Electrode potentials vary with temperature and concentrations of solutions therefore this experiment must be carried out under standard conditions (i.e. at a temperature of 298K and using a 1.0 mol dm –3 Copper Sulphate (aq)) (4). The Iron (II) Ammonium Sulphate (aq) concentration will be varied, as I want to see the effect of varying the concentration of solutions.
In this experiment Iron will be acting as the negative terminal as it has a more negative electrode potential. Thus an oxidation reaction will be taking place and the electrons produced will be supplied to the reduction reaction which is occurring in the less negative Copper half-cell (4).
Method (Adapted from SS 3.2 – Simple electrochemical cells)
-
Make up a set of various concentrations of Iron (II) Ammonium Sulphate by completing the following steps:
-
Make up a Iron (II) Ammonium Sulphate (aq) at a concentration of 0.1 mol dm-3 .
-
Accurately weigh out 3.92 grams of hydrated Iron (II) Ammonium Sulphate(s) using a balance and place it in a 100cm3 beaker.
-
Using a 50 cm3 measuring cylinder measure out 25cm3 of Sulphuric Acid (aq) into the beaker and using a glass rod stir the acid and solid together until the entire solid has dissolved.
-
Transfer the contents of the beaker through a small funnel into a 100cm3 volumetric flask.
-
Rinse the beaker and glass rod twice using small quantities of Sulphuric Acid (aq) and transfer all the washings to the volumetric flask.
-
Rinse the funnel with a small amount of Sulphuric Acid (aq) and transfer that into the volumetric flask.
-
Add Sulphuric Acid (aq) to the volumetric flask until it is about 1 cm below the graduation mark.
- Slowly add the acid using a clean dropping pipette until the bottom of the meniscus is touching the graduation mark.
- Stopper the flask and invert it several times.
-
Pipette 9 cm3 of the already made up Iron (II) Ammonium Sulphate (aq) into 10 cm3 volumetric flask.
-
Using a dropping pipette add Sulphuric Acid (aq) to the volumetric flask until the bottom of the meniscus is touching the graduation mark.
-
Stopper the flask and invert it several times. This will make up a solution with the concentration of 0.09 mol dm-3.
-
Repeat points j through to l varying the volume of 0.1 mol dm-3 Iron Ammonium Sulphate added to the volumetric flask to make up solutions at the concentrations of 0.08 mol dm-3, 0.07 mol dm-3, 0.06 mol dm-3, 0.05 mol dm-3, 0.04 mol dm-3 , 0.03 mol dm-3, 0.02 mol dm-2 and 0.01 mol dm-3.
-
Make up a solution of Copper Sulphate solution 0.1 mol dm-3 by completing the following steps :
-
.Accurately weigh out 2.5 grams of hydrated Zinc Sulphate using a balance and place it in a 100cm3 beaker
-
Pour about 25cm3 of distilled water into the beaker and using a glass rod stir the water and solid together until the entire solid has dissolved
-
Transfer the contents of the beaker through a small funnel into a 50cm3 volumetric flask
- Rinse the beaker and glass rod twice using small quantities of distilled water and transfer all the washings to the volumetric flask
- Rinse the funnel with a small amount of distilled water and transfer that into the volumetric flask
- Add distilled water to the volumetric flask until it is about 1 cm below the graduation mark
-
Clean a strip of Iron (s) using emery paper, rinse it in distilled water (l) and dry it using a paper towel.
-
Clean a strip of Copper (s) using emery paper, rinse it in distilled water (l) and dry it using a paper towel.
-
Using a measuring cylinder measure out 10cm3 of 1.0 mol dm-3 Copper Sulphate (aq) and place it in a boiling tube.
-
Using a measuring cylinder, measure out 10cm3 of 0.1 mol dm-3 Iron (II) Ammonium Sulphate (aq) and place it in a second boiling tube.
-
Place the Copper (s) strip in the boiling tube that contains Copper Sulphate (aq) and the Iron (s) strip in the boiling tube that contains Iron (II) Ammonium Sulphate (aq).
-
Take a piece strip of filter paper and soak it in saturated Potassium Nitrate (V) (aq), use this as the salt (ion) bridge to connect the solutions in the two boiling tubes.
- Attach crocodile clips to the two strips of metal and using wire connect them both to a high-resistance voltmeter.
- Record the voltage of the Iron-Copper cell, and note which side is positive and which is negative.
-
Repeat the experiment using varying concentrations of Iron (II) Ammonium Sulphate (aq) and a fresh salt (ion) bridge each time.
How the apparatus for this experiment should be set up
Justifications
A fresh piece salt (ion) bridge will be used for each new experiment to ensure that it does not contain any ions from the previous experiment that could cause the experiment to appear inaccurate.
Saturated Potassium Nitrate (aq) was used for the salt bridge rather than a piece of wire to avoid further metal/ion potentials in the circuit. It was also chosen because it did not react with either of the solutions in the half-cells.
