In order to speed up this process cold water was poured around the bottom of the outside of the conical flask making absolutely sure no additional water got into or out of the flask.
The solution was then poured into a 250cm3 flask using a funnel and made up to 250cm3 using deionised water making our stock solution. The funnel and any used/dirty apparatus were rinsed out using deionised waster as well. Using a pipette 25cm3 of the stock solution was extracted and place in a clean conical flask with the addition of 20cm3 of 1M sulphuric acid which had been measured in a clean measuring cylinder.
Using a funnel 0.02 mol dm-3 potassium manganate VII solution was poured into a burette up to 0.00cm3 mark and titrated until a permanent overall pink-ish colour change occurred. This reading was noted and the titration portion of the experiment was carried out two more times with all readings noted and the average of them used in the final calculations.
Results
Using the last two readings as they where the closest I can get my average:
18.2 + 18.4 = 18.6
2
Calculations:
Amount of MnO4- used: 18.6 cm3
Weight of Iron wool: 1.40g
Molar concentration of MnO4- = 0.02M
Reaction equation:
Fe + 2H+ Fe2+ + H2
MnO4- + 8H+ + 5Fe2+ Mn2+ + 4H2O + 5Fe3+
The concentration of Fe2+ in mol.dm-3
*The values have been changed to dm-3 form to fit in with equation (multiplying cm3 by 10-3)
Amount of MnO4- = Volume of MnO4- used X Concentration of MnO4-
= (18.8x10-3) X 0.02M
=3.72x10-4 mol dm-3
The ratio, according to the balanced equation above, of MnO4- to Fe2+ is 1:5 therefore multiply the molar quantity of MnO4- by 5 to give the molar quantity of Fe2+.-
3.72x10-4 X 5 = 1.86x10-3 moles of Fe2+
1.86x10-3 = 0.0744 mol dm-3 of Fe2+
25.0x10-3
Concentration in g of Fe2+
(Molecular mass of Iron is 55.845)
0.0744 mol dm-3 of Fe2 X 55.845 = 4.154868 g.dm-3 Fe2+
The above amount gives us the g. dm-3 Fe2+ in 1 liter to work out the amount in 250cm3:
4.154868/4= 1.038717 g of Iron in a 250cm3 sample
The percentage this value against the original mass of Iron Wool used:
(1.038717/1.40)x 100 = 74.19407143
the percentage purity of the experimental sample of Iron wool was:
74.20%
Conclusion
The general aim of the practical was to establish the percentage purity of an iron wool sample using a titration procedure. The reaction that occurred is known as a redox reaction and from the experiment it was found that 18.60cm3 of MnO4- was needed for the iron wool and acid solution to reach equilibrium. Using this reading along with a series of calculation and balanced reaction equations the percentage purity of the Iron wool was calculated to be 74.20%. We can assume this result to be the correct percentage purity of Iron wool however from other experiments along the same lines the percentage purity calculated was lower than anticipated. This therefore means experimental anomalies occurred. There can be numerous reasons for the anomalies, simple numerical conversions or calculation errors could be a source. These are easily rectified by double checking all conversions and calculation with any chemistry text book from A-level onwards. An error is more likely to occur during the procedure of the experiment. A viable reason could have been during the heating of the iron wool and Sulphuric acid. The procedure called for full dissolution of the two by heating on a hot plate, if this didn’t occur and remnants of the iron wool remained after cooling to room temperature and indeed through out the titration the concentration of the solution would have been different than assumed if full dissolution occurred. This in turn would affect the amount of Potassium Manganate VII needed to reach equilibrium giving unreliable reading used in the final calculations. This could have been avoided if either by leaving the solution on the hotplate for a longer period of time or to increase the temperature of the hotplate because the collision theory tells us that when molecules, especially in substances in a liquid form, get more heat energy they move faster providing more useful collision with each other as a faster rate therefore possibly quicken the pace at which dissolution would have occurred.
