chemistry report final

Introduction:

The laboratory method used in the experiment is titration. Titration is a method used in measuring the amount of an analytical reagent necessary to react quantitatively with the sample. Acid-base titrations are important for counting concentrations of acids and bases. The point at which the equivalence point has been reached is called the theoretical end point. The titration is detected by some physical change, such as colour change, to indicate the end point. The indicator solution used in this experiment is phenolphthalein which is a colourless in 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 much indicator will necessitate an indicator blank. The indicator should give a clear visual change in the liquid being titrated once the reaction between the standard solution and the substance is practically complete. The indicators can change colour because their ions have colours that are different from the undissociates molecule (Tan, 2010). The acid and base titrations used the Arrhenius theory. This theory stated that acid are substance which produce hydrogen ions in 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 litres

Objective :

One of the purposes of this experiment is to demonstrate the basic laboratory technique of titration towards the students. This experiment enables a hand-on practice done by 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 any experimental error. Moreover, this experiment also teach the students on the correct steps to calculate percentage of substance, molarity of solution and other calculations involve throughout the experiment.

Apparatus :

Burette with stand Pipette1-L volumetric flask with stopper250 mL Erlenmeyer flasksRetort stand with clamp

Reagents :Sodium hydroxide pelletsPotassium hydrogen phthalate (KHP)PhenolphthaleinUnknown vinegar

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 the phenolphthalein 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.

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 Solution

Particulars 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 0.10

Part B : Analysis of a Vinegar Solution

Particulars 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

Calculation : (Part A)

Equation for the reaction between NaOH and KHP.

KCO2C6H4CO2H + NaOH KCO2C6H4CO2Na + H2O

Given that the reaction between NaOH and KHP is of 1 : 1 stoichiometry.

Let A : KHP and B : NaOH

MAVA 1

MBVB 1

MAVA = MBVB

Trial 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 mass

0.625 g = no of mole for KHP X 204.2 g mol-1

No. of mole for KHP = 0.625 g

204.2 g mol-1

No. of mole for KHP = 0.00306 mol

Molarity of KHP = No. of mole

Volume of KHP

Molarity of KHP = 0.00306 mol

0.1 L

Molarity of KHP = 0.0306 mol L-1

From 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

Trial 2 :





204.2 g mol-1



Volume of KHP


0.1 L



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

Trial 3 :





204.2 g mol-1



Volume of KHP


0.1 L



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-1

Calculation : (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- ion

1 mole of CH3COOH supplies 1 mol of H3O+ ion

NaOH (aq) + CH3COOH (aq) CH3COONa (aq) + H2O (l)

Trial 1 :

Moles = volume in litre X molarity

= 0.0369 L X 0.10 mol L-1

=0.00369 mol

Number of mole for NaOH = 0.00369 mol

Number of mole for NaOH = Number of mole for CH3COOH

So, the number of mole for CH3COOH = 0.00369 mol

Molarity for CH3COOH = Number of mole

Volume

Molarity for CH3COOH = 0.00369 mol

0.005 L

= 0.738 mol L-1

Molarity of vinegar is 0.738 mol L-1

This 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 g

Mass of vinegar = 1.010 kg = 1010 g

% by weight = mass of acetic acid

Mass of vinegar

= 44.28

1010

= 4.38 %

Trial 2 :


= 0.0362 L X 0.10 mol L-1

=0.00362 mol





Volume


0.005 L

= 0.724 mol L-1




= 0.724 mol L X 60.0

= 43.44 g



Mass of vinegar

= 43.44

1010

= 4.30 %

Trial 3 :


= 0.0365 L X 0.10 mol L-1

=0.00365 mol





Volume


0.005 L

= 0.730 mol L-1




= 0.730 mol L X 60.0

= 43.80 g



Mass of vinegar

= 43.80

1010

= 4.34 %

Discussion

The acid and base titrations used the Arrhenius theory. This theory stated that acid are substance which produce hydrogen ions in solution and bases are substance which produce hydroxide ion in solution.

