1. Titration – simple mole ratio

The following experiment was carried out. a) An indigestion tablet was weighed and

crushed. b) The tablet was transferred into pre-weighed

beaker and then reweighed. c) About 50cm3 of water was used to dissolve

the tablet and the liquid from the resulting saturated solution was filtered and transferred into a 250 cm3 volumetric flask.

d) The flask was filled to the mark with tap water and labelled TD1.

e) 25cm3 of the tablet solution TD1 was pipetted into a conical flask.

f) Three drops of methyl orange indicator. g) The solution was titrated against TD2

(Hydrochloric acid 0.05mol.dm-3) from the burette until the solution went red and stayed red for one minute. This process was repeated to gain concordant results.

Results Mass of Beaker = 72.00g, Mass of beaker + tablet = 73.97 g Titration results

Initial Volume (cm3) Final Volume (cm3)

Trial 1 2.75 26.20

Trial 2 22.25 45.90

Trial 3 5.30 28.80

Questions Some indigestion tablets are hydroxide compounds and some are carbonate compounds.

Assuming the tablet only has magnesium hydroxide as its neutralisation agent:

1) Calculate the mass of active ingredient in the tablet.

2) Calculate the percentage of the tablet that is active ingredient (and not filler compounds).

2. Titration – moderate mole ratio

MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq) → Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)

TB1 is Fe(NH4)2(SO4)2.6H2O powder Mr = 392.1

TB2 will be made by you.

TB3 is 1.0mol.dm-3 sulfuric acid

TB4 is potassium manganate(VII) solution of unknown concentration

The following procedure was carried out

a) 4.03 g of TB1 was weighed in a beaker, dissolved in 100cm3 of TB3, then transfer to a 250cm3 volumetric flask. The flask was filled up to the mark with tap water and thoroughly mixed. This was labelled TB2.

b) 25.0 cm3 of the TB2 solution was pipetted into a clean conical flask.

c) The burette was filled with TB4 solution and a titration was carried out until the first pale pink colour remained for 15 seconds.

Titration results

Initial Volume (cm3) Final Volume (cm3)

Trial 1 6.10 28.50

Trial 2 1.20 23.10

Trial 3 3.05 25.00

Questions 1) Calculate the concentration of TB4.

2) The concentration of TB4 should be 0.00956 mol dm-3. Calculate the percentage error in your results.

3. Water of crystallisation in washing soda, Na2CO3.xH2O

The following procedure was carried out.

a) A burette was filled with the acid, TC2 (0.50 moldm-3 HCl).

b) 40.65 cm3 of TC2 was run into a 250 cm3 volumetric flask and made up to the mark with tap water then inverted a few times to ensure thorough mixing; this was labelled TC3.

c) 3.09 g of the washing soda TC1, was weighed into a beaker.

d) Tap water was added to the beaker to dissolve the crystals. The solution was transferred to a 250 cm3 volumetric flask and made up to 250 cm3 with tap water and inverted several times to ensure thorough mixing. This solution was labelled TC4.

e) 25.0 cm3 of TC4 was pipetted into a conical flask and a few drops of the indicator provided was added.

f) TC4 was titrate against TC3 from a burette until the end point was reached.

Titration results

Initial Volume (cm3) Final Volume (cm3)

Trial 1 3.25 34.80

Trial 2 7.25 38.65

Trial 3 4.25 35.70

Questions 1) Determine the value of x in Na2CO3.xH2O.

4. Part 1 The enthalpy change in the displacement reaction between zinc and copper(II) sulphate. **The specific heat capacity of any dilute salt solution is approximated to that of pure water**

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

The following procedure was carried out.

a) 50 cm3 of 0.2 mol dm-3 copper (II) sulphate was poured into the polystyrene cup.

b) 1.00 g (an excess) of powdered zinc was weighed out.

c) The initial temperature of the solution in the cup was recorded as 19.5 oC and then the powdered zinc was added.

d) The solution was stirred gently and constantly with the thermometer until the temperature stopped changing. The final temperature was recorded as 29.0 oC.

