Experiment 3 Acid and Base Titration

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EXPERIMENT 3: ACID AND BASE

EXPERIMENT 3: ACID AND BASE TITRATITRATIONTION Objectives:

Objectives: T

To determine the concentration of sodiuo determine the concentration of sodium hydroxide solution through titration technique usingm hydroxide solution through titration technique using hydrochloric acid and sulfuric acid

hydrochloric acid and sulfuric acid

Concepts: Concepts: 1.

1. TTo determio determined the concenned the concentratitration of acid and bason of acid and base solutioe solution through tin through titrattration with stion with standardandard solution.

solution. 2.

2. TTo appo apply tly the cohe correct rrect technitechnique ique in tn titratitration.ion. 3.

3. TTo carry ouo carry out acid bast acid base titre titration usation using phenoing phenolphthalphthalein as ilein as indicatndicatoror..

Introduction: Introduction:

Acid base titration involves a neutralization reaction in

Acid base titration involves a neutralization reaction in which an acid is react with an which an acid is react with an equivalentequivalent amount of base. For the neutralization of

amount of base. For the neutralization of hydrochloric acid with sodium hydroxide:hydrochloric acid with sodium hydroxide:

HCl + NaOH→ H

HCl + NaOH→ H22O + NaClO + NaCl

 Neutralization occurs when acid and

 Neutralization occurs when acid and bases exist in comparable stoichiometrybases exist in comparable stoichiometry, for instance the, for instance the amount of the hydrochloric acid (mole) is equivalent with the amount of sodium hydroxide amount of the hydrochloric acid (mole) is equivalent with the amount of sodium hydroxide (mole). The endpoint of titration can be determined using indicator.

(mole). The endpoint of titration can be determined using indicator.

Apparatus: Apparatus: V

Volumetrolumetric ic flask flask 250 250 mLmL Filter funnel Filter funnel Erlenmeyer flask  Erlenmeyer flask  Beaker Beaker 1 1 250 250 mLmL Burette Burette 1 1 50 50 mLmL Pipette Pipette 1 1 25 25 mLmL

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Chemical Reagent Chemical Reagent a) a) 100 mL 1.000 × 10100 mL 1.000 × 10-2-2M HCl solutionM HCl solution b) b) 100 mL 1.000 × 10100 mL 1.000 × 10-2-2M HM H 2 2SOSO44solutionsolution c)

c) 10 mL 10 mL C soC solution lution contaicontaining Nning NaOH (aOH (with with pipettpipette)e) d)

d) PhePhenolnolphtphthalhalien ien solsolutiutionon

Methods: Methods: 1.

1. 10 mL of C solution are put in volumetric flask, dilutes with 10 mL of C solution are put in volumetric flask, dilutes with distilled water to the mark anddistilled water to the mark and mix thoroughly. T

mix thoroughly. Transfer the solution to ransfer the solution to the clean beakethe clean beakerr. Labelled solution . Labelled solution as C. Rinse aas C. Rinse a flask with water twice.

flask with water twice. 2.

2. Burettes are clean and rinse with Burettes are clean and rinse with 5 mL HCl solution (1.000×10-2 M) twice. Place 25 mL of 5 mL HCl solution (1.000×10-2 M) twice. Place 25 mL of  HCl solution (1.000 × 10

HCl solution (1.000 × 10-2-2) to burette using funnel.) to burette using funnel.

3.

3. Pipettes are clean and rinse twice using C Pipettes are clean and rinse twice using C solution. Pipette 25 mL of C solution in threesolution. Pipette 25 mL of C solution in three Erlenmeyer flask. Adds 2 of phenolphthalein

Erlenmeyer flask. Adds 2 of phenolphthalein indicator.indicator. 4.

4. The initial volumes are recorded to The initial volumes are recorded to the nearest two decimal point. Titrate C solution with HClthe nearest two decimal point. Titrate C solution with HCl from the burette to colourless solution end point.

from the burette to colourless solution end point. Record the final volume reading Record the final volume reading andand calculate the used acid volume.

calculate the used acid volume.

 Note: phenolphthalein colour will change from magenta (base) to colourless (acid)  Note: phenolphthalein colour will change from magenta (base) to colourless (acid)

5.