In order to dilute down the Iron (II) Ammonium Sulphate (aq) into various solutions of different concentrations a certain amount had to be removed from the original solution and placed in a volumetric flask, which was filled up to the graduation mark with Sulphuric Acid (aq). If two volumes of Sulphuric Acid (aq) and Iron (II) Ammonium Sulphate (aq) were added together they might not have produced 10cm3 of solution as the solutions contained different densities thus the required concentration of solution would not have been produced. For this reason the solutions were made up into the graduation mark in 10cm3 volumetric flasks.
Modifications
I was unable to complete this experiment as I ran out of time, thus there are no modifications.
After looking at the three preliminary experiments I found that a redox titration between Potassium Manganate (VII) (aq) and Iron (II) Ammonium Sulphate (aq) was the most accurate in determining the concentrations of the Iron (II) Ammonium Sulphate (aq). The colorimetry experiment did not work out as I had expected it to and I did not have enough time to complete the Electrochemical cells, and for these reasons I have chosen to use a redox titration to find out how much Iron (II) there is in Spinach Oleracea.
Experiment Four - How temperature effects a titration between Oxalic acid (aq) and Potassium Manganate (aq)
Aim
To find out how temperature effects a titration between Oxalic Acid (aq) and Potassium Manganate (VII) (aq).
Background Theory
Spinach Oleracea contains a large concentration of Iron (II) but much of it is locked up in insoluble ethandioate (oxalate) complexes (2).
Potassium Manganate (VII) oxidises the oxalate. The reaction is slow at room temperature and so will be carried out at 70oc. the reaction is catalysed by Mn2+ ions which is a product of the reaction. As the Potassium Manganate (VII) is added the pink colour will remain until several millimetres have been added, the pink colour will then disappear as more is added until the end point is reached. At this point, one excess drop of Potassium Manganate will turn the solution pink (3).
5C2O42- (aq) + 2MnO4- (aq) +16H+ (aq) 10CO2 (g) + 8H2O (l) + 2Mn2+(aq)
A redox reaction is involved in this experiment, and by giving the reactants and products oxidation states it is possible to predict which chemical is the oxidising agent and which is the reducing agent.
5C2O42- (aq) + 2MnO4- (aq) +16H+ (aq) 10CO2 (g) + 8H2O (l) + 2Mn2+(aq)
The half reaction involving a loss of electrons is known as the oxidation reaction and the reaction that involves a gain in electrons is known as a reduction reaction (6)
Oxalic acid occurs naturally at in quite a large number of plants, it combines with iron to form less soluble salts known as oxalates. Oxalates also occur naturally in plants. The oxalic acid binds with important nutrients making the inaccessible to the body (10).
Method (Adapted from source 3)
-
Make up a solution of 0.1 mol dm-3 Oxalic Acid (aq) by completing the following steps:
a. Accurately weigh out 3.15 grams of hydrated Oxalate (s) using a balance and transfer it into a 100cm3 beaker.
-
Using a 50 cm3 measuring cylinder measure out approximately 25cm3 of 1.0 mol dm-3 Sulphuric Acid (aq) and 50 cm3 distilled water (l) and pour it into the beaker.
- Using a glass rod stir the solid and liquid together until the entire solid has dissolved.
-
Transfer the contents of the beaker through a small funnel into a 250 cm3 volumetric flask.
-
Rinse the beaker and glass rod twice using small quantities of Sulphuric Acid (aq) and transfer all the washings through the funnel to the volumetric flask.
-
Rinse the funnel with a small amount of Sulphuric Acid (aq) and transfer it into the volumetric flask.
-
Add Sulphuric Acid (aq) to the volumetric flask until it is about 1 cm below the graduation mark.
- Slowly add the acid using a clean dropping pipette until the bottom of the meniscus is touching the graduation mark.
- Stopper the flask and invert it several times.
-
Using a measuring cylinder measure out 30 cm3 of Oxalic Acid (aq) and place it in a 100cm3 beaker and place this in an ice bucket with a thermometer and wait for it to cool to 0 o c.
-
Again using a measuring cylinder measure out 30 cm3 of Oxalic Acid (aq) and transfer it to another beaker and place this in a heated water bath set to a temperature of 60oc, place a thermometer in the solution to monitor the temperature.
-
Again using a measuring cylinder measure out 30 cm3 of Oxalic Acid (aq) and transfer it to another beaker and place this in a heated water bath set to a temperature of 70oc, place a thermometer in the solution to monitor the temperature.
- Leave the volumetric flask out on a workbench in order for it to cool down to room temperature.
-
Using a 25cm3 pipette, transfer 20cm3 of the 0 oc Oxalic Acid (aq) into a conical flask. Leave the rest of the solution in the ice bucket to maintain its temperature.
- Place the conical flask on a white tile.
-
Titrate the Potassium Manganate (VII) (aq) into the Oxalic Acid (aq) in small quantities while swirling the conical flask until the first trace of a pink colour appears.
-
Continue to add the Potassium Manganate (VII) (aq) to the conical flask drop by drop, swirling between each drop, until the solution remains a pink colour for 30 seconds.
