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FACULTY OF RESOURCE SCIENCE AND TECHNOLOGY
DEPARTMENT OF CHEMISTRY

STK 1094 – Analytical Chemistry 1

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EXPERIMENT NO :
1

TITLE OF EXPERIMENT :
ACID – BASE TITRATIONS

DATE OF EXPERIMENT
:
27 SEPTEMBER 2013

GROUP MEMBERS &
MATRIX NUMBERS

:

1. DELVINO DISONNEO ANAK DALIN (44903)
2. GOH CHIA HONG (44927)
3.LINDA CHONG (41889)
4.LIM WHYE KIT LEONARD(41874)
5.

LAB FACILITATOR
:

REPORT DUE DATE
:
4 OCTOBER 2013

INTRODUCTION
The laboratory method used in the experiment 1 is titration. The technique of titration finds many applications, but is especially useful in the analysis of acidic and basic substances. Titration involves measuring the exact volume of a solution of known concentration that is required to react with a measured volume of a solution of unknown concentration, or with a weighed sample of unknown solid.

A solution of accurately known concentration is called a standard solution. In titration, standard solutions are now usually expressed in terms of molarity, M :

In many cases it is possible to prepare a standard solution by accurate weighing of the solute, followed by precise dilution to an exactly known volume in a volumetric flask.

One of the most common standard solutions used in acid-base titration analysis, however, cannot be prepared in this manner. Solutions of sodium hydroxide are commonly used in titration analysis of samples containing an acidic solute. Although sodium hydroxide is a solid, it is not possible to prepare standard sodium hydroxide solutions by mass. Solid sodium hydroxide is usually of questionable purity. Sodium hydroxide reacts with carbon dioxide from the atmosphere and is also capable of reacting with the glass of the container in which it is provided. For there reasons, sodium hydroxide solutions are generally prepared to be approximately a given concentration. They are then standardized by titration of a weighed sample of a primary standard acidic substance.

By measuring how many mL of the approximately prepared sodium hydroxide are necessary to react completely with a weighed sample of a primary standard acidic substance, the concentration of the sodium hydroxide solution can be calculated. Once prepared, however, the concentration of a sodium hydroxide will change with time ( for the same reasons stated above ). As a consequence, sodium hydroxide solutions must be used relatively quickly. In titration analysis, there must be some means of knowing when enough titrant has been added to react exactly and completely with the sample being titrated.

In an acid-base titration analysis, there should be a sudden change in pH when the reaction is complete. For example, if the sample being titrated is and acid, then the titrant to be used will be basic. When one excess drop of titrant is added, the solution being titrated will suddenly become basic. There are various natural and synthetic dyes, called indicators, which exist in different coloured forms at different pH values. A suitable indicator can be chosen that will change at a pH value consistent with the point at which the titration reaction is complete. The indicator to be used in this experiment is phenolphthalein, which is colourless in acidic solutions, but changes to a pink form at basic pH.

OBJECTIVES
This experiment is carried out to demonstrate the basic laboratory technique of titration toward the students. It also gives a chance for students to have a hands-on practice about titration. This will help the students to know how to handle the materials, apparatus and the best way to conduct the experiment. By carrying out this experiment, it also teach the student about the correct step to calculate the percentage of substance, molarity of solution and other calculations. Moreover, it also helps to remind the student to be careful to avoid from making any error.

PROCEDURES
Part A : Preparation of the Sodium Hydroxide Solution
1. A 500mL volumetric flask and stopper is cleansed and rinsed. The flask is then labeled “Approx. 0.1 M NaOH”. About 450 mL of distilled water is put into the flask. 2. Approximately 2.0 g of sodium hydroxide pellets is weighed out and transferred into a beaker with 50mL of distilled water. The pellets are dissolved and the solution is transferred into the 500mL volumetric flask. The flask is then stoppered and shook to mix the solution. 3. Additional distilled water is then added into the flask up until the mark on the neck of the flask when all the sodium hydroxide pellets have dissolved. The flask is stoppered and shook thoroughly to mix the solution.

Part B : Standardization of the Sodium Hydroxide Solution
1. The burette is set up in the burette clamp. The burette is rinsed and cleaned with the freshly prepared sodium hydroxide solution.

2. Three 250mL Erlenmeyer flasks are cleaned with water, and are then rinsed with distilled water. The three flasks are labeled as 1, 2 and 3.