5. The titrThe titration aration are repeate repeated until ted until the difhe different ferent volume of volume of acid is iacid is in the range of n the range of 0.30 for t0.30 for threehree experiments.

experiments. 6.

6. Concentration of NaOH solution (that was in flask) and Concentration of NaOH solution (that was in flask) and the concentration of C are the concentration of C are calculated.calculated. 7.

7. Steps 3 and 7 are repeated by replace HCl with HSteps 3 and 7 are repeated by replace HCl with H22SOSO44..

Result: Result:

Titration between HCl and NaOH Titration between HCl and NaOH

1.

1. VVolume of HCl olume of HCl need for need for the titration:the titration: Trial 1: 1.4 mL Trial 1: 1.4 mL Trial 2: 1.0 mL Trial 2: 1.0 mL Trial 3: 1.0 mL Trial 3: 1.0 mL

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2.

2. ConcConcententratration of Nion of NaOH by uaOH by usinsing each trg each trial vial volumolume:e: Trial 1: 0.0056M Trial 1: 0.0056M Trial 2: 0.004 M Trial 2: 0.004 M Trial 3: 0.004 M Trial 3: 0.004 M A

Aververage oage of cof concencentntraratition Non NaOaOH soH solulutitionon = 0.= 0.0050056 + 0.6 + 0.00004 + 0.04 + 0.00404 3

3 = 0.0045 M = 0.0045 M Titration between H

Titration between H22SOSO44and NaOHand NaOH

1.

1. Volume of HVolume of H22SOSO44 need for the titration:need for the titration:

Trial 1: 1.1 mL Trial 1: 1.1 mL Trial 2: 0.6 mL Trial 2: 0.6 mL Trial 3: 0.7 mL Trial 3: 0.7 mL 2.

2. ConcentConcentration ration of of NaOH NaOH by by using using each each trial trial volumevolume:: Trial 1: 0.0088M Trial 1: 0.0088M Trial 2: 0.0048 M Trial 2: 0.0048 M Trial 3: 0.0056 M Trial 3: 0.0056 M A

Aververage age of of concconcententratration ion NaONaOH H solsolutiutionon = = 0.00.0088 088 + + 0.00.0048 048 + + 0.00.0056056 3 3 = 0.0064 M = 0.0064 M Data Analysis Data Analysis 1.

1. Chemical equation for the reaction between hydrochloric acid and sodium hydroxideChemical equation for the reaction between hydrochloric acid and sodium hydroxide HCl + NaOH →NaCl + H HCl + NaOH →NaCl + H22OO Trial 1: Trial 1: n =MV n =MV n HCl = 0.1 × 1.4 n HCl = 0.1 × 1.4 = 0.14 mol = 0.14 mol

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From the equation 1 mol HCl

From the equation 1 mol HCl need 1 mol NaOH to need 1 mol NaOH to form 1 mol NaCl and 1 mol form 1 mol NaCl and 1 mol of water.of water. If the mol of HCl is 0.14M, so the mole of NaOH also 0.14 M

If the mol of HCl is 0.14M, so the mole of NaOH also 0.14 M

Concentration

Concentration of of NaOH, NaOH, M M = = n n / / VV = 0.14/ 25 = 0.14/ 25

= 0.0056 M = 0.0056 M

St

Stanandadard rd dedeviviatatioionn = 0.= 0.000056 56 - - 0.0.00004545  100100 0.0045 0.0045 = 24.44 = 24.44 Trial 2: Trial 2: n =MV n =MV n HCl = 0.1 × 1.0 n HCl = 0.1 × 1.0 = 0.10 mol = 0.10 mol

From the equation 1 mol HCl

From the equation 1 mol HCl need 1 mol NaOH to need 1 mol NaOH to form 1 mol NaCl and 1 mol form 1 mol NaCl and 1 mol of water.of water. If the mole of HCl is 0.10 mol, so the mol of NaOH is also 0.10 mol.

If the mole of HCl is 0.10 mol, so the mol of NaOH is also 0.10 mol.

