Trial Paper 3

7/27/2019 Trial Paper 3

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1. You are provided with the following

FB1 is an unknown concentration of HCl

FB2 is aqueous 0.100 mol dm-3

Na2CO3

You are required to determine the concentration, in mold dm-3

of HCl

Dilution of FB1

(a) By using a pipette, remove 25.0 mol dm-3

of HCl into the 250 cm3

volumetric flask ,

labeled FB3.

Make up the contents of the flask to the 250 cm3

mark with distilled water. Place the

stopper in the flask and mix the contents thoroughly by slowly inverting the flask a

number of times.

Titration

Filled the burette with FB3.

Pipette 25.0 cm3

of FB2 into a conical flask. Then add methyl orange.

Perform a rough (trial) titration and sufficient further titrations to obtain reliable results.

Record your titration results in the space below. Make certain that your recorded results

show the precision of your working.

Rough 1 2

Initial Volume of burette/cm3

5.0 8.50 5.55

Final Volume of burette/cm3

32.0 35.10 32.25

Volume of FB3 used/cm3

(b) From your titration results obtain a volume of FB3 to be used in your calculations. Show

clearly how you obtained this volume.


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(ii)

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(g) A student was asked to boiled the reaction mixture in the conical flask when the end

point was obtained. Comment on this statement why this step was considered.

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(h) Describe, briefly how a pure dry sample of sodium chloride could be obtained having

carried out the titration.

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(i) The precision of the pipette used is + 0.1 cm3

The precision of the 250 cm3

volumetric flask is + 1cm3

Determine the cumulative error of preparing 250 cm3

of HCl.

(j) The smallest division of the burette is 0.10 cm3. State the precision of the burette and the

error included in each reading of the burette.

(k) Hence, calculate the % error of the volume of the HCl used in the titration.


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1. You are provided with the following

FB4 is an unknown concentration of HCl

FB5 is aqueous 0.75 mol dm-3

Ca(OH)2

You are required to determine the concentration, in mold dm-3

of HCl

a Stand an insulated cup in a beaker for support.

b Using a pipette and safety filler, transfer 20 cm3

(or 25 cm3

) of the sodium hydroxide solution into

the cup, and measure the steady temperature.

c Using the burette, add a small portion (3 5 cm3) of dilute hydrochloric acid to the solution in the

cup, noting down the actual volume reading. Stir by swirling the cup and measure the highest

temperature reached.

d Immediately add a second small portion of the dilute hydrochloric acid, stir, and again measure the

highest temperature and note down the volume reading.

e Continue in this way until there are enough readings to decide the maximum temperature reached

during this experiment. You will need to add at least 40 cm3

of the acid.

fPlot a graph of temperature against the volume of acid added, and use extrapolation of the two

sections of the graph to deduce the maximum temperature reached without heat loss.

g Filled the empty columns with suitable values.h Use your results to calculate the concentration of the hydrochloric acid and the enthalpy change of

neutralization.

Unknown HCl

0.75 mol dm-3

Ca(OH)2


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Table of results

vol of HCl(cm3) temperature(

oC)

0.00 23.0

5.00 27.0

8.00 29.0

12.00 30.5

16.00 32.5

20.00 32.5

24.00 31.5

28.00 30.5

32.00 30.0

36.00 29.5

40.00 29.0

(a) From the graph, determine the volume of acid that is needed for complete neutralization?

(b) Calculate the number of moles of Ca(OH)2 in 25.0 cm3.

(c) Write a balanced equation for the reaction between aqueous HCl and aqueous Ca(OH)2.

.

(d) Hence, determine the number of moles of HCl that is needed to react completely withCa(OH)2.

(e) Calculate the molarity of unknown HCl solution.


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(f) Determine the maximum temperature change from the graph.

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(g) From the volume of HCl needed for complete neutralization, calculate the heat given out

during the reaction. .( Given Q=mcT the density of the solution is assumed to be1.00 g cm

-3and the specific heat capacity of the solution is 4.18 J g

oC

-1)

(h) Hence, calculate the enthalpy change of neutralization of this reaction.

(i) The enthalpy of neutralisation of a strong acid/base is constant. Why is the value constant?

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(j) Experimental results for HCl are usually a little less negative than the literature value.Suggest two reasons for this.

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(k) Suggest a reason why the enthalpies of neutralisation for the reactions involving sodiumcarbonate is always less negative than for strong acids and bases.

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(l) Suggest two improvements to obtain better results of this experiment.

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(m) Supposing the volume of the Ca(OH)2(aq) is doubled would this increase in volume doubledthe maximum temperature. Explain your reasons.

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(n) Supposing the concentration of Ca(OH)2(aq) is now doubled would this increase inconcentration doubled the maximum temperature? Explain your reasons.