Acid-base indicator is a weak organic acids or bases that dissociate slightly in aqueous solutions to form ions (Tan, 2010). The indicator that we used in this experiment is the phenolphthalein, where the ph range is around 8.2 to 10.0. The colour change from

colourless to a light pink. The following equilibrium is established between the indicator, HIn and 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 neither the 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 hydroxide with distilled water to be used in the experiment. In order to make sure that the sodium hydroxide pellets dissolved completely, we have to shake the flask thoroughly. Some chemicals 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 is easily contaminated. Thus if a solution of sodium hydroxide is prepared by weighing the sodium hydroxide, the concentration of the solution may not be precisely the intended concentration. Potassium hydrogen phthalate on the other hand, has a lesser tendency to absorb 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. Potassium hydrogen phthalate is a primary standard. This means that carefully prepared solutions of known 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 Solution by using titration. In this part, we need to prepare the KHP solution by adding distilled water to the KHP. Since the KHP need to dissolve completely, so we have to swirl the Erlenmeyer flask thoroughly. The mass of KHP that we weighed for three of the trials are 0.625 g for the first trial, 0.634 g for the second trial, and 0.627g for third trial. For the first trial, the volume of NaOH used is 30.40 mL, for the second trial is 30.30 mL and for the third trial is 30.20 mL. From the mass and volume that we get, we can calculate the molarity of the NaOH which is approximately 0.10 M.

NaOH is a strong base while KHP is a weak acid. NaOH will dissociate completely in water to form a Na+ ion and OH- ion. Meanwhile, KHP will dissociate partial in water to form a low concentration of H+ ion. The general equation of acid and base reaction is :

HA (acid) + MOH ( base) H2O + MA

The 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 acid and it contains acetic acid. Hence, when reacted with NaOH which is a strong base it will produce a basic solution. NaOH will ionise completely in the water to form high concentration of OH- ion. Meanwhile, acetic acid is a weak acid. It will dissociate partially in the water to form low concentration on H+ ions. 1 mole of NaOH supplies 1 mol of OH- ion while 1 mole of CH3COOH supplies 1 mol of H3O+ ion. The overall reaction shows that 1 mole of NaOH reacts with 1 mole of CH3COOH to form 1 mole of CH3COONa and 1 mole of 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 the three trials. The volume of NaOH used are 36.90 mL for the first trial, 36.20 mL for the second trial and 36.50 mL for the third trial. Since the molarity of the NaOH solution is 0.1 M, we can calculate the molarity of vinegar which are 0.74 M, 0.72 M and 0.73 M respectively. 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 acetic acid in vinegar is 4.34%. The following shows the ‘acid-base titration curve’.

“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 because ethanoic acid, CH3COOH, is a weak acid. As sodium hydroxide, NaOH solution is added to CH3COOH, pH of the solution increases slowly. There is a sharp increase in pH, from approximately pH 6.5 to pH 10.5 at the equivalence point. This sharp increase is due to the excess of around one drop of NaOH added from the burette. The pH at the equivalence point is the pH of the salt solution formed, which is around pH 8.5. It is because when acid neutralized , the solution remain basic since the acid’s conjugate base remain in solution. This is because the salt formed, CH3COONa undergo hydrolysis. CH3COONa is a salt of a weak acid-strong base. The conjugate base CH3COO- ion undergo hydrolysis to produce alkaline solution.

CH3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq)

At buffered region, the PH changes as the bases added. Enough base is added for half of the acid to be converted to the conjugate base. Hence, the product’s concentration of H + is equal to the Ka value of the acid.

Impractically, sometimes a very small difference occurs, this shows the titrations error. The indicator and the experimental conditions should be selected so that the difference between the visible point and the theoretical point is so small as possible. Accuracy will indicates the closeness of the measurement to its true or accepted value and expressed by the error. The term error refers to the absolute value of the numerical difference between the known value and the experimental value. The closer the percent error is to zero, then the more accurate your experimental value (i.e., the more closely the experimental value agrees with 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, the

difference between one trial with another trial is very small, around 0.1 to 0.4 mL difference in 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. This might be either due to systemic error or random error. In the experiment, some error such as difference in the rate of swirling between each trial, and error while weighing the KHP on electronic balance will affect the expected result. Another error is when recording an incorrect initial volume of NaOH solution, such as recording the initial volume as 0.00 mL if the level of solution was actually higher than the 0.00 mL on the burret. The excess NaOH solution above the 0.00 mL mark would result in more NaOH solution delivered than is actually recorded based on the endpoint. Because an incorrectly low volume of NaOH delivered will be recorded, the resulting calculated molar concentration of acetic acid will be incorrectly low as well. Thus, correct technique is essential for obtaining good data and accurate and precise results in this experiment.

