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A2 Practical Worksheets

Investigation of the reaction between manganate(VII) ions and ethanedioate ions

Investigation of the reaction between manganate(VII) ions and ethanedioate ions

 

Ethanedioic acid is a diprotic acid of formula HOOC-COOH. On ionisation it forms -OOC-COO- ions (C2O42-). These ions are oxidised by manganate(VII) ions and the object of this practical is to find the mole ratio of ethanedioate to manganate(VII) and hence suggest the oxidation product of the ethanedioate ion.

The oxidation is slow at room temperature and the solution must be heated before titrating. Once the reaction has begun it is catalysed by the Mn2+ produced by the reduction of the manganate(VII) ions. (Note that we have already studied the rate of this autocatalysed reaction using a colorimeter see previous practical).

 

Procedure:

Calculation:

HYDROLYSIS OF ESTERS (2)

Hydrolysis of an Ester (1) (Methyl Benzoate)

 

Procedure:

 

1 Transfer 15 drops of the class sample of methyl benzoate to a 100 cm3 conical flask..

 

2 Measure out 10 cm3 of bench sodium hydroxide solution in a measuring cylinder and add this to the flask. Add a small spatula of anti-bumping granules.

 

3 Heat the solution gently and maintain a very gentle boiling until all the oily drops of ester have disappeared.

 

4 Cool the flask under running water from the tap and then carefully add bench dilute hydrochloric acid. You will need to add several cm3 to neutralise the excess sodium hydroxide before you observe a permanent precipitate of the insoluble benzoic acid forming.

 

5 Filter your precipitate under reduced pressure using a Buchner funnel and wash your product with cold water.

 

6 Note down your observations and write equations for:

a) The hydrolysis of the methyl benzoate

b) The precipitation of benzoic acid from the sodium salt of the acid as HCl is added

 

 

HYDROLYSIS OF ESTERS (2)

The preparation of soap

To prepare your own small sample of soap proceed as follows

1) Place 10cm3 of 2M NaOH in a 100cm3 beaker and add from a teat pipette 1cm3 of castor oil

2) Place the beaker on a tripod and gauze and heat gently with a small bunsen flame stirring continuously with a glass rod.

3) When the mixture starts to boil, adjust the flame so that gentle boiling is maintained. Keep stirring for 5 minutes.

4) Add 10cm3 of distilled water and 6 spatula measures of sodium chloride. Heat to boiling and boil for one minute with stirring.

5) Place the beaker in an ice/water bath and leave to cool stirring occasionally.

6) When cold, filter the contents of the beaker using a Buchner funnel and filter pump. Wash once with cold water and remove as much water as possible.

7) Lift the filter paper plus soap out of the Buchner funnel using a microspatula and place it on a larger filter paper. Dry the soap as much as possible.

8) Test the lathering properties of your soap. Place a spatula measure in a clean test tube followed by about a 2cm depth of distilled water. Stopper and shake the tube.

Do you get a lather?

N.B. Your soap is still too alkaline to risk on your hands or face. Check the pH with a few drops of universal indicator solution. Put the soap you have left into the labelled beaker provided.

 

After clearing away read the section on detergents in the H&H. Then using reference books, try to find answers to the following questions.

1) What is meant by the term detergent?

2) What fat or oil is usually used to make household soap? What is the chemical mane of ordinary soap?

3) Castor oil was used here because it hydrolyses very easily. What is the natural source of castor oil?

4) Castor oil is mainly the ester formed from glycerol and ricinoleic acid. Name the sodium salt of this acid. From the initials suggest what will known product contains this soap.

5) Ricinoleic acid is 12-hydroxyoctadec-9-eneoic acid. Attempt to write its structure.

6) Adding salt to the hydrolysed ester is a standard part of the soap making procedure. What is the purpose of the salt?

The reaction between a carboxylic acid and an alcohol is reversible and also very slow

Preparation of an Ester - Methyl Benzoate

 

The reaction between a carboxylic acid and an alcohol is reversible and also very slow.

A strong acid (e.g. concentrated sulphuric acid) catalyses the reaction effectively.

