1. Introduction: The laboratory method used in the experiment is titration. Titration is a method usedin measuring the amount of an analytical reagent necessary to react quantitatively with thesample. Acid-base titrations are important for counting concentrations of acids and bases. Thepoint at which the equivalence point has been reached is called the theoretical end point. Thetitration is detected by some physical change, such as colour change, to indicate the endpoint. The indicator solution used in this experiment is phenolphthalein which is a colourlessin acidic solution. When the reaction reached the endpoint, the solution becomes pale pink.One or two drops of phenolphthalein are sufficient for the titration. Addition of too muchindicator will necessitate an indicator blank. The indicator should give a clear visual changein the liquid being titrated once the reaction between the standard solution and the substanceis practically complete. The indicators can change colour because their ions have colours thatare different from the undissociates molecule (Tan, 2010). The acid and base titrations usedthe Arrhenius theory. This theory stated that acid are substance which produce hydrogen ionsin solution and bases are substance which produce hydroxide ion in solution. In titration,standard solutions are now usually expressed in terms of molarity, M. Molarity, M =moles of solute/volume of solutions in litresObjective : One of the purposes of this experiment is to demonstrate the basic laboratorytechnique of titration towards the students. This experiment enables a hand-on practice doneby the students themselves. Thus, the students will know how to handle the materials,apparatus and the correct way to conduct the experiment as well as being careful to avoid anyexperimental error. Moreover, this experiment also teach the students on the correct steps tocalculate percentage of substance, molarity of solution and other calculations involvethroughout the experiment.Apparatus :Burette with standPipette1-L volumetric flask with stopper250 mL Erlenmeyer flasksRetort stand with clampReagents :Sodium hydroxide pelletsPotassium hydrogen phthalate (KHP)PhenolphthaleinUnknown vinegar
2. Procedure :Part A : Preparation of the Sodium Hydroxide Solution1. 1-L volumetric flask and stopper are cleaned and rinsed. The flask is labelled with “Approx. 1.0 M NaOH”. About 500 mL of distilled water is put into the flask.2. Approximately 4.0 g of sodium hydroxide pellets are weighed out and transferred to the 1-L flask. Stopper is put and the flask is shaken to dissolve the sodium hydroxide.3. Additional distilled water is added to the bottle until the mark on the neck of the flask when all the sodium hydroxide pellets have dissolved. Stopper is put and the flask is shaken thoroughly to mix.Part B : Standardization of the Sodium Hydroxide Solution1. The burette is set up in the burette clamp. The burette is rinsed and filled with the freshly prepared sodium hydroxide solution.2. Three 250-mL Erlenmeyer flasks are cleaned with water and then rinsed with distilled water. They are labelled as 1, 2 and 3.3. The bottle of dried KHP is removed from the oven. When the KHP is completely cool, three samples of KHP between 0.6 and 0.8 g are weighed, where on for each of the Erlenmeyer flasks. The exact weight of each KHP sample is recorded to the nearest mg (±0.001 g).4. 100 mL of distilled water is added to KHP sample 1. Then, 2 to 3 drops of phenolphthalein indicator solution is added. The flask is swirled to dissolve the KHP sample completely.5. The initial reading of the NaOH solution in the burette is recorded to the nearest 0.02 mL.6. NaOH solution is added from the burette to the sample in the Erlenmeyer flask, the flask is swirled constantly during the addition.7. When the titration is approaching the endpoint, NaOH is added one drop at a time, with constant swirling, until one single drop of NaOH causes a permanent pale pink colour that does not fade on swirling. The reading of the burette is recorded to the nearest 0.02 mL.8. Step 4 to 7 is repeated with the other 2 KHP samples.9. Given that the molecular mass of KHP is 204.2, the number of moles of KHP in samples 1, 2 and 3 is calculated.10. From the number of moles of KHP present in each sample, and from the volume of NaOH solution used to titrate the sample, the molar concentration (M) of NaOH in the titrant solution is calculated. The reaction between NaOH and KHP is 1:1 stoichiometry.Part C : Analysis of a Vinegar SolutionVinegar is a dilute solution of acetic acid and can be effectively titrated with NaOH using thephenolphthalein endpoint.1. Three Erlenmeyer flasks are cleaned and labelled as samples 1, 2 and 3.2. The 5-mL pipette is rinsed with small portions of the vinegar solution and the rinsing is discarded.3. Using the pipette, 5-mL of the vinegar solution is pipetted into each of the Erlenmeyer flasks. About 100 mL of distilled water and 2 to 3 drops of phenolphthalein indicator solution is added to each flask.