- Repeat this experiment 3 times in order to gain an idea of how the temperature is effecting the reaction.
-
Repeat steps 5 –11 for each of the solutions created (i.e.- room temperature 60oc and 70oc). Record the data in a suitable table.
Modifications
I found after my first titration that it was taking a large volume of Potassium Manganate (VII) (aq) to react with the 20cm3 of Oxalic Acid (aq), and I was running out of Potassium Manganate (VII) (aq) in the 50 cm3 burette. For this reason I thought it would be necessary to lower the volume of Oxalic Acid (aq) being used in the reaction first to 10 cm3 and then to 5 cm3 as I realised that 10 cm3 was still taking too much Potassium Manganate (VII) (aq) to reach the end point of the reaction.
I also found that the temperatures of the solutions under test were changing once the Potassium Manganate (VII) (aq) was being added. The solutions cooler than the Potassium Manganate (VII) (aq) were becoming warmer and those warmer were becoming cooler. To overcome this I found it necessary to reheat or cool the solutions every so often. I found that the best method was to add 10cm3 of Potassium Manganate (VII) (aq) and then reheat or cool the solutions to there original temperatures, add a further 5cm3 and again reheat or cool the solutions and continue to reheat or cool the solutions after each 5 cm3 volume of Potassium Manganate (VII) (aq) had been added.
Justifications
The solutions that I was required to make myself were made up into volumetric flasks as these could provide a more accurate measurement for the volume of sulphuric acid that needed to be added to make the solution at the correct concentration.
The solution of Oxalic Acid (aq) that was used in this experiment was made to a concentration of 0.1 mol dm-3 as this is the same concentration as the Iron (II) Ammonium Sulphate (aq) that was used in earlier experiments as this would keep the experiment constant when compared to the others.
The solutions were keep in the water baths or ice buckets before and during experiments so that the solution remained at a constant temperature.
As the volume of Sulphuric Acid (aq) added to the Iron (II) Ammonium Sulphate (aq) had to be in excess to supply H+ ions I felt it would not be necessary to have an exact volume so I used a measuring cylinder to measure the quantity required instead of the more accurate equipment like a pipette which I used to measure out the volumes of the other solutions.
Three values within 0.1 cm3 of each other need to recorded in this experiment, as an average titre needs to be obtained. If points within 0.1 cm3 of each other are used the average results will be more accurate. This will also ensure that any random results are noted and not included in the equation that I will use to work out the average.
Experiment Five - How much Iron (II) can be extracted from 15 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq)
Aim
To find out how much Iron (II) can be extracted from 15 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq) and then use this data to work out how much Iron (II) is present in 100 grams of Spinach Oleracea..
Method (Adapted from SS1.2 - How much iron in spinach?)
-
Using a balance accurately weigh out 15 grams of Spinach Oleracea.
-
Using scissors cut the leaf into small equal sized pieces and transfer it to a 250 cm3 beaker.
-
Using a 50 cm3 measuring cylinder measure out 50 cm3 of Sulphuric Acid (aq) and add it to the beaker.
- Using a tripod and Bunsen burner, gently boil the spinach for 5 minutes, stirring occasionally using a glass rod.
- Leave the mixture to cool and filter it using a Butcher Funnel and vacuum filtration.
-
Wash the beaker and glass rod twice using a small volume of Sulphuric Acid (aq) to collect up any remaining Iron (II).
-
Wash the residue in the funnel once with a little Sulphuric Acid (aq) and collect the filtrate.
-
Transfer the filtrate and washings into a 100 cm3 volumetric flask and add Sulphuric Acid to the volumetric flask until it is about 1 cm3 below the graduation mark.
-
Using a clean dropping pipette add Sulphuric Acid (aq) to the volumetric flask until the bottom of the meniscus is touching the graduation mark.
- Stopper the flask and invert it several times.
-
Full a 50 cm3 burette with 0.01 mol dm-3 Potassium Manganate (VII) (aq).
-
Using a 20 cm3 pipette transfer 20 cm3 of the spinach extract solution to a conical flask.
-
Using a 50 cm3 measuring cylinder, measure out 50 cm3 of Sulphuric Acid (aq) and transfer it to the conical flask.
-
Heat the solution over a Bunsen burner monitoring the temperature with a thermometer until it reaches 70 oc.
- Place the conical flask containing the hot solution on a white tile.
-
Titrate the Potassium Manganate (VII) (aq) into the hot spinach extract solution until a pink colour appears.
-
Continue to add the Potassium Manganate (VII) (aq) drop by drop while swirling the conical flask until the pink colour persists for 30 seconds.
- Repeat the experiment until 3 results within 0.1 of each other have been recorded.
Modifications
I found that the 20 cm3 of Spinach Extract solution that I used took to much Potassium Manganate (VII) (aq) for it to reach the end point of the reaction and this was taking too long. For this reason I have decided to titrate the Potassium Manganate (VII) (aq) into 10 cm3 of spinach extract solution and 25 cm3 of Sulphuric Acid (aq).