In order to make methyl benzoate, benzoic acid and methanol are required:

 

C6H5COOH + CH3OH C6H5COOCH3 + H2O

 

In order to improve the yield of the ester a large excess of methanol is used and water is excluded from the mixture as much as possible. Under these conditions the yield of ester will not be 100% even with perfect experimental technique.

Procedure:

 

1 Weigh out about 7.5g of benzoic acid into a small dry plastic beaker using a top pan balance. Record the exact mass of solid used to 2 d.p.

 

2 Add 25 cm3 of methanol and stir to dissolve the solid. Then add 2 cm3 of conc. H2SO4 and stir.

 

3 Clamp a 100 cm3 quickfit flask in a spring clamp and add a spatula of anti-bumping granules. Use a funnel to pour the reaction mixture into the flask.

 

4 Put a water condenser vertically, directly into the mouth of the flask. Turn on the water supply and have your apparatus checked before proceeding.

 

5 Reflux the mixture gently for 30 minutes. (During this time you should be able to draw a good diagram under the title to show the apparatus used for reflux.)

 

6 Remove the heat and leave the apparatus to cool sufficiently so that you can handle the condenser safely (1 minute if you are careful of the hot flask below). Arrange the apparatus above the flask for distillation and get it checked (methanol is highly flammable). Heat gently to distil off the excess methanol. You will need to collect 12 - 15 cm3 and so arrange to collect the distillate in a measuring cylinder so that you can keep an eye on the volume.

 

7 Without delay pour the residue from the quickfit flask into a separating funnel which is half full of water (be careful, it is hot). Add 3 cm3 of trichloroethane, stopper the funnel, and shake gently. The trichloroethane dissolves the ester and together they form the lower layer in the funnel.

(This procedure speeds up the isolation of the ester which is immiscible with water but is so similar in density that it forms an emulsion which only separates very slowly; the ester/trichloroethane layer separates more quickly).

 

8 Remove the stopper from the separating funnel, run the lower layer into a small beaker and discard the upper aqueous layer.

 

9 Return the organic layer to the separating funnel and add an approximately equal volume of sodium hydrogencarbonate solution. Shake gently releasing the pressure occasionally. Allow the layers to separate. Collect the lower organic layer and discard the aqueous layer.

 

10 Repeat step 9 using an equal volume of water for the washing instead of sodium hydrogencarbonate solution. Transfer the organic layer to a dry sample tube and add a spatula of anhydrous calcium chloride. Stir the mixture with a microspatula until the liquid is no longer cloudy. The sample can be left overnight as it dries at this stage.

 

Producing the Pure Product:

 

The samples of ester prepared are too small for individual distillations and so they are collected together for a combined distillation at this stage.

The crude product is dry but mixed with trichloroethane. Thus the initial distillation, using a water cooled condenser, isolates the trichloroethane which has a boiling point of 73C.

The ester boils at 196C and hence an air condenser is used in the final stage.

 

Questions:

 

a) Calculate the percentage yield of the product, methyl benzoate.

This is the percentage of actual product produced compared to the expected amount of product (knowing the amount of starting material used and the reaction equation).

Proceed as follows:

Calculate the no. of moles of benzoic acid used for the whole class.

Use the reaction equation to deduce the theoretical no. of moles of product expected for the whole class. Convert this to the mass of product expected for the class.

% yield = Mass produced x 100

Mass expected

 

b) Give reasons why the percentage yield is less than 100%

You should consider the experimental procedure carefully to identify where your

product may have been lost. Is there any other reason for the poor yield?

c) What was the purpose of the sodium hydrogencarbonate in step 9?

 

d) Why was an air condenser used in the final distillation instead of a water condenser?

Investigation of the Kinetics of a Reaction

Investigation of the Kinetics of a Reaction

 

The reaction to be studied involves the oxidation of iodide ions in acidic solution by hydrogen peroxide:

 

H2O2 + 2H+ + 2I- I2 + 2H2O

 

The rate equation will be of the form: Rate = k [H2O2]x [I-]y [H+]z

 

In order to find the initial rate of reaction, for various set-ups, as the concentration of one of the reactants is varied, a small but accurately measured amount of sodium thiosulphate will be added each time along with 10 drops of starch indicator.