3. The bottle of dried KHP is removed from the oven. Three samples of KHP between 0.6 and 0.8 g are
weighed one for each of the Erlenmeyer flasks when the KHP is completely cool. The exact weights of each KHP samples are recorded to the nearest mg (±0.001 g).

4. 100mL of distilled water is added to KHP sample 1. 2-3 drops of phenolphthalein indicator solution are then added. The KHP sample is swirled to dissolve completely.

5. the initial reading of the NaOH solution in the burette is recorded to the nearest 0.02mL.

6. NaOH solution from the burette is added to the sample 1 in the Erlenmeyer flasks, and during the addition the flask is swirled constantly.

7. NaOH solution is added one drop at a time with constant swirling until one single drop of NaOH will 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. Steps 4-7 are repeated by using the other 2 KHP samples. 9. With the given molecular mass of KHP which is 204.2, the number of moles of KHP in samples 1, 2 and 3 are calculated. 10. From the number of moles of KHP present in each samples, and from the volume of NaOH solution used to titrate the samples, the molar concentrations (M) of NaOH in the titrant solution is calculated. The reaction between NaOH and KHP is of 1:1 stoichiometry.

Part C : Analysis of a Vinegar Solution

Vinegar 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 labeled as samples 1, 2 and 3. 2. The 5mL pipette is rinsed with small portions of the vinegar solution and the rinsing is discarded. 3. 5 mL of the vinegar solution is pipetted into each of the Erlenmeyer flasks using the pipette. About 100mL of distilled water and 2-3 drops of phenolphthalein indicator solution is then added into each flask.

4. The burette is refilled with the NaOH solution and the initial reading of the burette is recorded to the nearest 0.02mL. Sample 1 of vinegar is titrated in the same manner as in the standardization until one drop of NaOH causes the appearance of the pale pink colour.

5. The final reading of the burette is recorded to the nearest 0.02mL.

6. The titration for the other two vinegar samples is repeated.

7. The molar concentration of the vinegar solution is calculated based on the volume of vinegar sample taken, and on the volume and average concentration of NaOH titrant used.

8. The percent by mass of acetic acid in the vinegar solution is calculated given that the formula mass of acetic acid is 60.0 and the density of the vinegar solution is 1.01 g/mL.

MATERIAL
Sodium hydroxide pellets
Potassium hydrogen phthalate (KHP)
Phenolphthalein
Unknown vinegar

APPARATUS
Burette with stand
Pipette
500ml volumetric flask with stopper
250ml Erlenmeyer flasks
Retort stand with clamp

RESULTS
PART B : Standardization of the Sodium Hydroxide Solution

Particulars
Trial 1
Trial 2
Trial 3
Mass of KHP taken (g)
0.617
0.619
0.622
Final burette reading (mL)
30.00
31.20
31.2
Initial burette reading (mL)
0.00
0.00
0.00
Volume of NaOH used (mL)
30.00
31.2
31.2
Molarity of NaOH solution
0.10
0.10
0.10
Average molarity of NaOH solution
0.10

PART C : Analysis of a Vinegar Solution

Particulars
Trial 1
Trial 2
Trial 3
Volume of vinegar solution used (mL)
5.00
5.00
5.00
Final burette reading (mL)
37.4
36.8
37.0
Initial burette reading (mL)
0.00
0.00
0.00
Volume of NaOH used (mL)
37.4
36.8
37.0
Molarity of NaOH solution
0.10
0.10
0.10
Molarity of vinegar solution
0.748
0.736
0.740
% mass of acetic acid in vinegar
4.44
4.38
4.40
Average molarity of vinegar solution
0.742
Average % mass of acetic acid in vinegar
4.40