Justifications
Only 15 grams of Spinach Oleracea were used in this experiment as I wanted to see how much Potassium Manganate (VII) (aq) it would take to reach an end point, before deciding whether to use 10 cm3 or 20 cm3 in the final experiments.
The spinach was boiled in Sulphuric Acid (aq) as this broke the spinach up completely and it formed a mush, and thus all the Iron (II) would be extracted from it.
The solution was boiled to a temperature of 70 oc as this is hot enough to break up the oxalate complexes around the Iron (II). Although my preliminary experiment to show this did not work, I knew from other experiments and text books that this was needed for the reaction to occur.
As the volume of Sulphuric Acid (aq) added to the Iron (II) Ammonium Sulphate (aq) had to be in excess to supply H+ ions I felt it would not be necessary to have an exact volume so I used a measuring cylinder to measure the quantity required instead of the more accurate equipment like a pipette which I used to measure out the volumes of the other solutions.
Three values within 0.1 cm3 of each other need to recorded in this experiment, as an average titre needs to be obtained. If points within 0.1 cm3 of each other are used the average results will be more accurate. This will also ensure that any random results are noted and not included in the equation that I will use to work out the average.
Experiment Six - How much Iron (II) can be extracted from 20 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq)
Aim
To find out how much Iron (II) can be extracted from 20 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq) and then use this data to work out how much Iron (II) is in 100grams of Spinach Oleracea.
Method (Adapted from SS1.2 - How much iron in spinach?)
-
Using a balance accurately weigh out 20 grams of Spinach Oleracea.
-
Using scissors cut the leaf into small equal sized pieces and transfer it to a 250 cm3 beaker.
-
Using a 50 cm3 measuring cylinder measure out 50 cm3 of Sulphuric Acid (aq) and add it to the beaker.
- Using a tripod and Bunsen burner, gently boil the spinach for 5 minutes, stirring occasionally using a glass rod.
- Leave the mixture to cool and filter it using a Butcher Funnel and vacuum filtration.
-
Wash the beaker and glass rod twice using a small volume of Sulphuric Acid (aq) to collect up any remaining Iron (II).
-
Wash the residue in the funnel once with a little Sulphuric Acid (aq) and collect the filtrate.
-
Transfer the filtrate and washings into a 100 cm3 volumetric flask and add Sulpheric Acid (aq) to the volumetric flask until it is about 1 cm3 below the graduation mark.
-
Using a clean dropping pipette add Sulphuric Acid (aq) to the Volumetric flask until the bottom of the meniscus is touching the graduation mark.
- Stopper the flask and invert it several times.
-
Full a 50 cm3 burette with 0.01 mol dm-3 Potassium Manganate (VII) (aq).
-
Using a 10 cm3 pipette transfer 10 cm3 of the spinach extract solution to a conical flask.
-
Using a 50 cm3 measuring cylinder, measure out 25 cm3 of Sulphuric Acid (aq) and transfer it to the conical flask.
-
Heat the solution over a Bunsen burner monitoring the temperature with a thermometer until it reaches 70 oc.
- Place the conical flask containing the hot solution on a white tile.
-
Titrate the Potassium Manganate (VII) (aq) into the hot spinach extract solution until a pink colour appears.
-
Continue to add the Potassium Manganate (VII) (aq) drop by drop while swirling the conical flask until the pink colour persists for 30 seconds.
- Repeat the experiment until 3 results within 0.1 of each other have been recorded.
-
Repeat the experiment using a second sample of Spinach Oleracea.
Modifications
Again I found that the volume of spinach extract solution I had purposed to use was using up too much Potassium Manganate (VII) (aq), thus I decided to decrease the volume of the spinach extract solution to be used in each titration.
Justifications
20 grams of Spinach Oleracea were boiled to create this solution in order to make the final part of the experiment, where I work out the volume of Iron (II) in 100 grams of Spinach Oleracea easier as I would simply have to multiply the volume of Iron (II) in 20 grams by 5.
The spinach was boiled in Sulphuric Acid (aq) as this broke the spinach up completely and it formed a mush, and thus all the Iron (II) would be extracted from it.
The solution was boiled to a temperature of 70 oc as this is hot enough to break up the oxalate complexes around the Iron (II). Although my preliminary experiment to show this did not work, I knew from other experiments and text books that this was needed for the reaction to occur.
As the volume of Sulphuric Acid (aq) added to the Iron (II) Ammonium Sulphate (aq) had to be in excess to supply H+ ions I felt it would not be necessary to have an exact volume so I used a measuring cylinder to measure the quantity required instead of the more accurate equipment like a pipette which I used to measure out the volumes of the other solutions.
Three values within 0.1 cm3 of each other need to recorded in this experiment, as an average titre needs to be obtained. If points within 0.1 cm3 of each other are used the average results will be more accurate. This will also ensure that any random results are noted and not included in the equation that I will use to work out the average.