Also faults may lie in the using of the equipment, unnoticed bubbles may have collected in the pipette or burette giving false readings of how much iron solution was used or how much Potassium mangante VII solution was used. Interpretation of the “end point” may have caused errors. Correctly swirling the Iron solution while titrating would ensure the Potassium mangante molecules would correctly distribute around preventing a too sudden colour change to occur.
Bibliography
-
General Chemistry 5th edition- Ebbing, Houghton Mifflin Company 1996
-
Chemistry matters-Hart. Richard, Oxford University Press 1978
-
Chemistry in context 3rd edition- Hill, Graham et al, Thomas Nelson and Sons
1989
-
Chemistry 1-Brian Ratcliff et al, Cambridge University Press 2000
-
Calculations for A-level Chemistry 4th edition- Ramsden, E.N. , Nelson
Thornes 2001
Determining the number of molecules of water crystallization
in a sample of sodium carbonate crystal, Na2Co3xH2O
Introduction
The fundamental aim of this experiment is to determine how many molecules of water of crystallisation are found in a sample of sodium carbonate crystal. A titration method relevant to an acid-base reaction will be used to determine the quantity of hydrochloric acid needed to bring 25cm3 sodium carbonate solution to equilibrium. For two solutions to reach equilibrium means they are balanced in proportions to one other. Using the amount of hydrochloric acid and a balance experimental equation we can calculate the number of water molecules present.
The molecular equation for the experiment is:
Na2Co3(aq) + 2HCL(aq) 2NaCl(aq) + H2O(aq) + CO2(g)
The above equation indicates that the base, sodium carbonate will react with the acid, hydrochloric acid to form a salt, sodium chloride, water and carbon dioxide gas.
The Bronsted-Lowery theory on acid-base reaction states that:
“An acid is a proton donor A base is a proton acceptor”1 which means during a reaction when the acid and base get mixed a proton from the acid will transfer over to the base forming the end compound.
The final product of sodium carbonate and hydrochloric acid is actually carbonic acid (H2CO3(aq)) but this is a highly unstable compound and decomposes into water and carbon dioxide gas.
The end product of the reaction is a sodium chloride salt. Within a salt there is a set proportion of water in the crystal lattice which plays an essential role in the shape and occasionally the colour of the salt. This is known as water of crystalisation. Therefore in the lattice structure of sodium chloride each Na+ (cation) is surrounded by 5 water molecules producing a hydrated crystal.
Unlike a redox-titration, an indicator is needed in an acid-base titration. "Indicators are substances which change colour according to the hydrogen ions concentration of the solution to which they are added”, for example the experiments calls for the use of Methyl Orange indicator which in an alkali is a yellow colour however in an acid is red. A noticeable colour change using methyl orange should occur at around pH4. The point where the colour change occurs is know as the “end-point” and is when the solutions have reached equilibrium.
Method:
3.50g (approx.) of Sodium Carbonate crystals should be weighted out and deposited into a 250cm3 volumetric flask with 100cm3 deionised water. Crystals should then be fully dissolved and solution made up to the 250cm3 shaking thoroughly with a stopped in place. Using a pipette 25cm3 of stock sodium carbonate solution should be transferred into a clean conical flask and the burette filled with 0.209M hydrochloric acid. Add 3-4 drops of methyl orange indicator to the base solution and swirls (colour should be yellowish). Titrate the hydrochloric acid against the sodium carbonate solution until a noticeable (pink) colour change has occur making sure during titration the solution is swirled around to correctly distribute acid around base solution. Note down the amount of hydrochloric acid used and repeat titration twice.