Questions

1) Calculate the enthalpy change (kJmol-1) that took place with respect to CuSO4. c = 4.18J.oC-1.g-1.

2) Show that zinc is indeed the reactant that is in excess.

Part 2: The enthalpy change reaction between citric acid and sodium hydrogen carbonate.

C6H8O7(aq) + 3NaHCO3(s)→ C6H5O7Na3(aq) + 3CO2(g) + 3H2O(l)

The following procedure was carried out. a) 25 cm3 of 1 mol dm-3 citric acid was poured

into the polystyrene cup. b) 8.15 g of sodium hydrogen carbonate (an

excess) was weighed onto a piece of paper. c) The initial temperature of the solution in the

cup was recorded as 20.0 oC and then the powdered NaHCO3 was added.

d) The reaction frothed up, so the powdered solid was added slowly to prevent it coming over the top of the cup.

e) The final temperature reached was 4.0 oC.

Question

1) Calculate the enthalpy change (kJmol-1) that took place with respect to citric acid. c = 4.18J.oC-1.g-1.

2) Show that NaHCO3 is indeed the reactant that is in excess.

5. Measuring the enthalpy change of neutralisation The following procedure was carried out. a) A clean, dry Styrofoam cup was weighed at

5.70g.

b) 50cm3 of hydrochloric acid solution (1 mol.dm-3) was added to the cup.

c) The initial temperature of the solution was recorded at 18.0 oC.

d) About 50cm3 of sodium hydroxide solution (1 mol.dm-3) was added to a small beaker.

e) The initial temperature of this solution was recorded at 19.0 oC..

f) The two solutions’ temperatures were averaged.

g) The NaOH from the beaker was poured into the HCl in the cup.

h) The highest temperature reached by the solution was recorded at 25.5 oC, and the temperature change was calculated.

i) The cup was reweighed at 106.70g. Calculate the mass of solution used.

Results

Average Initial Temperature =

Temperature change =

Mass of solution =

Questions

1) Calculate the value for H for this reaction in kJmol-1 for hydrochloric acid.

2) Explain why the value gained for the H is unreliable by giving three reasons for errors throughout this practical.

3) Explain why the value for H for sodium hydroxide should be the same.

4) When a substance like ammonium chloride is put into water, the temperature decreases.

H[NH4Cl(s) NH4Cl(aq)] = 44.9kJmol-1. Mr(NH4Cl) = 53.5 Calculate the temperature decrease when 4.00g of NH4Cl is dissolved into 50cm3 of water.

6. Titration – Hard mole Ratio - Reaction of iodine with sodium thiosulphate

TE1 = unknown concentration potassium

iodate solution, 4.00gdm-3. TE2 = 0.1mol.dm-3 potassium iodide solution TE3 = 1.0mol.dm-3 sulfuric acid solution

TE4 = 0.0989mol.dm-3 sodium thiosulfate solution (standard solution)

Fresh starch solution (indicator)

The following procedure was followed a) 25cm3 of TE1 was pipetted into a flask. b) To this flask excess TE2, (25cm3) and

excess TE3 (2cm3) was added. c) The burette was filled with TE4 and the

initial volume recorded. d) TE4 was titrated into the flask until the

solution in the flask was a pale yellow. At

this point a few drops of starch was added to make the solution blue/black.

e) Titrating was continued until the colour of the solution went colourless. The final volume was recorded.

f) This titration was repeated until two concordant titre values were determined

Titration results

Initial Volume (cm3) Final Volume (cm3)

Trial 1 1.25 22.80

Trial 2 6.30 27.70

Trial 3 5.00 26.45

Questions M(KIO3) = 214 g.mol-1. 1) Calculate the concentration of TE1 (in g.dm-3).

2) Compare this value with that created with the impure KIO3 and determine the percentage purity of the solid potassium iodate.