Concentration

Concentration of of NaOH, NaOH, M M = = n n / / VV = 0.10/ 25 = 0.10/ 25

= 0.0040 Molar  = 0.0040 Molar 

St

Stanandadard rd dedeviviatatioionn = 0.= 0.000040 40 - - 0.0.00004545  100100 0.0045 0.0045 = 11.11 = 11.11 Trial 3: Trial 3: n =MV n =MV n HCl = 0.1 × 1.0 n HCl = 0.1 × 1.0 = 0.10 mol = 0.10 mol

From the equation 1 mol HCl

From the equation 1 mol HCl need 1 mol NaOH to need 1 mol NaOH to form 1 mol NaCl and 1 mol form 1 mol NaCl and 1 mol of water.of water. If the mole of HCl is 0.10M, so the mol of NaOH also 0.10 M

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Concentration

Concentration of of NaOH, NaOH, M M = = n n / / VV = 0.10/ 25 = 0.10/ 25

= 0.0040 Molar  = 0.0040 Molar  St

Stanandadard rd dedeviviatatioionn = 0.= 0.000040 40 - - 0.0.00004545  100100 0.0045

0.0045 = 11.11 = 11.11

2.

2. Chemical equation for the reaction between Sulfuric acid and Sodium hydroxide.Chemical equation for the reaction between Sulfuric acid and Sodium hydroxide. H H22SOSO44 + 2NaOH → Na+ 2NaOH → Na22SOSO44 + 2H+ 2H22OO Trial 1: Trial 1: n =MV n =MV n H n H22SOSO44 = 0.1 × 1.1= 0.1 × 1.1 = 0.11 M = 0.11 M

From the equation, 1 mole H

From the equation, 1 mole H22SOSO44 need 2 mole of the NaOH to produce 1 mole of Naneed 2 mole of the NaOH to produce 1 mole of Na22SOSO44

and 2

and 2 mole of water.mole of water. If got 0.11 mole of H

If got 0.11 mole of H22SOSO44, the reaction needed 0.22 mole of NaOH., the reaction needed 0.22 mole of NaOH.

Concentration

Concentration of of NaOH, NaOH, M M = = n n ∕ ∕ VV = 0.22/ 25 = 0.22/ 25 = 0.0088 M = 0.0088 M St

Stanandadard rd dedeviviatatioionn = 0.= 0.000088 88 - - 0.0.00006464  100100 0.0064 0.0064 = 37.5 = 37.5 Trial 2: Trial 2: n =MV n =MV n H n H22SOSO44 = 0.1 × 0.6= 0.1 × 0.6 = 0.06 mole = 0.06 mole

From the equation, 1 mole H

From the equation, 1 mole H22SOSO44 need 2 mole of the NaOH to produce 1 mole of Naneed 2 mole of the NaOH to produce 1 mole of Na22SOSO44

and 2

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If got 0.06 mole of H

If got 0.06 mole of H22SOSO44, the reaction needed 0.12 mole of NaOH., the reaction needed 0.12 mole of NaOH.

Co

Concncenentrtratatioion of n of NaNaOHOH, M, M = n ∕ = n ∕ VV = 0.12 / 25 = 0.12 / 25 = 0.0048 M = 0.0048 M

St

Stanandadard rd dedeviviatatioionn = 0.= 0.000048 48 - - 0.0.00006464  100100 0.0064 0.0064 = -25 = -25 Trial 3: Trial 3: n =MV n =MV n H n H22SOSO44 = 0.1 × 0.7= 0.1 × 0.7 = 0.07 mol = 0.07 mol

From the equation, 1 mole H

From the equation, 1 mole H22SOSO44 need 2 mole of the NaOH to produce 1 mole of Naneed 2 mole of the NaOH to produce 1 mole of Na22SOSO44

and 2

and 2 mole of water.mole of water. If got 0.07 mole of H

If got 0.07 mole of H22SOSO44, the reaction needed 0.14 mole of NaOH., the reaction needed 0.14 mole of NaOH.

Concentartion

Concentartion of of NaOH, NaOH, M M = = n n ∕ ∕ VV = 0.14/ 25 = 0.14/ 25 = 0.0056 M = 0.0056 M St

Stanandadard rd dedeviviatatioionn = 0.= 0.000056 56 - - 0.0.00006464  100100 0.0064

0.0064 = -12.5

= -12.5 The result is summarized below: The result is summarized below:

For the reaction between HCl with NaOH For the reaction between HCl with NaOH

T

Trriiaal l 11 TTrriiaal l 22 TTrriiaal l 33 Concentration of  Concentration of   NaOH (M)  NaOH (M) 0 0..00005566 00..00004400 00..00004400 S