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(o) If 0.75 mol dm-3

of sulfuric acid is used instead of HCl what would be the expected enthalpychange of neutralization of this reaction. Give your reasons.

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(p) The smallest division of the thermometer is 1.0oC. What is the error in the recorded

temperature?

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(q) The initial temperature of the acid and base were recorded. Equivalent volumes of the acidand base were added together and the maximum temperature is recorded. What is the errorin recording the maximum temperature change?( The smallest division of the thermometer is

1.0

o

C)

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(r) The room temperature increases from 20 oC to 25oC due to power failure. How would thisaffect the value of the enthalpy of neutralization obtained in the experiment? Give yourreasons.

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The main concern in this experiment is the heat loss. If possible a lid should be used. More reliable

results can be achieved using two polystyrene cups (one inside the other).

With abler or older students, it is possible to discuss the extrapolation of the cooling curve to

estimate the maximum temperature reached without heat loss. The link below gives an example of

how extrapolation is used to determine the maximum temperature reached.

To reinforce the theory involved here, an indicator could also be used to show that the end-point

really did occur at the highest temperature.

Health & Safety checked August 2008

Aim: The aim is to determine the concentrations of two acids and consequently the enthalpy of neutralisation for eachreaction.

Chemicals:

NaOH 1M (standardised)

HCl 2.0 M

CH3COOH 2.0 M

So the formulae would be:


10/13

NaOH + HCl --> NaCl + H20NaOH + CH3COOH --> NaCH3COO + H2O

So basically, I added 50 cm3 of NaOH to a polystyrene cup and then put the HCl in the burette. I added successiveportions of 5cm3 and saw the temperature until I reached 50cm3 of HCl

Did the same for CH3COOH.

Now we have to find the enthalpy of neutralisation values

Questions to answer

The enthalpy of neutralisation of a strong acid/base is constant. Why is the value constant?

Experimental results for HCL are usually a little less negative than the literature value. Suggest two reasons

for this.

Suggest a reason why the enthalpies of neutralisation for the reactions involving weak acids and weak bases

are always less negative than for strong acids and bases.

I think I have to use the formula: Q=mc/_\T but for the m = mass of NaOH + HCl but for the mass of HCl, do I use the

volume until it reaches the maximum temperature or do I use 50cm3?

1. We make the assumption that strong acids and strong alkalis are fully ionised in solution, and that the

ions behave independently of each other. For example, dilute hydrochloric acid contains hydrogen ions

and chloride ions in solution. Sodium hydroxide solution consists of sodium ions and hydroxide ions in

solution.

The equation for any strong acid being neutralised by a strong alkali is essentially just a reaction between

hydrogen ions and hydroxide ions to make water. The other ions present (sodium and chloride, for

example) are just spectator ions, taking no part in the reaction.

The full equation for the reaction between hydrochloric acid and sodium hydroxide solution is:

NaOH + HCl --> NaCl + H20

but what is actually happening is:

H+

+ OH---> H2O

If the reaction is the same in each case of a strong acid and a strong alkali, then the enthalpy changes

are similar. Even with 2M of HCl and 1M of NaOH in your experiment, since NaOH becomes the limiting

reagent, therefore the enthalpy change is the same.

2. There may be heat loss throughout the process causing the calculated value to be less exothermic.

However, if you draw a graph and extrapolate the actual heat when added then it shouldn't be that far

away from the literature value. Can't exactly think of another reason now, I'll add another one when it

comes to me..

3. Weak acids give less negative values since they only partially dissociate. For example CH3COOH, onlyaround 4% dissociates into its ions thus the enthalpy of neutralisation is less exothermic and the overall

reaction will therefore be more endothermic as more energy is needed to break the bonds.

Calculations

NaOH + HCl --> NaCl + H2O

Conc. 1.00 1.67


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Vol. 0.05 0.03

n 0.05 0.05 0.05 0.05

The initial temperature for NaOH was 23.7C

And the maximum temperature it reached was 32C

So the T = 23.7-32 = -8.3

So the concentration of HCl used was 1.67 M

Q = m x c x T

Q = [(50+30) 1000] x 4.18 x 8.3

Q = 2.780.05

Q = 55.5 kJmol-1

NaOH + CH3COOH --> NaCH3COO + H2O

Conc. 1.00 1.54

Vol. 0.05 0.0325

n 0.05 0.05 0.05 0.05

So the concentration of CH3COOH used was 1.54 M

Q = m x c x T

Q = [(50+32.5) 1000] x 4.18 x 8

Q = 2.76 0.05

Q = 55.2 kJmol-1

vol of acid(cm3)

temp(o

C)


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0.00 23.00

5.00 27.00

8.00 29.00

12.00 30.50

16.00 32.50

20.00 32.50

24.00 31.50

28.00 30.50

32.00 30.00

36.00 29.50

40.00 29.00


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Trial Paper 3