IO3– + 5I- + 6H+ 3I2 + 3H2O 3I2 + 6S2O3

2- 3S4O62- + 6I-

7. The enthalpy change of solution for sodium hydroxide In this experiment the temperature is measured over time and plotted on a graph; the regular method of determining the maximum observable T rise is just an approximation. Extrapolation of the cooling curve will give a correct value for the T rise. This makes an allowance for the heat loss to the surroundings.

The following procedure was carried out 1. A polystyrene cup was weighed at 2.77g.

2. Approximately 50cm3 of water was

added to the cup –the total mass was

recorded at 54.19g, and therefore mass

of water can be calculated.

3. The temperature of the water in the cup

was recorded for 2 minutes to determine

the baseline temperature.

4. 11 lumps of solid sodium hydroxide, from

the sealed container, was added to the

water at the 2 minute mark and stirred,

while continuing to time.

5. The cup was rewighed to determine the

mass of NaOH added.

6. Timing was continued but only once the

NaOH was all dissolved were the

temperatures recorded again.

7. The temperature of the water was

recorded over the next 10 minutes (total

of 12 minutes of timing).

Results

Mass of empty cup / g 2.77

Mass of cup plus water / g 54.19

Mass of cup, water and NaOH / g 54.72

Mass of water added / g

Mass of NaOH added / g

Time period / s

Temperature / oC

Δ temperature / oC

0 29.5 0

30 29.5 0

60 29.5 0

90 29.5 0

120 29.5 0

150

180 42.5 13.0

210 42.0 12.5

240 42.0 12.5

270 42.0 12.5

300 41.5 12.0

330 41.5 12.0

360 41.0 11.5

390 41.0 11.5

420 41.0 11.5

450 40.5 11.0

480 40.5 11.0

510 40.0 10.5

540 40.0 10.5

570 40.0 10.5

600 39.5 10.0

630 39.5 10.0

660 39.5 10.0

690 39.0 9.5

720 39.0 9.5

Graphing

Plot a graph of temperature vs time.

There should be two distinct regions/lines to the

graph.

The cooling section of the graph needs to be

extrapolated back to the time at which the solid

was added to determine a corrected ΔT for the

reaction.

Calculations

Find the enthalpy value for the reaction in kJmol-1.

8. The enthalpy change of solution for ammonium chloride The following procedure was followed a) 25cm3 of water was pipetted into a small beaker.

b) The initial temperature was recorded.

c) 3.00g of MA1 was weighed out and added to the water.

d) The highest temperature (final temperature) was recorded.

e) The experiment was repeated using approximately 4.00, 5.00, 6.00, 7.00, 8.00 9.00 and 10.00 g of MA1.

Results

mass/g T change /oC

2.99 7.0

3.25 8.0

4.49 11.0

4.93 12.0

5.50 13.0

6.50 15.0

mass/g T change /oC

7.05 16.5

7.27 16.5

8.00 13.0

8.01 17.0

8.47 15.0

9.52 18.0

Questions 1) Plot a graph of ΔT vs m for your data, and draw a trend line.

2) Determine (by extrapolation) the value for ΔT when the mass of MA1 is 11.00g.

3) Determine the energy released when the mass of MA1 is 11.00g, where c = 4.18JoC-1K-1.

4) Calculate the ΔHr (units kJmol-1) for ammonium chloride solid dissolving into water.

9. Determining the water of crystallisation by gravimetric analysis

Waters of crystallisation are water molecules trapped within the lattice of an ionic compound in a set ratio. E.g. CuSO4.5H2O has five waters of crystallisation for every one CuSO4. The mole ratio of the waters can be determined easily by the mass lost when a stable hydrated ionic compound loses all of its water molecules to become anhydrous. The following procedure was followed:

1. An empty crucible was weighed at 5.70 g. 2. 5.50 grams of hydrated magnesium sulphate were added. 3. The crucible was heated from beneath, carefully at first, then a little bit stronger for about a

minute. 4. The heat was removed and the crucible was allowed to cool a little bit, before being

weighed. 5. The crucible was reheated for another minute, cooled and reweighed. 6. This process was repeated until the mass dropped by less than 0.05g compared to the

previous heating. The final mass of the crucible and solid was 8.51 g. Results – draw an appropriate results table from the information above. Questions

