using eglish to report

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, NH_3, 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%.