Sttaannddaarrd d ddeevviiaattiioonn 2244..4444 1111..1111 1111..1111 For the reaction between H

For the reaction between H22SOSO44 with NaOHwith NaOH

T

Trriiaal l 11 TTrriiaal l 22 TTrriiaal l 33 Concentration of  Concentration of   NaOH (M)  NaOH (M) 0 0..00008888 00..00004488 00..00005566 S

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Discussion: Discussion:

Titration is a technique for determining either the concentration of a solution of unknown Titration is a technique for determining either the concentration of a solution of unknown molarity or the number of moles of a substance in a given sample. A chemical reaction is used molarity or the number of moles of a substance in a given sample. A chemical reaction is used for this purpose, and the reaction must be fast, be complete, and have a determinable end point. for this purpose, and the reaction must be fast, be complete, and have a determinable end point. The reactions of strong acids and bases generally meet these criteria, and acid-base titrations are The reactions of strong acids and bases generally meet these criteria, and acid-base titrations are among the most important examples of this technique.

among the most important examples of this technique.

In this experiment, the sample is hydrochloric acid and sulfuric acid as acid substance In this experiment, the sample is hydrochloric acid and sulfuric acid as acid substance and sodium hydroxide as base substance where the concentration of sodium hydroxide unknown. and sodium hydroxide as base substance where the concentration of sodium hydroxide unknown. Given that, the concentration of the both acid are 0.1 M.

Given that, the concentration of the both acid are 0.1 M.

An indicator is used as signal the point which the titration is stopped. An acid-base An indicator is used as signal the point which the titration is stopped. An acid-base indicator is a weak acid or base that has a different colour from its salt. At least one of them-the indicator is a weak acid or base that has a different colour from its salt. At least one of them-the indicator or its salt-must be intensely coloured so that it can be seen even in very dilute solution. indicator or its salt-must be intensely coloured so that it can be seen even in very dilute solution. The colour of the solution is thus different depending on the acidity or basicity of the solution it The colour of the solution is thus different depending on the acidity or basicity of the solution it is in, and

is in, and when the acidity owhen the acidity of a solution chanf a solution changes sufficientlyges sufficiently, a , a colour change will colour change will occuroccur.. In acid base reaction, the general equation is,

In acid base reaction, the general equation is,

HA + MOH

HA + MOH  HH22O + MAO + MA

Acid

Acid base base water water saltsalt

The end point is the neutral point. In end point of acid-base titration, there are produced salt and The end point is the neutral point. In end point of acid-base titration, there are produced salt and water that are neutral (pH7). For most strong acid-strong base reactions, ionic

water that are neutral (pH7). For most strong acid-strong base reactions, ionic equation is:equation is: H

H++ + + OHOH-- HH 2 2OO

In this experiment, the indicator that use is phenolphthalein. If we use base as titrant, and In this experiment, the indicator that use is phenolphthalein. If we use base as titrant, and acid as solution in the Erlenmeyer flask, at the end point, the solution in the Erlenmeyer flask  acid as solution in the Erlenmeyer flask, at the end point, the solution in the Erlenmeyer flask  will turn to light pink. If we use acid as titrant, and base as solution in the Erlenmeyer flask, at will turn to light pink. If we use acid as titrant, and base as solution in the Erlenmeyer flask, at the end point, the solution in the Erlenmeyer flask will turn to colourless. For this experiment, the end point, the solution in the Erlenmeyer flask will turn to colourless. For this experiment, the solution turn from the purple to co

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The chemical equation is, The chemical equation is,

The reaction between HCl and NaOH, The reaction between HCl and NaOH, HCl + NaOH →NaCl + H

HCl + NaOH →NaCl + H22OO

The reaction between H

The reaction between H22SOSO44and NaOHand NaOH

H

H22SOSO44 + 2NaOH → Na+ 2NaOH → Na22SOSO44 + 2H+ 2H22OO

The different between the reactions is the ionize of the acid. HCl is monoprotic acid, while The different between the reactions is the ionize of the acid. HCl is monoprotic acid, while H

H22SOSO44 is diprotic acid.is diprotic acid.