Results:
Because 2 last results were identical that value will be my reading:
Reaction equation:
Na2CO3 (s) + 2HCL 2NaCl(aq) + CO2(g) + H2O(l)
Calculations:
Concentration of HCL: 0.209M
Amount of HCL needed: 11.70 cm3
Mass of Sodium Carbonate crystal: 3.54g
Molecular mass: Na2Co3= 105.988 H2O= 18.0248
(The values have been changed to dm-3 form to fit in with equation (multiplying cm3 by 10-3))
Concentration of Na2CO3-
Amount of HCL= Volume used (dm-3) X Concentration of HCL
= (11.70x10-3) X 0.209
= 2.4453x10-3 mol of HCL
according to the reaction equation there is half as Na2CO3 much then HCL so to calculate the moles of Na2CO3 present:
2.4453x10-3 X 0.5= 1.22265x10-3 mol of Na2CO3
to change this into concentration:
Concentration (mol dm-3) = Amount of Na2CO3 (mol)
Volume (dm-3)
= 1. 22265x10-3
25.0x10-3
= 0.048906 mol dm-3 of Na2CO3
Concentration of Na2CO3 in g dm-3–
0.0484 (mol dm-3 of Na2CO3) X105.988= 5.183449128 g dm-3
Concentration of Na2Co3xH2O in g dm-3–
3.54 (mass of original sample) X 0.0484 mol = 0.1694 g
Calculation of x
The previous equations were based upon a 250 cm3 solution however the calculation of x needs to be based upon a 1000 cm3 (1 liter) solutions:
For this we multiply the original mass of sodium carbonate by (1000/250=) 4:-
3.54 x 4=14.16g of sodium carbonate in 1 liter of solution
of the 14.16g, 5.183449128g is Na2Co3 which means the mass of water will be (14.16-5.183449128=) 8.976550872g
to work out the ratio of water molecules the amount in grams of the Na2Co3 and water are divided by their molecular mass’ then ratio against each other:
Na2Co3= 5.183449128 H20= 8.976550872
105.988 18.0248
= 0.048906 = 0.498011122
For the ratio we divide by the lowest value=
Na2Co3= 0.048906 H20= 0.498011122
0.048906 0.048906
= 1 10.18302707 which is rounded to 10
The fixed percentage of water molecules located in the crystal lattice:
(number of water molecules/total number of moles present)-
(10/11) x 100 = 90.9%
Conclusion
The aim of this experiment was to see how many molecules of water of cryatallisation were in a sample of sodium carbonate crystal. Once sample was added to the ionized water it readily dissolved without a change in environment i.e. change in temperature. This was because within the structural crystal lattice of sodium carbonate are ions which when mixed with water, are attracted by its polar molecules causing the solid structure of the salt to dissolve forming an aqueous solution.
It took 11.70cm-3 of hydrochloric acid to neutralise/equilibrate 25.0cm-3 of the sodium carbonate base in the presence of the weak acid indicator methyl orange.
From these results along with a sequence of calculations it was found that for every molecule of sodium carbonate in a sodium carbonate crystal there is 10 molecules of water which gives the formula for sodium crystals, based on my experiment to be Na2Co310H2O.
Within the crystal lattice of a sodium carbonate salt there are 5 water molecules surrounding every sodium ion (Na+) and because there are two ions present there are 10 water molecules present. This corresponds with the formula of sodium carbonate crystals given by Hart which means that my experiment was successful in that it produced readings which amounted to the correct value for x. In the background research it was stated that a fixed percentage of the crystal lattice of the salt is made up of water giving the water of crystallisation therefore from the calculation done the percentage of water molecules present can be worked out. I found this to be 90.9%.
The variations in the experiments were kept to a minimum with the all base solutions for the three titrations deriving from the same stock solution and the same equipment i.e. burette and pipette we kept the same.
As with most titration experiments possibly carrying out more than three titrations could increase the accuracy of the results further possibly making the ratio calculations closer to 10 than my own.
Bibliography
-
General Chemistry 5th edition- Ebbing, Houghton Mifflin Company 1996
-
Chemistry matters-Hart. Richard, Oxford University Press 1978
-
Chemistry in context 3rd edition- Hill, Graham et al, Thomas Nelson and Sons
1989
-
Chemistry 1-Brian Ratcliff et al, Cambridge University Press 2000
-
Calculations for A-level Chemistry 4th edition- Ramsden, E.N. , Nelson
Thornes 2001
Ebbing, General Chemistry
1 Hill and Holman, Chemistry in Context, p.225
Hill and Holman, Chemistry in Context, p. 383
Hart, Chemistry Matters, p119
Hart, Chemistry Matters p124