The first small amount of iodine to be produced in each experiment will react with the thiosulphate before it has a chance to affect the indicator.

Only when all the thiosulphate has been used up will the starch turn blue/black.

Thus the same amount of iodine will have been produced in all experiments when the starch goes blue/black.

If the time taken for the appearance of this blue/black colour is t then the initial rate of reaction will be proportional to 1/t

 

Remind yourself of your redox chemistry by writing the equation for the reduction of iodine by thiosulphate: 2S2O32- S4O62- + 2e-

 

Determination of y, the order in iodide ions:

 

 

Expt.

Volume of H2SO4/cm3

Drops of starch

Volume of water/cm3

Volume of KI/cm3

Volume of Na2S2O3/cm3

1

5

10

0

5

1

2

5

10

1

4

1

3

5

10

2

3

1

4

5

10

3

2

1

5

5

10

4

1

1

 

Treatment of Results:

 

In the above experiment the initial volumes of acid and hydrogen peroxide have remained constant and only the initial concentration of potassium iodide has been varied by adjusting the volumes of water and KI in the combined volume of 5cm3.

Hence the rate equation can be simplified to:

Rate } [I-]y

Since the total volume is kept constant, the concentration of KI can be taken as proportional to its volume.

So the rate equation becomes: Rate } (initial volume of KI)y

 

Obtaining y:

If y = 1 then Rate
} (initial volume of KI) and a graph of rate plotted against initial volume of KI should be a straight line.

 

Plot a graph of rate against initial volume of KI to find out whether it is a straight line and whether y = 1

(If y = 2 which graph would you plot to demonstrate the relationship?)

Determination of x, the order in hydrogen peroxide:

 

Devise a method for obtaining x. Note carefully what you intend to do and consult with your teacher before proceeding.

Obtain sufficient results to determine x by plotting a graph as above

 

Errors:

 

What are the major sources of error in these experiments?

Which of your results are likely to be the least accurate and why?

 

Determination of the order in H+ ions:

 

The effect of H+ ions on the rate of this reaction is complicated when the concentration of the acid is high, but the situation is simpler at lower concentrations.

Determine the order in H+ when the concentration of acid is very low by carrying out the following experiments:

 

Mix 5cm3 of 0.1M KI solution, 1cm3 0.01M sodium thiosulphate solution, 1cm3 of starch solution and one drop of 0.25M sulphuric acid. Add 5cm3 of 0.1M hydrogen peroxide, start the clock and note the time for the blue/black colour to appear.

 

Repeat the above experiment, but this time use 2 drops of 0.25M sulphuric acid.

 

On the basis of these two results (not very scientific) deduce the order with respect to H+ ions. Explain your answer.

 

Extra:

 

Both the orders x and y can be obtained using logs:

E.g. for the value of y the expression is:

Rate } (initial volume of KI)y

 

Taking logs of both sides this becomes:

log(initial rate) = y log(initial vol.of KI) + constant

 

\ a graph of log(initial rate) plotted against log(initial vol.of KI) should give a straight line of gradient = y

 

 

THE REDUCTION OF A CARBONYL USING SODIUM BOROHYDRIDE

THE REDUCTION OF A CARBONYL USING

SODIUM BOROHYDRIDE

 

Sodium borohydride (sodium tetrahydridoborate(III), NaBH4) is the preferred reducing agent for converting carbonyls to alcohols. It has several advantages over hydrogen with a Ni catalyst:

The borohydride ion acts as a source of hydride ions (:H-). When the reaction is complete, water is added to:

a)      Destroy excess sodium borohydride and

b)      Provide a source of H+ to complete the formation of the alcohol

 

The reduction of diphenylethane-dione

This dione is used because it can be reduced in high yield to the diol and the product is a solid that can be easily separated from the solution of boron salts that is left.

 

C6H5COCOC6H5 + 4[H] C6H5CH(OH)CH(OH)C6H5

 

Procedure:

 

Write up:

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