A diprotic acid is an acid that yields two H

A diprotic acid is an acid that yields two H++ ions per acid molecule. Examples of diprotic acidsions per acid molecule. Examples of diprotic acids

are sulfuric acid, H

are sulfuric acid, H22SOSO44, and carbonic acid, H, and carbonic acid, H22COCO33. A diprotic acid dissociates in water in two. A diprotic acid dissociates in water in two

stages: stages:

(1) H

(1) H22X(aq) HX(aq) H++(aq)(aq)+ HX+ HX--(aq)(aq)

(2) HX (2) HX

--(aq)

(aq) HH++(aq)(aq)+ X+ X2-2-(aq)(aq)

A monoprotic acid is an acid that yields only one H

A monoprotic acid is an acid that yields only one H++ ions per acid molecule. Example of ions per acid molecule. Example of 

monoprotic acid is Hydrochloric acid. While H

monoprotic acid is Hydrochloric acid. While H22SOSO44 is a diprotic acid that’s yield two mole of His a diprotic acid that’s yield two mole of H++..

The mean, when the acid dissociate with NaOH, H

The mean, when the acid dissociate with NaOH, H22SOSO44 will need 1 mole of Hwill need 1 mole of H22SOSO44 while thewhile the

 NaOH needed 2 mole. That’s mean, we need less amount of acid if we use diprotic acid. That’s  NaOH needed 2 mole. That’s mean, we need less amount of acid if we use diprotic acid. That’s

why the volume of H

why the volume of H22SOSO44that has been use is less than that has been use is less than HCl.HCl.

In this experiment, there are some mistake like use more than acid volume to titrate In this experiment, there are some mistake like use more than acid volume to titrate solution C or NaOH. The volume has pass the end point, so the volume uses is much more than solution C or NaOH. The volume has pass the end point, so the volume uses is much more than needed. To overcome this problem, we can titrate slowly and shake the volumetric flask for about needed. To overcome this problem, we can titrate slowly and shake the volumetric flask for about 30 second when the solutions show changing in colour from dark purple to colourless. The other  30 second when the solutions show changing in colour from dark purple to colourless. The other  rea

reasonsons s is is the using the using of of volvolumeumetritric c flaflask sk thathat’t’s s had been had been useused d witwith h othother er solsolutiution. on. So, theSo, the concentration of newest solution will affect. To overcome this problem, we must make sure the concentration of newest solution will affect. To overcome this problem, we must make sure the volumetric flask is totally clean and dry. That’s problems had affect our result. That’s why our  volumetric flask is totally clean and dry. That’s problems had affect our result. That’s why our 

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result is not precise and accurate to the correct value. We can see that from the standard result is not precise and accurate to the correct value. We can see that from the standard deviation. But, we had seen the distribution of result is in range of 0.0045 M to 0.0065 M. If we deviation. But, we had seen the distribution of result is in range of 0.0045 M to 0.0065 M. If we refer to trial 2 and 3 from titration between HCl and solution C (NaOH), the result might be said refer to trial 2 and 3 from titration between HCl and solution C (NaOH), the result might be said as 0.004 M base on the standard deviation. The standard deviation is the lowest among others as 0.004 M base on the standard deviation. The standard deviation is the lowest among others and can be said as accepted result. But, for the titration using H

and can be said as accepted result. But, for the titration using H22SOSO44, the accepted value for , the accepted value for 

concentration of solution C (NaOH) is 0.0056 M. concentration of solution C (NaOH) is 0.0056 M.

Conclusion: Conclusion:

The concentration and mole of acid or base can be determined using titration process by a given The concentration and mole of acid or base can be determined using titration process by a given value for one of the substance. The accepted values of concentration NaOH using HCl is 0.0040 value for one of the substance. The accepted values of concentration NaOH using HCl is 0.0040 M, but the value of concentration NaOH using H

M, but the value of concentration NaOH using H22SOSO44 is 0.0056 M.is 0.0056 M.

Reference: Reference: 1.

1. http:/http://dwb.u/dwb.unl.edu/nl.edu/calculcalculatorsators/acti/activitievities/Dis/DiprotiproticAcid.hcAcid.htmltml 2.

2. http:/http://www/www.bioch.biochem.norem.northwesthwestern.edtern.edu/holmu/holmgren/Ggren/Glosslossary/Dary/Definitefinitions/ions/Def- Def-M/monoprotic_acid.html

M/monoprotic_acid.html 3.

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References

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