The Titration Of Vinegar And Bleach Solutions Biology Essay

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The aims of the experiments this time, is to find out the concentration of common food or household solution (vinegar and bleach solution in exact) by titration. And master the method of acid and base titration.

Vinegar is the ethanoic acid solution. The ethanoic acid is kind of organic compounds that the chemical formula is CH3COOH (Talk talk, 2010). The sodium hydroxide (NaOH) is a kind of base solutions that can neutralization with acid. The reaction equation is shown below (Lane, 2010a).

H+ + OH- → H2O

The main composition of bleach solution is an alkali metal hypochlorite. The alkali metal means the elements of Group I (FreePatentsOnline, 2010). Hypochlorite ion can react with iodide ion in an acid surrounding can provide iodine solution. There has two steps. The equations of this are shown below (Lane, 2010b).

First step: ClO- + 2I- + 2H+ → Cl- + I2 + H2O

Second step: I2(aq) + 2 S2O3 2-(aq) → 2 I-(aq) + S4O6 2-(aq)

Phenolphthalein is a colourless indicator that can appear a pink colour when meets base solution. The reaction is shown below (Beaumont, 2010).

Figure1. Theory of Phenolphthalein as an Indicator (Beaumont, 2010)

Method

Equipments and Medicals

Vinegar (1 dm3)

Bleach (a bottle)

Sodium hydroxide (0.1mol/ 1 dm3) (NaOH)

Phenolphthalein Indicator

Sodium thiosulphate solution (concentration: 0.1mol/ 1 dm3) (Na2S2O3)

Potassium Iodide (1mole) (KI)

Dilute Sulphuric Acid (H2SO4)

Test tubes; Funnel; Beaker; Burette; Conical flask; Pipette + pipette filler

Titration of Vinegar

Before starting the experiment, the phenolphthalein indicator was tested by acid and base solution. After pouring a little vinegar and NaOH solution each into a test tube, phenolphthalein indicator was added into each test tube. The colour changes were recorded.

The funnel was put on the top of the burette and pouring the NaOH solution into the funnel. Then, the level was adjusted to 0cm3 reading. Next, exactly 2.5cm3 vinegar was poured into the conical flask. 20cm3 water was poured into the same flask. After all these process, a few drops of phenolphthalein indicator was added into the flask. The flask with some white paper under it was put under the burette. As NaOH was starting added into the flask, the flask with mix solution was swirled. When the NaOH was down to 25cm3 level in burette, the burette was filled to the 0cm3 reading with some new NaOH using the same method that motioned before if the reaction did not come to the end point yet. As the experiment near ending point, some colour change appeared and disappeared by shaken the flask. The speed of adding NaOH was delay, one drop each time. Keep adding, until the colour occurs and not disappeared any more. Then, the end reading point was recorded and calculating the volume of NaOH for neutralization and takes a note.

The processes of experiment were repeated twice, and recording t he real initial and final burette reading.

Titration of Bleach

The burette was filled with sodium thiosulfate solution (Na2S2O3), and adjusts the reading to 0cm3. Then, 1cm3 bleach solution, 10cm3 potassium iodide solution (KI) and 10cm3 of dilute sulphuric acid (H2SO4) were add conical flask. The flask was moved under the burette with a white paper under it, the solution in the burette was released into the flask and it was swirled continuously. The starch indicator solution was added into flask just before the end-point that the colour of the solution in the flask was fades to pale yellow. After a blue-black colour appears, the Na2S2O3 solution was added dropwise and do not stop until the colour turned to colourless.

The processes of experiment were repeated twice, and recording t he real initial and final burette reading.

Result

The observation of indicator tests is shown in table 1.

Solution

Observation (Colour changes)

NaOH(aq)+Phenolphthalein

From transparent to pink.

CH3COOH(aq)+Phenolphthalein

No change

Table 1 Indicator Experiment

The statistics in the Table1 is from experiment (A), the Titration of Vinegar.