3. 4. The burette is refilled with the NaOH solution and the initial reading of the burette is recorded to the nearest 0.02 mL. Sample 1 of the vinegar is titrated in the same manner as in the standardization until one drop of NaOH causes the appearance of the pink color.5. The final reading of the burette is recorded to the nearest 0.02 mL.6. The titration is repeated for the other two vinegar samples.7. Based on the volume of vinegar sample taken, and on the volume and average concentration of NaOH titrant used, the molar concentration of the vinegar solution is calculated.8. Given that the formula mass of acetic acid is 60.0, and the density of the vinegar solution is 1.01 g/mL, the percent by mass of acetic acid in the vinegar solution is calculated.Result :Part A : Standardization of The Sodium Hydroxide SolutionParticulars Trial 1 Trial 2 Trial 3Mass of KHP taken (g) 0.625 0.634 0.627Final burette reading (mL) 30.40 30.30 30.20Initial burette reading (mL) 0.00 0.00 0.00Volume of NaOH used (mL) 30.40 30.30 30.20Molarity of NaOH solution 0.10 0.10 0.10Average molarity of NaOH solution 0.10Part B : Analysis of a Vinegar SolutionParticulars Trial 1 Trial 2 Trial 3Volume of vinegar solution used (mL) 5.00 5.00 5.00Final burette reading (mL) 36.90 36.20 36.50Initial burette reading (mL) 0.00 0.00 0.00Volume of NaOH used (mL) 36.90 36.20 36.50Molarity of NaOH solution 0.10 0.10 0.10Molarity of vinegar solution 0.74 0.72 0.73% mass of acetic acid in vinegar 4.38 4.30 4.34Average molarity of vinegar solution 0.73Average % mass of acetic acid in vinegar 4.34
4. Calculation : (Part A)Equation for the reaction between NaOH and KHP.KCO2C6H4CO2H + NaOH KCO2C6H4CO2Na + H2OGiven that the reaction between NaOH and KHP is of 1 : 1 stoichiometry.Let A : KHP and B : NaOHMAVA 1MBVB 1MAVA = MBVBTrial 1 :For KHP , volume given is 100 mL (0.1 L) , and molecular mass given is 204.2 g mol-1 :Mass = number of mole X molecular mass0.625 g = no of mole for KHP X 204.2 g mol-1No. of mole for KHP = 0.625 g 204.2 g mol-1No. of mole for KHP = 0.00306 molMolarity of KHP = No. of mole Volume of KHPMolarity of KHP = 0.00306 mol 0.1 LMolarity of KHP = 0.0306 mol L-1From the data collected,MAVA = MBVB(0.0306 mol L-1)(0.1 L) = MB (0.0304 L)MB = (0.0306 mol L-1)(0.1 L) (0.0304 L)MB = 0.10 mol L-1
5. Trial 2 :For KHP , volume given is 100 mL (0.1 L) , and molecular mass given is 204.2 g mol-1 :Mass = number of mole X molecular mass0.634 g = no of mole for KHP X 204.2 g mol-1No. of mole for KHP = 0.634 g 204.2 g mol-1No. of mole for KHP = 0.00310 molMolarity of KHP = No. of mole Volume of KHPMolarity of KHP = 0.00310 mol 0.1 LMolarity of KHP = 0.0310 mol L-1From the data collected,MAVA = MBVB(0.0310 mol L-1)(0.1 L) = MB (0.0303 L)MB = (0.0310 mol L-1)(0.1 L) (0.0303 L)MB = 0.10 mol L-1
6. Trial 3 :For KHP , volume given is 100 mL (0.1 L) , and molecular mass given is 204.2 g mol-1 :Mass = number of mole X molecular mass0.627 g = no of mole for KHP X 204.2 g mol-1No. of mole for KHP = 0.627 g 204.2 g mol-1No. of mole for KHP = 0.00307 molMolarity of KHP = No. of mole Volume of KHPMolarity of KHP = 0.00307 mol 0.1 LMolarity of KHP = 0.0307 mol L-1From the data collected,MAVA = MBVB(0.0307 mol L-1)(0.1 L) = MB (0.0302 L)MB = (0.0307 mol L-1)(0.1 L) (0.0302 L)MB = 0.10 mol L-1Calculation : (Part B)Vinegar contains acetic acid. When vinegar is neutralized :NaOH Na+ + OH-CH3COOH + H2O H3O+ + CH3COO-1 mole of NaOH supplies 1 mol of OH- ion1 mole of CH3COOH supplies 1 mol of H3O+ ionNaOH (aq) + CH3COOH (aq) CH3COONa (aq) + H2O (l)
7. Trial 1 :Moles = volume in litre X molarity = 0.0369 L X 0.10 mol L-1 =0.00369 molNumber of mole for NaOH = 0.00369 molNumber of mole for NaOH = Number of mole for CH3COOHSo, the number of mole for CH3COOH = 0.00369 molMolarity for CH3COOH = Number of mole VolumeMolarity for CH3COOH = 0.00369 mol 0.005 L = 0.738 mol L-1Molarity of vinegar is 0.738 mol L-1This means 1 litre of vinegar contains 0.738 mol of acetic acid.Mass = Molarity X molar mass = 0.738 mol L X 60.0 = 44.28 gMass of vinegar = 1.010 kg = 1010 g% by weight = mass of acetic acid Mass of vinegar = 44.28 1010 = 4.38 %