I had to repeat this experiment as I ran out of Potassium Manganate (VII) (aq) and when I tried to create a new solution of it I got my calculations incorrect and thus created a solution of 0.025 mols dm-3 for Potassium Manganate (VII) (aq). By redoing the experiment I had results that I could compare to future experiments.
I created a second spinach extract solution using 20 grams of Spinach Oleracea. This was done so that I could compare my results from the two solutions and thus be able to detect whether one of the solutions had been contaminated. I would be able to do this, as both solutions should have created similar average titres.
Experiment Seven - How much Iron (II) can be extracted from 20 grams of Spinach Oleracea when boiled in Distilled Water(l)
Aim
To find out how much Iron (II) can be extracted from 20 grams of Spinach Oleracea when boiled in distilled water (l), and then use this data to work out how much Iron (II) is in 100grams of Spinach Oleracea.
Method (Adapted from SS1.2 - How much iron in spinach?)
-
Using a balance accurately weigh out 20 grams of Spinach Oleracea.
-
Using scissors cut the leaf into small equal sized pieces and transfer it to a 250 cm3 beaker.
-
Using a 50 cm3 measuring cylinder measure out 50 cm3 of distilled water (l) and add it to the beaker.
- Using a tripod and Bunsen burner, gently boil the spinach for 10 minutes, stirring occasionally using a glass rod.
- Leave the mixture to cool and filter it using a Butcher Funnel and vacuum filtration.
-
Wash the beaker and glass rod twice using a small volume of distilled water (l) to collect up any remaining Iron (II).
-
Wash the residue in the funnel once with a little distilled water (l) and collect the filtrate.
-
Transfer the filtrate and washings into a 100 cm3 volumetric flask and add distilled water (l) to the volumetric flask until it is about 1 cm3 below the graduation mark.
-
Using a clean dropping pipette add distilled water (l) to the Volumetric flask until the bottom of the meniscus is touching the graduation mark.
- Stopper the flask and invert it several times.
-
Full a 50 cm3 burette with 0.01 mol dm-3 Potassium Manganate (VII) (aq).
-
Using a 10 cm3 pipette transfer 10 cm3 of the spinach extract solution to a conical flask.
-
Using a 50 cm3 measuring cylinder, measure out 25 cm3 of Sulphuric Acid (aq) and transfer it to the conical flask.
-
Heat the solution over a Bunsen burner monitoring the temperature with a thermometer until it reaches 70 oc.
- Place the conical flask containing the hot solution on a white tile.
-
Titrate the Potassium Manganate (VII) (aq) into the hot spinach extract solution until a pink colour appears.
-
Continue to add the Potassium Manganate (VII) (aq) drop by drop while swirling the conical flask until the pink colour persists for 30 seconds.
- Repeat the experiment until 3 results within 0.1 of each other have been recorded.
-
Repeat the whole experiment using a second sample of Spinach Oleracea.
Modifications
I originally planed to boil the Spinach Oleracea in 50 cm3 of distilled water (l) for 10 minutes but I found that the water all boiled away leaving the Spinach Oleracea burning at the bottom of the beaker. For this reason I decided to boil the Spinach Oleracea in 100 cm3 for 10 minutes. This would allow the Spinach Oleracea to boil for the required time thus allowing as much Iron (II) as possible to leave the Spinach Oleracea.
Justifications
20 grams of Spinach Oleracea were boiled to create this solution in order to make the final part of the experiment, where I work out the volume of Iron (II) in 100 grams of Spinach Oleracea easier as I would simply have to multiply the volume of Iron (II) in 20 grams by 5.
The spinach was boiled in distilled water (l) as this is how the Spinach would be prepared if it was to be eaten. I also wanted to see how this affected the volume of Iron (II) that I was able to extract from the Spinach Oleracea.
The solution was boiled to a temperature of 70 oc as this is hot enough to break up the oxalate complexes around the Iron (II). Although my preliminary experiment to show this did not work, I knew from other experiments and text books that this was needed for the reaction to occur.
The Spinach Oleracea was boiled for 10 minutes in the distilled water (l), which is longer than when it was boiled in the Sulphuric Acid (aq) and this is because the Spinach Oleracea will take longer to release the Iron (II) under these conditions.
As the volume of Sulphuric Acid (aq) added to the Iron (II) Ammonium Sulphate (aq) in the conical flask had to be in excess to supply H+ ions I felt it would not be necessary to have an exact volume so I used a measuring cylinder to measure the quantity required instead of the more accurate equipment like a pipette which I used to measure out the volumes of the other solutions.
Three values within 0.1 cm3 of each other need to recorded in this experiment, as an average titre needs to be obtained. If points within 0.1 cm3 of each other are used the average results will be more accurate. This will also ensure that any random results are noted and not included in the equation that I will use to work out the average.
I created a second spinach extract solution using 20 grams of Spinach Oleracea. This was done so that I could compare my results from the two solutions and thus be able to detect whether one of the solutions had been contaminated. I would be able to do this, as both solutions should have created similar average titres.
