hawaii.

Titration

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