8. Trial 2 :Moles = volume in litre X molarity = 0.0362 L X 0.10 mol L-1 =0.00362 molNumber of mole for NaOH = 0.00362 molNumber of mole for NaOH = Number of mole for CH3COOHSo, the number of mole for CH3COOH = 0.00362 molMolarity for CH3COOH = Number of mole VolumeMolarity for CH3COOH = 0.00362 mol 0.005 L = 0.724 mol L-1Molarity of vinegar is 0.724 mol L-1This means 1 litre of vinegar contains 0.724 mol of acetic acid.Mass = Molarity X molar mass = 0.724 mol L X 60.0 = 43.44 gMass of vinegar = 1.010 kg = 1010 g% by weight = mass of acetic acid Mass of vinegar = 43.44 1010 = 4.30 %
9. Trial 3 :Moles = volume in litre X molarity = 0.0365 L X 0.10 mol L-1 =0.00365 molNumber of mole for NaOH = 0.00365 molNumber of mole for NaOH = Number of mole for CH3COOHSo, the number of mole for CH3COOH = 0.00365 molMolarity for CH3COOH = Number of mole VolumeMolarity for CH3COOH = 0.00365 mol 0.005 L = 0.730 mol L-1Molarity of vinegar is 0.730 mol L-1This means 1 litre of vinegar contains 0.718 mol of acetic acid.Mass = Molarity X molar mass = 0.730 mol L X 60.0 = 43.80 gMass of vinegar = 1.010 kg = 1010 g% by weight = mass of acetic acid Mass of vinegar = 43.80 1010 = 4.34 %
10. Discussion The acid and base titrations used the Arrhenius theory. This theory stated that acidare substance which produce hydrogen ions in solution and bases are substance whichproduce hydroxide ion in solution. Acid-base indicator is a weak organic acids or bases that dissociate slightly inaqueous solutions to form ions (Tan, 2010). The indicator that we used in this experiment isthe phenolphthalein, where the ph range is around 8.2 to 10.0. The colour change fromcolourless to a light pink. The following equilibrium is established between the indicator, HInand its conjugate base (In-) (Tan, 2010). HIn (aq) + H2O (l) H3O+ (aq) + In- (aq) acid (colour A) conjugate base (colour B) Equivalence point is the point where there are equal amounts (in moles) of H3O+ (aq)and OH- (aq) in the titration flask. At this point, the neutralisation is completed and neitherthe acid nor the alkali is in excess. The solution only consists of salt and water. H3O+ (aq) + OH- (aq) 2H2O (l) The end point of an acid-base titration is the point where the indicator changes colour.A suitable indicator must be chosen so that the end point coincides with the equivalent point. In part A, we prepared the Sodium Hydroxide solution by adding sodium hydroxidewith distilled water to be used in the experiment. In order to make sure that the sodiumhydroxide pellets dissolved completely, we have to shake the flask thoroughly. Somechemicals can be purchased in a pure form and remain pure over a long period or time.However, sodium hydroxide absorbs moisture from the air and often appears wet, thus it iseasily contaminated. Thus if a solution of sodium hydroxide is prepared by weighing thesodium hydroxide, the concentration of the solution may not be precisely the intendedconcentration. Potassium hydrogen phthalate on the other hand, has a lesser tendency toabsorb water from the air and when dried will remain dry for a reasonable period of time.Potassium hydrogen phthalate may be purchased in pure form at reasonable cost. Potassiumhydrogen phthalate is a primary standard. This means that carefully prepared solutions ofknown concentration of potassium hydrogen phthalate may be used to determine, by titration,the concentration of another solution such as sodium hydroxide. In part B, we have to conduct the standardization of the Sodium Hydroxide Solutionby using titration. In this part, we need to prepare the KHP solution by adding distilled waterto the KHP. Since the KHP need to dissolve completely, so we have to swirl the Erlenmeyerflask thoroughly. The mass of KHP that we weighed for three of the trials are 0.625 g for thefirst trial, 0.634 g for the second trial, and 0.627g for third trial. For the first trial, the volumeof NaOH used is 30.40 mL, for the second trial is 30.30 mL and for the third trial is 30.20mL. From the mass and volume that we get, we can calculate the molarity of the NaOHwhich is approximately 0.10 M.