Concept Map
Risk Assessments
Bibliography
- Understanding Biology for advanced level (Third Edition)
Glenn and Susan Toole
Stanley Thornes Publishers –Cheltenham (1995)
Canadian Health Authority
- www.chemistry.eku.edu/GODBEy/Homepage_files/CHE_112…/oxred_oxalate.htm
- Chemical Ideas (Second Edition)
G. Bulon, J. Holman, J. Lazonby, G. Pilluis, D.Waddington
Heinemann
The Bath Press-Bath (2000)
- El 2.1 How much Iron is in a sample of an Iron compound?
Salters Advanced Chemistry (2000)
- Understanding Chemistry for Advanced Level (Third Edition)
Ted Lister and Janst Renshaw
Stanley Thornes Publishers-Cheltenham (2000)
- Chemistry (fourth Edition)
Raymond Chang
Van Hoffman Press Inc.-America (1991)
- A new introduction to Biology (AS AQA Biology Specification A)
B. Indge, M. Rowland, M. Baker
Hodder and Stoughton Eductional-London (2000)
- Chemistry
Ann and Patrick Fullick
Heinemann- Oxford (1994)
ARS Phytochemical Database
Implementing
Experiment One - How accurate is a Potassium Manganate titration in determining the concentration of Iron (II) Ammonium Sulphate (aq)
Experiment A -
[Iron (II) Ammonium Sulphate (aq)] = 0.1 mol dm-3
[Potassium Manganate (VII) (aq)] = 0.1 mol dm-3
Average titre =1.95 + 2.05 + 1.95 + 1.90 + 2.00
5
= 1.97 cm3
Experiment B-
[Iron (II) Ammonium Sulphate (aq)] = 0.1 mol dm-3
[Potassium Manganate (VII)(aq)] = 0.01 mol dm-3
Average titre = 19.85 + 19.70 +19.80 + 19.95 +19.90
5
= 19.84 cm3
Experiment Two - How accurate is a Calibration Graph in determining the concentration of Iron (II) Ammonium Sulphate (aq)
Experiment A – I was unable to produce data for this experiment, as the concentrations I had proposed to use were too concentrated to be able to dissolve the solid fully in a standard volume of Sulphuric Acid (aq). I was able to dissolve them by heating the Iron (II) Ammonium Sulphate (s) with Sulphuric Acid (aq) in a beaker over a Bunsen burner but as the solution cooled, the Iron (II) Ammonium solidified and separated from the acid.
Experiment B –
Experiment Three - How accurate is an Electrochemical Cell in determining the concentration of Iron (II) Ammonium Sulphate (aq)
Unfortunately I was unable to gain data for this experiment as I ran out of time before I was able to complete it.
The results should have shown that as the concentration of Iron (II) Ammonium Sulphate (aq) increased the voltage being passed between the solutions would have increased. This would have created a graph showing that an increase in concentration produces an increase in the voltage produced by the Electrochemical Cell. From this I would have been able to work out the concentration of Iron (II) in Spinach Oleracea by recording the voltage from the Electrochemical Cell produced by combining Spinach Extract (aq)/ Fe(s) and Cu2+ (aq) / Cu (s) and then reading the concentration of the solution off the graph.
Experiment Four - How temperature effects a titration between Oxalic acid (aq) and Potassium Manganate (aq)
[Oxalic Acid (aq)] = 0.1 mol dm-3
[Potassium Manganate (aq)] = 0.01 mol dm-3
In this experiment I noticed that there were bubbles of gas produced during the titration. By looking at the chemical equation of the reaction that was occurring here it is possible to deduce that the gas being produced was Carbon Dioxide.
Average Titre at 0 oc = 30.00 + 29.85 + 29.60
3
= 29.82cm3
Average Titre at 25 oc = 25.45 + 25.25 + 25.20
3
= 25.30cm3
Average Titre at 60 oc = 23.25 + 21.00 + 20.50
3
= 21.58 cm3
Average Titre at 70oc = 21.10 + 20.00 + 21.00
3
= 20.70 cm3
Experiment Five - How much Iron (II) can be extracted from 15 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq)
Experiment A – preformed using a spinach extract solution created by boiling 15 grams of spinach in Sulphuric acid (aq) and making it up to a 100cm3 solution by adding Sulphuric Acid (aq).
[Potassium Manganate (VII) (aq)] = 0.01 mol dm-3
20 cm3 of Spinach extract solution mixed with 50cm3 Sulphuric Acid (aq)
Spinach extract solution heated to 70 oc before titration
Average titre = 23.80 + 17.70 + 24.00
3
= 21.83 cm3
Experiment B - preformed using a spinach extract solution created by boiling 15 grams of spinach in Sulphuric acid (aq) and making it up to a 100cm3 solution by adding Sulphuric Acid (aq).
[Potassium Manganate (VII)(aq)] = 0.01 mol dm-3
10 cm3 of Spinach extract solution mixed with 20 cm3 Sulphuric Acid (aq)
Spinach extract solution heated to 70 oc before titration
Average titre = 18.20 + 19.80 + 18.90
3
= 18.97 cm3
Experiment Six - How much Iron (II) can be extracted from 20 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq)
Experiment A - preformed using a spinach extract solution created by boiling 20 grams of spinach in Sulphuric acid (aq) and making it up to a 100cm3 solution by adding Sulphuric Acid (aq).