Post-lab questions

Give the definition of indicators

An acid-base indicator (also known as pH indicator) such as litmus paper or phenolphthalein is a water soluble dye that changes colour due to the concentration of hydrogen ion in the solution to which the indicator is added. It is used to indicate the completion of a chemical reaction or to indicate the presence of acid or alkali or the degree of reaction between two or more substances. The following equilibrium is established between the indicator, HIn and its conjugate base (In-) (Tan, 2010).

HIn (aq) + H2O (l) H3O+ (aq) + In- (aq)

acid (colour A) conjugate base (colour B)

Suppose a NaOH solution were to be standardized against pure solid primary standard grade KHP. If 0.4538g of KHP requires 44.12 mL of the NaOH to reach a phenolphthalein endpoint, what is the molarity of the NaOH solution.

Given that for KHP, mass = 0.4538 g, molar mass = 204.2

for NaOH, volume = 44.12 mL (0.04412 L)

Number of mole for KHP = mass, g

Molar mass, g mol-1

Number of mole for KHP = 0.4538 g

204.2 g mol-1

Number of mole for KHP = 0.002219 mol

The number of mole for KHP = number of mol for NaOH

Hence, the number of mole for NaOH = 0.002219 mol

Molarity of NaOH = number of mole, mol

Volume, L

Molarity of NaOH = 0.002219 mol

0.04412 L

Molarity of NaOH = 0.05040 mol L-1

Commercial vinegar is generally 5.0+/- 0.5% acetic acid by weight. Assuming this to be the true value for your sample, by how much were you in error in your analysis?

% Error = 4.50 - 4.34

4.50

= 3.56 %

Conclusion

A titration is a valid form of finding the concentration of a substance within a solution. In this experiment, a sample of vinegar was analyzed via titration with a standard 0.10 M NaOH solution. The vinegar’s molar concentration was determined to be 0.73 M, and its mass percent concentration of acetic acid was determined to be 4.34 %, which gives a percent difference of 4% compared to the manufacturer’s reported acetic acid content of 5.0+/- 0.5%.

Reference Tan,Y.T., Ashy Kumren. (2011) Chemistry for Matriculation. Selangor, Malaysia : Oxford Fajar.Tan,Y.T., Loh,W.L., Kathirasan Muniandy. (2010) Ace ahead chemistry volume 1. Selangor, Malaysia : Oxford Fajar.Tan,Y.T., Loh,W.L., Kathirasan Muniandy. (2010) Ace ahead chemistry volume 2. Selangor, Malaysia : Oxford Fajar.Retrived from www.chemguide.co.uk/physical/acidbaseeqia/theories.html.

http://www.chemguide.co.uk/physical/acidbaseeqia/theories.html

FACULTY OF RESOURCE SCIENCE AND TECHNOLOGY

DEPARTMENT OF CHEMISTRY

STK 1094 - Analytical Chemistry

EXPERIMENT NO : 1

TITLE OF EXPERIMENT : ACID – BASE TITRATIONS

DATE OF EXPERIMENT : 27 SEPTEMBER 2012

GROUP MEMBERS &

MATRIC NUMBERS:

RAHMAH FADILAH SJAM’UN (39663)

RASYIDAH BINTI RAMLEE (38458)

UMMI SYAHIDA ZAMRI (39213)

RAFIZA SHAFINA BINT ROWTHER NEINE (38444)

ADITA LIA (39664)

LAB FACILITATOR :

REPORT DUE DATE : 11 OCTOBER 2012

chemistry report final

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