11. NaOH is a strong base while KHP is a weak acid. NaOH will dissociate completely inwater to form a Na+ ion and OH- ion. Meanwhile, KHP will dissociate partial in water toform a low concentration of H+ ion. The general equation of acid and base reaction is : HA (acid) + MOH ( base) H2O + MAThe equation for the reaction of potassium hydrogen phthalate with sodium hydroxide is: KCO2C6H4CO2H + NaOH KCO2C6H4CO2Na + H2O In part C, we have to analyse of a vinegar solution. Vinegar solution is a weak acidand it contains acetic acid. Hence, when reacted with NaOH which is a strong base it willproduce a basic solution. NaOH will ionise completely in the water to form highconcentration of OH- ion. Meanwhile, acetic acid is a weak acid. It will dissociate partially inthe water to form low concentration on H+ ions. 1 mole of NaOH supplies 1 mol of OH- ionwhile 1 mole of CH3COOH supplies 1 mol of H3O+ ion. The overall reaction shows that 1mole of NaOH reacts with 1 mole of CH3COOH to form 1 mole of CH3COONa and 1 moleof H2O. This experiment involves the reaction between strong base and weak acid.Half equation :NaOH Na+ + OH-CH3COOH + H2O H3O+ + CH3COO-Overall equation :NaOH (aq) + CH3COOH (aq) CH3COONa (aq) + H2O (l) The volume of vinegar solution that we used in this experiment is 5.00 mL for thethree trials. The volume of NaOH used are 36.90 mL for the first trial, 36.20 mL for thesecond trial and 36.50 mL for the third trial. Since the molarity of the NaOH solution is 0.1M, we can calculate the molarity of vinegar which are 0.74 M, 0.72 M and 0.73 Mrespectively. The percentage mass of acetic acid in vinegar are 4.38% in the first trial, 4.30%in the second trial and 4.34% in the third trial. So, the average percentage mass of the aceticacid in vinegar is 4.34%. The following shows the „acid-base titration curve‟.