[Potassium Manganate (VII) (aq)] = 0.01 mol dm-3
10 cm3 spinach extract solution and 50 cm3 Sulphuric Acid
Spinach extract solution heated to 70oc before titration
Average titre = 24.00 + 22.40
2
= 23.20 cm3
Experiment B - preformed using a spinach extract solution created by boiling 20 grams of spinach in Sulphuric acid (aq) and making it up to a 100cm3 solution by adding Sulphuric Acid (aq).
[Potassium Manganate (VII) (aq)] = 0.01 mol dm-3
5 cm3 spinach extract solution and 25 cm3 Sulphuric Acid
Spinach extract solution heated to 70oc before titration
Average titre = 15.70 + 15.60 + 15.65
3
= 15.65 cm3
Experiment C - preformed using a spinach extract solution created by boiling 20 grams of spinach in Sulphuric acid and making it up to a 100cm3 solution by adding Sulphuric Acid (aq).
[Potassium Manganate (VII) (aq)] = 0.025 mol dm-3
5 cm3 spinach extract solution and 25 cm3 Sulphuric Acid
Spinach extract solution heated to 70oc before titration
Average Titre = 7.15 + 6.95 + 7.40
3
= 7.17 cm3
Experiment D - preformed using a different spinach extract solution but also created by boiling 20 grams of spinach in Sulphuric acid and making it up to a 100cm3 solution by adding Sulphuric Acid (aq).
[Potassium Manganate (VII) (aq)] = 0.025 mol dm-3
5 cm3 spinach extract solution and 25 cm3 Sulphuric Acid
Spinach extract solution heated to 70oc before titration
Average Titre = 7.30 + 7.30 + 7.20
3
= 7.27cm3
Experiment Seven - How much Iron (II) can be extracted from 20 grams of Spinach Oleracea when boiled in Distilled Water(l)
Experiment A - preformed using a spinach extract solution created by boiling 20 grams of spinach in Distilled Water (l) and making it up to a 100cm3 solution by adding Sulphuric Acid (aq).
[Potassium Manganate (VII) (aq)] = 0.025 mol dm-3
5 cm3 spinach extract solution and 25 cm3 Sulphuric Acid
Spinach extract solution heated to 70oc before titration
Average Titre = 2.10 + 2.15 + 2.05
3
= 2.10 cm3
Experiment B – preformed using a different spinach extract solution also created by boiling 20 grams of spinach in Distilled Water (l) and making it up to a 100cm3 solution by adding Sulphuric Acid (aq).
[Potassium Manganate (VII) (aq)] = 0.025 mol dm-3
5 cm3 spinach extract solution and 25 cm3 Sulphuric Acid
Spinach extract solution heated to 70oc before titration
Average Titre = 2.10 + 2.05 + 2.20 + 2.25 +2.35
5
= 2.19 cm3
Analysing Evidence and Drawing Conclusions
Experiment One - How accurate is a Potassium Manganate titration in determining the concentration of an Iron (II) Ammonium Sulphate (aq)
Experiment A –
5Fe2+ (aq) + MnO4 (aq) + 8H+ (aq) 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)
The equation above shows that 1 mol of MnO4 reacts with 5 mol of Fe2+, and by using the equation below it is possible to work out the concentration of the Iron (II) Ammonium Sulphate.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.1 mol dm –3 x 1.97 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.000197 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) is 1:5. Therefore to work out the mols of Iron (II) Ammonium Sulphate (aq) used in the titration I need to multiply the moles of Potassium Manganate (VII) (aq) by 5.
0.000197 mol dm-3 x 5 = 0.000985 mol dm-3
Now that I know the mols of Iron (II) Ammonium Sulphate (aq) present in the titration I can work out the concentration of it.
Moles = Concentration x Volume
Concentration = Moles
Volume
Concentration = 0.000985
10
Concentration = 0.0000985 mol dm-3
This answer tells me how many mols are present in 1000cm3 but as I only used 100cm3 solution I need to multiply my results by 1000.
0.0000985 x 1000 = 0.0985 mols dm-3
This answer is very close to the expected results of 0.1 mol dm-3 that I was expecting for this experiment. From this I can conclude that this experiment is accurate in determining the concentration of Iron (II) Ammonium Sulphate (aq). However I chose to repeat this experiment using a more diluted concentration of Potassium Manganate (VII) (aq) to see if I can produce more accurate results, as it was difficult to take readings of the 0.1 mol dm-3 Potassium Manganate (VII) (aq) in the burette.
Experiment B –
5Fe2+ (aq) + MnO4 (aq) + 8H+ (aq) 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l)
The equation above shows that 1 mol of MnO4 reacts with 5 mol of Fe2+, and by using the equation below it is possible to work out the concentration of the Iron (II) Ammonium Sulphate.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 19.84 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.0001984 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) is 1:5. Therefore to work out the mols of Iron (II) Ammonium Sulphate (aq) used in the titration I need to multiply the moles of Potassium Manganate (VII) (aq) by 5.