1. The formula for the hydrous magnesium sulphate is MgSO4.xH2O, Determine x from the above data.

Answers 1. Titration

1. 2HCl + Mg(OH)2 MgCl2 + 2H2O 2. E.g. (23.45 + 23.65 + 23.50) / 3 =

23.53cm3 = 0.0235dm3. 3. n =cV = 0.05mol.dm-3 x 0.0235dm3 =

0.00118mol 4. Use the mole ratio. 0.00118mol / 2 =

0.000588mol

5. 0.000588mol per 25cm3, so times by 10 for the whole flask: 0.00588mol

6. m = nM = 0.00588mol x 58gmol-1 = 0.34g

7. % = 0.35g / 1.97g x (100/1) = 17.4%

2. Titration.

1. mol0103.0gmol1.392

g03.4

M

mn

1

2. The total 250cm3 flask has 0.0103mol, so the 25cm3 pipette has a tenth of that: 0.00103mol

3. n(Fe2+) : n(MnO4-) = 5:1. Divide by five. mol10x05.2

5

mol00103.0 4

4.

33 cm93.21cm925.212

)95.2190.21(titreAverage

5. mol10x35.9dm02193.0

mol10x05.2

V

nc 3

3

4

.

6. %20.21

100

00956.0

)00935.000956.0(error%

.

3. Water of crystallisation

1. 3

3

33 moldm0813.0

cm250

cm65.40moldm50.0

2. From titres that are concordant: 33 dm03147.0cm47.313

45.3140.3155.31

3. n=cV = 0.0813mol.dm-3 x 0.03147dm3 = 0.00256mol

4. Na2CO3:HCl = 1:2 so 0.00256mol /2 = 0.00128mol of sodium carbonate

5. 0.00128mol per 25cm3. Times by 10 for the 250cm3 flask = 0.0128mol

6. m = nM = 0.0128mol x 106gmol-1 = 1.36g

7. I added a total of 3.09g of hydrated sodium carbonate, and have just shown that 1.36g of it

is anhydrous sodium carbonate. The rest of the mass is water.

3.09g – 1.36g = 1.70g of H2O.

First Second Third

Final reading / cm3 28.50 23.10 25.00

Initial reading / cm3 6.10 1.20 3.05

Titre / cm3 22.40 21.90 21.95

8. mol0944.0gmol18

g70.1OHmol0128.0

gmol106

g36.1CONa

12132

OH7.CONatherefore736.7mol0128.0

mol0944.0OH1

mol0128.0

mol0128.0CONa 232232

Data booklets suggest that the actual value is 10 waters of crystallisation.

Experiment 4 Part 1

(1) E = mcΔT, m = 50g, c = 4.18, ΔT= 9.5oC

E = 50 x 4.18 x 9.5

E = 1985.5 J

ΔH = -E/n, n = cV n = 0.2 x 50/1000 = 0.01 mol

ΔH = -1.9855kJ/0.01 mol

ΔH = -198.55 kJ mol-1 or -200 kJ mol-1.

(2) n = m/M, M = 65.4 so n = 1/65.4

n = 0.015. As it is a 1:1 mole ratio and 0.015 > 0.01 then Zn is in excess.

Part 2

(1) E = mcΔT, m = 25g, c = 4.18, ΔT= -16 oC

E = 25 x 4.18 x -16

E = -1672 J

ΔH = -E/n, n = cV n = 1 x 25/1000 = 0.025 mol

ΔH = -(-1.672)kJ/0.025 mol

ΔH = 66.88 kJ mol-1 or 70 kJ mol-1.

(2) n = m/M, M = 23+1+12+48 = 84 so n = 8.15/84

n = 0.097. As it is a 3:1 mole ratio and 0.0977 > 3 x 0.025 then it is in excess.

