1 LAB REPORT ON ACID-BASE TITRATIONS
Aim:
To determine the weight-to-volume percent of
acetic acid present in commercial vinegar and the weight-to-volume percent of
ammonia present in window cleaners. Acid-base titrations can be conducted using
the above samples against standardized aqueous NaOH and standardized aqueous
HCl solutions respectively.
2. Introduction:
Acetic acid, commonly known as ethanoic acid CH3COOH, is found
in commercial vinegar. It is a weak organic acid that gives vinegar its
characteristic pungent smell, sour taste and slight acidity. Being an acid,
vinegar is used as a mild disinfectant in cleaning agents as well as a food
preservative. Ammonia, NH3, is used in household and industrial
cleaning agents due to its ability to solubilize grease. In
this experiment, we shall use acid-base titration to determine the
weight-to-volume percentage of acetic acid and ammonia. An acid-base titration
is a process of obtaining quantitative information of a sample using an
acid-base reaction by reacting with a certain volume of reactant whose
concentration is known. A suitable indicator for determining the equivalence
point is used to indicate the end point of an acid-base titration. To achieve
accurate results, multiple titrations will be carried out until duplicate
determinations agree to within 0.05mL of each other.
4. Results & Calculations:
4.1 Standardization of NaOH
Molarity of KHC8H4O4
is 0.05044 moldm-3 = 
= 0.05044 moldm-3)
|
Titration
|
1
|
2
|
3
|
|
Final volume of titrant/cm3
|
5.00
|
10.00
|
14.80
|
|
Initial volume of titrant/cm3
|
0.00
|
5.00
|
10.00
|
|
Volume of titrant used/cm3
|
5.00
|
5.00
|
4.80
|
|
|
|
|
|
KHC8H4O4 (aq) +
NaOH (aq)
KHC8H4O3
- Na+ (aq) + H2O (l)
Average volume of NaOH titrate used = ½ (5.00 + 5.00)
= 5.00 cm3
No. of moles of KHC8H4O4 reacted
= 0.01 × 0.05044
= 5.044 x 10-4 mol
From the equation, mole ratio of NaOH ≡ KHC8H4O4
is 1:1,
No. of moles of NaOH reacted
=
5.044 x 10-4 mol
Concentration of NaOH = moles of
NaOH/ volume of NaOH
=
5.044 x 10-4 / 
= 0.1009 moldm-3
Molarity of NaOH is: 0.1009moldm-3
4.2 Standardization of HCl
|
Titration
|
1
|
2
|
3
|
|
Final volume of titrant/cm3
|
10.80
|
10.75
|
10.70
|
|
Initial volume of titrant/cm3
|
0.00
|
0.00
|
0.00
|
|
Volume of titrant used/cm3
|
10.80
|
10.75
|
10.70
|
|
|
|
|
|
HCl (aq) + NaOH (aq)
NaCl
(aq) + H2O (l)
Average NaOH titrate used = ½ (10.70
+ 10.75)
= 10.73 cm3
No. of moles of NaOH reacted
= (10.73 ÷ 1000) × 0.1009
= 1.082 x 10-3 mol
From equation, mole ratio of NaOH ≡ HCl is 1:1,
no. of moles of HCl reacted
=
1.082 x 10-3 mol
Hence concentration of HCl = 1.082 x 10-3 (1000 ÷ 10.00)
= 0.1082 moldm-3
Molarity of HCl is: 0.1082 moldm-3
4.3 Determination of acetic acid in vinegar
|
Titration
|
1
|
2
|
3
|
|
Final volume of titrant/cm3
|
12.15
|
12.20
|
12.20
|
|
Initial volume of titrant/cm3
|
0.00
|
0.00
|
0.00
|
|
Volume of titrant used/cm3
|
12.15
|
12.20
|
12.20
|
|
|
|
|
|
CH3COOH (aq) + NaOH (aq)
CH3COO-Na+
(aq) + H2O (l)
Average volume of NaOH titrate used = ½ (12.20 + 12.20)
= 12.20 cm3
No. of moles of NaOH reacted =
(12.20 ÷ 1000) × 0.1009
=
1.231 x 10-3 mol
From equation, mole ratio of NaOH ≡ CH3COOH is
1:1,
no. of moles of CH3COOH reacted
= 1.231 x 10-3
mol
No. of moles of CH3COOH in volumetric flask (diluted) = (100 ÷ 10) × 1.231 x 10-3
= 1.231 x 10-2 mol
No. of mole of CH3COOH in 25mL (undiluted) = 1.231 x 10-2 mol
Molarity of acetic acid = (1.231 x 10-2 mol) ÷(25 x 10-3)=0.4924
moldm-3
Molar mass of CH3COOH = 2(12.01) + 4(1.0079) + 2(16.00)
= 60.0516 g mol-1
Therefore, weight-volume percentage of CH3COOH in vinegar
= (0.4924 x 60.0516 ÷ 1000) × 100% = 2.96%
Percentage of acetic acid in the vinegar is: 2.96%
Indicator used: phenopthalein
4.4 Determination of ammonia in window cleaner
|
Titration
|
1
|
2
|
3
|
|
Final volume of titrant/cm3
|
4.85
|
9.65
|
4.85
|
|
Initial volume of titrant/cm3
|
0.00
|
4.85
|
0.00
|
|
Volume of titrant used/cm3
|
4.85
|
4.80
|
4.85
|
|
|
|
|
|
NH3 (aq) + HCl (aq)
NH4+Cl-
(aq)
Average volume of HCl titrate used = ½ (4.85 + 4.85)
= 4.85 cm3
No. of moles of HCl reacted = 0.1082 x (4.85 x10-3)
=
5.248 x 10-4 mol
From equation, mole ratio of HCl ≡ NH3 is 1:1,
no. of moles of NH3 reacted = 5.248 x 10-4 mol
No. of moles of NH3 in volumetric flask (diluted) = (100 ÷ 10) ×
5.248 x 10-4 mol
= 5.248 x 10-3 mol
No. of mole of NH3 in 10mL (undiluted) = 5.248 x 10-3 mol
Concentration of NH3 = (5.248 x 10-3) ÷
(10 x 10-3) = 5.248 x 10-1 moldm-3
Molar mass of NH3 = 14.01 + 3 (1.0079)
= 17.0337 g mol-1
Therefore, weight-volume percentage of NH3
in window cleaner
= (17.0337 x 5.248 x 10-1)/ 103 × 100%
= 0.894%
Percentage of ammonia in the window cleaner is: 0.894%
Indicator used: methyl red
5. Discussion:
The main reaction in the series of acid-base titrations is neutralization.