CALCULATION
Part B : Standardization of the Sodium Hydroxide Solution
Trial 1 :
Molecular mass of KHP is 204.2 g mol-1, its volume is given by 100mL (0.1L) Mass = number of mole × molecular mass
0.617 g = no of mole of KHP × 204.2 g mol-1
No. of mole for KHP =
No. of mole for KHP = 0.00302 g mol-1
Molarity of KHP =
=
= 0.0302 mol L-1
From the data collected,
MaVa = MbVb
(0.0302 mol L-1)(0.1 L) = Mb (0.0300 L)
Mb =
= 0.10 mol L-1
Trial 2 :
Molecular mass of KHP is 204.2 g mol-1, its volume is given by 100mL (0.1L)
Mass = number of mole × molecular mass
0.619 g = no of mole of KHP × 204.2 g mol-1
No. of mole for KHP =
No. of mole for KHP = 0.00303 g mol-1
Molarity of KHP =
=
= 0.0303 mol L-1
From the data collected,
MaVa = MbVb
(0.0303 mol L-1)(0.1 L) = Mb (0.0312 L)
Mb =
= 0.10 mol L-1
Trial 3 :
Molecular mass of KHP is 204.2 g mol-1, its volume is given by 100mL (0.1L) Mass = number of mole × molecular mass
0.622 g = no of mole of KHP × 204.2 g mol-1
No. of mole for KHP =
No. of mole for KHP = 0.00305 g mol-1
Molarity of KHP =
=
= 0.0305 mol L-1
From the data collected,
MaVa = MbVb
(0.0305 mol L-1)(0.1 L) = Mb (0.0312 L)
Mb =
= 0.10 mol L-1

Calculation Part C : Analysis of a Vinegar Solution
Half equations :
NaOH Na+ + OH-
CH3COOH + H2O H3O+ + CH3COO-
Overall equation :
NaOH (aq) + CH3COOH (aq) CH3COONa (aq) + H2O (l)

Trial 1 :
Moles = volume in litre × molarity
No. of mole of NaOH = 0.0374 L × 0.1 mol L-1
= 0.00374 mol
From the overall equation, 1 mol of NaOH = 1 mol of CH3COOH
So, the number of mole for CH3COOH = 0.00374 mol
Molarity for CH3COOH =
=
= 0.748 mol L-1
Mass of CH3COOH = no. of mol × molar mass
= 0.00374 mol × 60.0 amu
= 0.224 g
Given density of vinegar solution is 1.01 g/mL
Density =
Mass of vinegar solution = Density × Volume
=1.01 g/mL × 5.00 mL
= 5.05 g
% by weight = × 100× 100%
= × 100%
= 4.44 %

Trial 2 :
Moles = volume in litre × molarity
No. of mole of NaOH = 0.0368 L × 0.1 mol L-1
= 0.00368 mol
From the overall equation, 1 mol of NaOH = 1 mol of CH3COOH
So, the number of mole for CH3COOH = 0.00368 mol
Molarity for CH3COOH =
=
= 0.736 mol L-1
Mass of CH3COOH = No of mole × molar mass
= 0.00368 mol × 60.0 amu
= 0.221 g
Given density of vinegar solution is 1.01 g/mL
Density =
Mass of vinegar solution = Density × Volume
=1.01 g/mL × 5.00 mL
= 5.05 g
% by weight = × 100× 100%
= × 100%
= 4.38 %

Trial 3 :
Moles = volume in litre × molarity
No. of mole of NaOH = 0.0370 L × 0.1 mol L-1
= 0.00370 mol
From the overall equation, 1 mol of NaOH = 1 mol of CH3COOH
So, the number of mole for CH3COOH = 0.00370 mol
Molarity for CH3COOH =
=
= 0.740 mol L-1
Mass of CH3COOH = No of mole × molar mass
= 0.0037 × 60.0 amu
= 0.222 g
Given density of vinegar solution is 1.01 g/mL
Density =
Mass of vinegar solution = Density × Volume
=1.01 g/mL × 5.00 mL
= 5.05 g
% by weight = × 100%
= × 100%
= 4.40 %

DISCUSSION
In part A, we need to dissolve completely the sodium hydroxide pellets in distilled water through shaking the volumetric flask with a stopper. Sodium hydroxide can be easily contaminated as it absorb the moisture from the air. Potassium hydrogen phthalate on the other hand, has a lower tendency to adsorb water from the air and when dried will remain dry for a reasonable time period of time. Potassium hydrogen phthalate is a primary standard, which means that carefully prepared solutions of known concentration of potassium hydrogen phthalate may be used to determine, by titration, the concentration of another solution, in this case, it is sodium hydroxide solution.