Answers to Experiment 5

1) E = 101g x 4.2J oC-1 g-1 x 7oC = 2970J =

2.97kJ

2) n = cV = 1mol.dm-3 x 0.050dm3 = 0.05mol

3) DH = E / n = 2.97kJ / 0.05mol = 59.4kJmol-1.

4) Possible errors: The measured volume of HCl was not exactly 50cm3, so the molar amount is wrong. The total amount of HCl may not have been fully neutralised, if there was

insufficient NaOH. The temperature recorded to the nearest half degree, over such a small temperature rise has a large error.

5) If equal molar amounts of HCl and NaOH were used, then the calculation is identical.

6) molgmol

g

M

mn 0748.0

5.53

00.41

JkJnHE 336036.3

Cg

J

mc

ET o16

2.450

3360

Answers to Experiment 6 The working is as it should be, but the numbers are different to yours because your titration will have gained different titre values.

1) From titres that are concordant: 33 dm02147.0cm47.213

45.2140.2155.21

2) n=cV = 0.0989mol.dm-3 x 0.02147dm3 = 0.00212mol of TE4

3) S2O32- : IO3

- = 6:1 = so 0.00212mol /6 = 0.000354mol of potassium iodate TE1

4) 0.000354mol per 25cm3. 3

3dm.mol0142.0

dm025.0

mol000354.0

V

nc

5) 0.0142mol.dm-3 x 214gmol-1 = 3.03gdm-3

6) %7.751

100x

gdm00.4

gdm03.33

3

Experiment 7

Mass of empty cup / g 2.77

Mass of cup plus water / g 54.19

Mass of cup, water and NaOH / g 54.72

Mass of water added / g 51.42

Mass of NaOH added / g 0.53

Time period / s Temperature / oC oC

0 29.5 0

30 29.5 0

60 29.5 0

90 29.5 0

120 29.5 0

150

180 42.5 13

210 42 12.5

240 42 12.5

270 42 12.5

300 41.5 12

330 41.5 12

360 41 11.5

390 41 11.5

420 41 11.5

450 40.5 11

480 40.5 11

510 40 10.5

540 40 10.5

570 40 10.5

600 39.5 10

630 39.5 10

660 39.5 10

690 39 9.5

720 39 9.5

0

2

4

6

8

10

12

14

0 100 200 300 400 500 600 700 800

E = mcΔT = 51.42g x 4.18J g-1 oC-1 x 13.4oC = 2880J = 2.88kJ

n = m/M = 0.53g / 40.0gmol-1 = 0.013mol

ΔH = E/n = 2.880kJ / 0.013mol = 217kJ Experiment 8

mass/g T change

2.99 7.0

3.25 8.0

4.49 11.0

4.93 12.0

5.50 13.0

6.50 15.0

7.05 16.5

7.27 16.5

8.00 13.0

8.01 17.0

8.47 15.0

9.52 18.0 (2) 20.8 oC (3) E = mcΔT

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

16.0

18.0

20.0

0.00 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00

tem

p c

han

ge/d

eg

C

Mass/g

T change

Linear (T change)

E = 25 x 4.18 x 20.8 E = 2173.6 J released. (4) n = m/M n = 11.00/ 53.5 n = 0.206 (5) ΔH = -E/n ΔH = -2.1736/0.206 ΔH = -10.55 kJ mol-1 (-10.6) Experiment 9 Results Table

Mass (g)

Crucible 5.70

Crucible + hydrous magnesium sulphate 11.20

Crucible + anhydrous magnesium sulphate 8.51

(1) Mass of anhydrous magnesium sulphate = 8.51 – 5.70 = 2.81 g Mass of water = 11.2 – 8.51 = 2.69g (2) n(MgSO4) = m/M = 2.81/120.3 = 0.0233, n(H2O) = m/M = 2.69/18 = 0.1494 (3) Ratio = 0.0233/0.0233 : 0.1494/0.0233 = 1 : 6.4 MgSO4.6H2O.