Trial 1

Trial 2

Trial 3

Trial 4

Initial burette reading

10.0cm3

0cm3

0cm3

0cm3

Final burette reading

>25.0cm3

17.7cm3

(25+10.4)cm3

(25+14.2)cm3

Volume of NaOH for neutralisation

(discarded statistics)

17.7cm3

35.4cm3

39.2cm3

Volume of vinegar in the flask

(discarded statistics)

1.0cm3

2.5cm3

2.5cm3

Table2. Titration of Vinegar Statistics

When the NaOH (aq) was added more than 30cm3, there has a pink colour appears but vanished as shaking the flask. The pink colour retained as the end-point was added.

The statistics in the Table2 is from experiment (B), the Titration of Bleach.

Trial 1

Trial 2

Trial 3

Initial burette reading

0cm3

7.0cm3

15.0cm3

Final burette reading

7.1cm3

16.0cm3

22.6cm3

Volume of Na2S2O3 for neutralisation

7.1cm3

9.0cm3

7.6cm3

Volume of bleach in the flask

1.0cm3

1.0cm3

1.0cm3

Volume of KI solution in the flask

10.0cm3

10.0cm3

10.0cm3

Volume of H2SO4 in the flask

10.0cm3

10.0cm3

10.0cm3

Table3. Titration of Bleach Statistics

The mixture in the flask appears a dark brown colour with a unique smell of I2. The colour stared to fading to paled yellow as the Na2S2O3 (aq) was added into the flask. After adding the starch indicator, the colour of mixture turns into a dark black-blue. When reached the end-point, the final mixture is colourless.

Discussion

A) Calculate the concentration of H+ in vinegar.

a) Volume of vinegar in the flask is 2.5cm3

Average Volume of NaOH (aq) = (35.4+39.2) / 2= 37.3cm3

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

Concentration: 0.1 mol dm-3 x mol dm-3 (x is the volume should be solved)

Volume: 37.3 cm3 2.5 cm3

x = (0.1 - 37.3 -10-3) / (2.5 -10-3)

≈ 1.49 mol dm-3

The concentration (a) of vinegar is 1.49mol dm-3.

b) Volume of vinegar in the flask is 1cm3

Average Volume of NaOH (aq) = 17.7cm3

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

Concentration: 0.1 mol dm-3 x mol dm-3 (x is the volume should be solved)

Volume: 17.7 cm3 1.0 cm3

x = (0.1 - 17.7 -10-3) / (1.0 -10-3)

≈ 1.77 mol dm-3

The concentration (b) of vinegar is 1.77mol dm-3.

The average of concentration (a) and (b) is

(1.49+ 1.77) / 2 = 1.63 mol dm-3

∴The concentration of vinegar is about 1.63mol dm-3.

B) Calculate the concentration of ClO- in bleach solution.

Average volume of Na2S2O3 (aq) = (7.1+9.0+9.6) / 3 = 8.6 cm3

NaOCl (aq) + 2KI (aq) + H2SO4 (aq) → NaCl (aq) + I2 (aq) + K2SO4 (aq) + H2O (aq)

Concentration: 1 mol dm-3 x mol dm-3

Volume 10 cm3 a cm3

2Na2S2O3 (aq) + I2 (aq) → Na2S4O6 (aq) + 2NaI (aq)

Concentration: 0.1 mol dm-3 x mol dm-3

Volume 8.6 cm3 a cm3

n (KI) : n (I2) = 2 : 1

n (Na2S2O3) : n (I2) = 2 : 1

n (Na2S2O3): n (KI) = 1 : 1

n (NaOCl) : n (KI) = 1 : 2

n (Na2S2O3) = n (KI) = c - v = 1-10-10-3 = 0.01 mol, n (NaOCl) = 0.005 mole

∴Concentration of NaOCl = n / v = 0.005 / (8.6-10-3) ≈ 0.58 mol dm-3

Some non-accurate operation may affect the result. The judgment by using eyes about end-point may not accurate. Another factor that may influence the result is numerical reading not read by one person. This may cause the reading mistakes.

Conclusion

In conclusion, throw the experiments. The concentration of vinegar was found as about 1.63mol dm-3 by using the titration. Meanwhile, the concentration of ClO- in bleach solution is approximately 0.58 mol dm-3 by using the titration. The reactions are acid-base neutralization processes.

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