During neutralization, the acid and base reacts to form a salt and water (i.e.
H+ + OH- H2O).
As the all the reactants are colourless, an appropriate indicator must be added
to indicate the colour change at the end point.
|
For Experiment 3.3, the indicator phenolphthalein was chosen as it was a
weak acid-strong base titration, as acetic acid is a weak acid and NaOH is a
strong base. At equivalence point, the resulting solution contains only CH3COONa.
Thus salt hydrolysis occurs and the pH at equivalence point is > 7.
Phenolphthalein changes colour at a pH range of 8.2 – 10.0.
The (pH)end point falls within the pH for colour change of
phenolphthalein in the titration. The colour of phenolphthalein changes
explicitly from pink to colourless and the colour change is over a narrow pH
range. This further enhances its suitability as an indicator for this
titration.
Salt hydrolysis that raises equivalence point to > 7 pH :
CH3COO- + H2O
CH3COOH
+ OH-
|
 *
The pH titration curve of weak acid (CH3COOH) and strong base
(NaOH)
|
|
However, for Experiment 3.4 a different indicator was chosen. Methyl red
was used instead as it was a case of a strong acid-weak base titration, as
HCl is a strong acid and NH3 is the weak base. At equivalence
point, the resulting solution contains only NH4Cl. Thus salt
hydrolysis occurs and the pH at equivalence point is < 7. Methyl red
changes colour over a pH range of 4.8 – 6.0.
The (pH)end point falls within the pH range for colour change
of methyl red in this titration. The colour of methyl red changes explicitly
from red to yellow and the colour change is over a narrow pH range. This
further enhances its suitability as an indicator for this titration.
Salt hydrolysis that lowers equivalence point to < 7 pH :
NH4Cl + H2O
NH3
+ H3O+
|
 *
The pH titration curve of weak base (NH3) and strong acid
(HCl)
|
Strong acid-weak base and weak acid-strong
base titrations are employed in this experiment instead of weak acid-weak base
titrations as the hydrolysis of a salt of a weak acid-weak base is incomplete.
The hydrolysis of the conjugate base results in an acidic solution and at the same
time, hydrolysis of the conjugate acid of the salt gives rise to a basic
solution. Hence, the resulting solution at equivalence point can be neutral,
acidic or basic depending on which hydrolysis occurs at a greater extent. This
means that calculations have to be done preceding the titration in order to
select the appropriate indicators. Therefore, weak acid-weak base titration is
not commonly used to determine the quantitative information of an unknown
sample.
While titrating, the conical flask has to
be swirled constantly throughout to ensure that the contents are mixed evenly.
Also, some titrant may drip onto the sides of the conical flask and may not
react with the solution in the conical flask. This reduces the accuracy of the
results as extra titrant would be used to achieve the equivalence point. To
prevent this, deionised water can be used to wash down the unreacted titrant
when nearing the equivalence point and the conical flask should be swirled
before continuing with the titration. To achieve consistent results, one
important factor is to add in the titrant in a drop wise manner before the
equivalence point.
Deionised water was used not only for
diluting the sample but also for washing the apparatus, instead of using tap
water. This is because deionized water is pH neutral and will not negatively
affect the results of the acid-base titrations. This is important as acid-base
titrations are pH sensitive reactions.
Multiple titrations were carried out until duplicate determinations agree
to within 0.05mL of each other. This ensures that the results are not once-off
outliers. By averaging the two most consistent and accurate results, the
probability of incurring random errors was accounted for and decreased.
Other factors that would affect the accuracy of the results include
parallax error when taking the readings and making sure that no air bubbles are
present within the pipette and burette while titrating.
6. Conclusion:
Using acid-base chemistry, repeated titrations of vinegar sample and window
cleaner sample was carried out against standardized solutions of NaOH and HCl
respectively. Through stoichiometric relations, the concentration and
weight-to-volume percentage of acetic acid in vinegar and ammonia in window
cleaner can be found. The percentage of acetic acid in the vinegar sample
studied was deduced to be 3.96%. The percentage of ammonia in the window
cleaner sample is 0.894%.