In Part B, we have to conduct the standardization of sodium hydroxide solution by using titration. We are required to prepare the KHP solution by adding distilled water to the KHP. We have to ensure the KHP dissolve completely by swirling the Erlenmeyer flask thoroughly. The mass of KHP that we weighed for three trials are 0.617g, 0.619g and 1.622g respectively. For the first trial, the volume of NaOH used is 30.00mL, 31.2mL for the second trial while for the third trial, it is 31.2mL. The molarity of NaOH is then calculated to be 0.10M, using the mass and volume that we get. NaOH which is a strong base will dissociate completely in water to form a Na+ ion and OH- ion. At the same time, KHP which is a weak acid will dissociate partially in water releasing low concentration of H+ ion. General acid base equation :

HA (acid) + BOH (base) H2O + BA
Equation of the reaction between potassium hydrogen phthalate (KHP) and sodium hydroxide is:
KCO2C6H4CO2H + NaOH KCO2C6H4CO2Na + H2O
In part C, vinegar solution is an acid which contain acetic acid. As a weak acid, vinegar will produce a basic solution when react with NaOH (strong base). 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 5.00mL for the three trials. The volume of NaOH used are 37.4mL for the first trial, 36.8mL for the second trial, as for the third trial, it is 37.0mL. Since the molarity of NaOH has not changed (maintained at 0.10M), we can deduce the molarity of vinegar which are 0.748M, 0.736M and 0.740M respectively. The percentage mass of acetic acid in vinegar are also calculated and the values are obtained. For the first trial, the percentage of acetic acid in vinegar is 4.44%, 4.38% of percentage mass of acetic acid in vinegar is calculated for the second trial, as for the third trial, the obtained value is 4.40%. “The titration curve of CH3COOH (acetic acid) with NaOH” pH

Volume of NaOH added (mL)
Indicator: phenolphthalein

Based on the acid-base titration curve above, the pH start at around pH 2.0 and not pH 1.0 because ethanoic acid is a weak acid. When sodium hydroxide is added to ethanoic acid, pH of the solution increases gradually. A sharp increase in pH can be seen at around pH 6.5 to pH10.5 at the equivalence point. This is due to the excess of about one drop of NaOH added from the burette. The pH at the euivalence point is the pH of the salt solution formed, which is around pH 8.5. Ethanoic acid dissociates in water incompletely to form ethanoic ion and hydroxide ion. The ethanoic ion undergo hydrolysis to produce alkaline solution.

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

After the experiment is done, we have used a method to determine the deviation of our result from the actual result, to ensure that our experimental results are both precision and accurate. The percentage error is calculated using the formula below:

Known value – experimental value X 100% = % error Known value

The closer the percentage error is to zero, the more accurate the experimental value. For our experiment, our percentage error calculated ranged between 1.33% and 2.67%, which proves that our results are considered accurate enough.

However, in every experiment, some small errors are impossible to be avoided, this might be either caused by systematic error or random error. There are few aspect that could have been affecting the accuracy of this experiment, such as the different rate of swirling for each trial, and slight error due to wind and humidity when taking the weight of KHP on the electronic balance. When reading the volume of NaOH from the burette, the positon of eye level may be differed by 0.02mL, the slight difference in reading the volume may result in deviation of the experimental results from the actual results.s Thus, correct technique is essential for obtaining good data and accurate results in this experiment. Continuos practice, too, can help to improve our skills in conducting the experiment in a more standard and proper way.

CONCLUSION
The titration can be used to find the concentration of a substance within a solution. In this experiment, a sample of vinegar was analyzed via titration with a standard 0.10M NaOH solution. The molar concentration of vinegar was calculated to be 0.742M, its mass percent concentration of acetic acid was 4.40%, which is 2.22% different from the manufacturer’s acetic acid content of 5.0±0.5%.

POST-LAB QUESTIONS
1. Give the definition of indicators.

Indicators ( in acid-base titration ) 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.

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

Mass of KHP given = 0.4538g
Molar mass = 204.2
Volume of NaOH = 0.04412L

Number of mole =

Number of mole for KHP = = 0.002219 mol

The number of mole for KHP is equal to the number of mole for NaOH. the number of mole for NaOH is 0.002219 mol.

Molarity =

Molarity of NaOH = = 0.05040 mol L-1

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

% of error = × 100%
= × 100%
= 2.22%

REFERENCES
Tan.Y.T., Loh.W.L., Kathirasan Muniandy. (2011) Ace ahead chemistry volume 1. Selangor, Malaysia ; Oxford Fajar. Tan.Y.T.,Ashy Kumren. (2011) Chemistry for Matriculation, Selangor, Malaysia : Oxford Fajar.

Cite this Chemistry report Essay

Chemistry report Essay. (2016, Jun 02). Retrieved from https://graduateway.com/chemistry-report/

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