12. “The titration curve of CH3COOH (weak acid) with NaOH (strong base)” pH pH ~ 9 phenolphthalein methyl orange Volume of NaOH added ( mL) Based on the curve above, the pH starts at around pH 2.0 and not pH 1.0 becauseethanoic acid, CH3COOH, is a weak acid. As sodium hydroxide, NaOH solution is added toCH3COOH, pH of the solution increases slowly. There is a sharp increase in pH, fromapproximately pH 6.5 to pH 10.5 at the equivalence point. This sharp increase is due to theexcess of around one drop of NaOH added from the burette. The pH at the equivalence pointis the pH of the salt solution formed, which is around pH 8.5. It is because when acidneutralized , the solution remain basic since the acid‟s conjugate base remain in solution. Thisis because the salt formed, CH3COONa undergo hydrolysis. CH3COONa is a salt of a weakacid-strong base. The conjugate base CH3COO- ion undergo hydrolysis to produce alkalinesolution.CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq) At buffered region, the PH changes as the bases added. Enough base is added for halfof the acid to be converted to the conjugate base. Hence, the product‟s concentration of H+ isequal to the Ka value of the acid. Impractically, sometimes a very small difference occurs, this shows the titrationserror. The indicator and the experimental conditions should be selected so that the differencebetween the visible point and the theoretical point is so small as possible. Accuracy willindicates the closeness of the measurement to its true or accepted value and expressed by theerror. The term error refers to the absolute value of the numerical difference between the
13. known value and the experimental value. The closer the percent error is to zero, then themore accurate your experimental value (i.e., the more closely the experimental value agreeswith the known value). Accuracy calculated as follows : Precision is obtained through measurement of replicate samples. In our experiments,we had three trials each for both experiment A and experiment B titration. For all the trial, thedifference between one trial with another trial is very small, around 0.1 to 0.4 mL differencein titration. Hence, we can say that the experiment done by our group is precise. However, naturally, in every experiment, there might be a small error done. Thismight be either due to systemic error or random error. In the experiment, some error such asdifference in the rate of swirling between each trial, and error while weighing the KHP onelectronic balance will affect the expected result. Another error is when recording anincorrect initial volume of NaOH solution, such as recording the initial volume as 0.00 mL ifthe level of solution was actually higher than the 0.00 mL on the burret. The excess NaOHsolution above the 0.00 mL mark would result in more NaOH solution delivered than isactually recorded based on the endpoint. Because an incorrectly low volume of NaOHdelivered will be recorded, the resulting calculated molar concentration of acetic acid will beincorrectly low as well. Thus, correct technique is essential for obtaining good data andaccurate and precise results in this experiment.Post-lab questionsGive the definition of indicators An acid-base indicator (also known as pH indicator) such as litmus paper orphenolphthalein is a water soluble dye that changes colour due to the concentration ofhydrogen ion in the solution to which the indicator is added. It is used to indicate thecompletion of a chemical reaction or to indicate the presence of acid or alkali or the degree ofreaction between two or more substances. The following equilibrium is established betweenthe indicator, HIn and its conjugate base (In-) (Tan, 2010). HIn (aq) + H2O (l) H3O+ (aq) + In- (aq) acid (colour A) conjugate base (colour B)
14. Suppose a NaOH solution were to be standardized against pure solid primary standardgrade KHP. If 0.4538g of KHP requires 44.12 mL of the NaOH to reach aphenolphthalein endpoint, what is the molarity of the NaOH solution.Given that for KHP, mass = 0.4538 g, molar mass = 204.2for NaOH, volume = 44.12 mL (0.04412 L)Number of mole for KHP = mass, g Molar mass, g mol-1Number of mole for KHP = 0.4538 g 204.2 g mol-1Number of mole for KHP = 0.002219 molThe number of mole for KHP = number of mol for NaOHHence, the number of mole for NaOH = 0.002219 molMolarity of NaOH = number of mole, mol Volume, LMolarity of NaOH = 0.002219 mol 0.04412 LMolarity of NaOH = 0.05040 mol L-1Commercial vinegar is generally 5.0+/- 0.5% acetic acid by weight. Assuming this to bethe true value for your sample, by how much were you in error in your analysis?% Error = 4.50 - 4.34 4.50 = 3.56 %ConclusionA titration is a valid form of finding the concentration of a substance within a solution. In thisexperiment, a sample of vinegar was analyzed via titration with a standard 0.10 M NaOHsolution. The vinegar‟s molar concentration was determined to be 0.73 M, and its masspercent concentration of acetic acid was determined to be 4.34 %, which gives a percentdifference of 4% compared to the manufacturer‟s reported acetic acid content of 5.0+/- 0.5%.
16. FACULTY OF RESOURCE SCIENCE AND TECHNOLOGY DEPARTMENT OF CHEMISTRY STK 1094 - Analytical Chemistry EXPERIMENT NO : 1 TITLE OF EXPERIMENT : ACID – BASE TITRATIONSDATE OF EXPERIMENT : 27 SEPTEMBER 2012 RAHMAH FADILAH SJAM‟UN (39663)GROUP MEMBERS & RASYIDAH BINTI RAMLEE (38458)MATRIC NUMBERS : UMMI SYAHIDA ZAMRI (39213) RAFIZA SHAFINA BINT ROWTHER NEINE (38444) ADITA LIA (39664)LAB FACILITATOR :REPORT DUE DATE : 11 OCTOBER 2012