0.0001984 mol dm-3 x 5 = 0.000992 mol dm-3
Now that I know the mols present in the Iron (II) (aq) and the volume used in the titration I can work out the concentration of it.
Moles = Concentration x Volume
Concentration = Moles
Volume
Concentration = 0.000992
10
Concentration = 0.0000992 mol dm-3
This answer tells me how many mols are present in 1000cm3 but as I only a 100cm3 solution I need to multiply my results by 1000.
0.0000992 x 1000 = 0.0992 mols dm-3
This result is more accurate than the one above suggesting that my idea to decrease the concentration of the Potassium Manganate (VII) (aq) was the correct idea.
Experiment Two - How accurate is a Calibration Graph in determining the concentration of an Iron (II) Ammonium Sulphate (aq)
As the graph on the previous page shows, experiment two was unsuccessful. Although it produced results these results were not what is expected when a Calibration Graph is produced.
The reasons these anomalous results occurred could be due to the low concentrations of the Iron (II) Ammonium Sulphate (aq) that I used. I was forced into using these concentrations, as the concentrations I originally choose to use were extremely difficult to produce solutions for. I did try to make up the stronger solutions but the solid would not dissolve in the required volume of Acid unless the solution had been heated, but once the solution cooled it solidified again.
Experiment Three - How accurate is an Electrochemical Cell in determining the concentration of Iron (II) Ammonium Sulphate (aq)
Unfortunately I was unable to gain data for this experiment as I ran out of time before I was able to complete it.
The results should have shown that as the concentration of Iron (II) Ammonium Sulphate (aq) increased the voltage being passed between the solutions would have increased. This would have created a graph showing that an increase in concentration produces an increase in the voltage produced by the Electrochemical Cell. From this I would have hopefully been able to work out the concentration of Iron (II) in Spinach Oleracea by recording the voltage from the Electrochemical Cell produced by combining Spinach Extract (aq)/ Fe(s) and Cu2+ (aq) / Cu (s) and then reading the concentration of the solution off the graph.
Experiment Four - How temperature effects a titration between Oxalic acid (aq) and Potassium Manganate (aq)
C2O42- (aq) + MnO4 (aq) + 8H+ (aq) 2CO2 (g) + Mn2+ (aq) + 4H2O (l)
The equation above shows that 1 mol of MnO4 reacts with 1 mol of C2O42-, and by using the equation below it is possible to work out the concentration of the Oxalic Acid (aq). The experiment was preformed at a range of temperatures, and I will be trying to find the best temperature to perform this titration (i.e. the temperature that produces the most accurate results).
At a temperature of 0oc
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 28.82 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.0002882 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Oxalic Acid is 1:1. Therefore the mol of Potassium Manganate (VII) (aq) used is equal to those used of Oxalic Acid (aq). Now that I know the mols present in the Oxalic Acid (aq) and the volume used in the titration I can work out the concentration of it.
Moles = Concentration x Volume
Concentration = Moles
Volume
Concentration = 0.0002882 mol dm-3
5
Concentration = 0.00005764 mol dm-3
This answer tells me how many mols are present in 1000cm3 but as I only a 100cm3 solution I need to multiply my results by 1000.
0.00005764 x 1000 = 0.05764 mols dm-3
At a temperature of 25 oc
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 25.30 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.000253 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Oxalic Acid is 1:1. Therefore the mol of Potassium Manganate (VII) (aq) used is equal to those used of Oxalic Acid (aq). Now that I know the mols present in the Oxalic Acid (aq) and the volume used in the titration I can work out the concentration of it.
Moles = Concentration x Volume
Concentration = Moles
Volume
Concentration = 0.000253 mol dm-3
5
Concentration = 0.0000506 mol dm-3
This answer tells me how many mols are present in 1000cm3 but as I only a 100cm3 solution I need to multiply my results by 1000.
0.0000506 x 1000 = 0.0506 mols dm-3
At a temperature of 60 oc
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 21.58 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.0002158 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Oxalic Acid is 1:1. Therefore the mol of Potassium Manganate (VII) (aq) used is equal to those used of Oxalic Acid (aq). Now that I know the mols present in the Oxalic Acid (aq) and the volume used in the titration I can work out the concentration of it.
Moles = Concentration x Volume
Concentration = Moles
Volume
Concentration = 0.0002158 mol dm-3
5
Concentration = 0.00004316 mol dm-3
This answer tells me how many mols are present in 1000cm3 but as I only a 100cm3 solution I need to multiply my results by 1000.
0.00004316 x 1000 = 0.04316mols dm-3
At a temperature of 70 oc
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 20.70 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.000207 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Oxalic Acid is 1:1. Therefore the mol of Potassium Manganate (VII) (aq) used is equal to those used of Oxalic Acid (aq). Now that I know the mols present in the Oxalic Acid (aq) and the volume used in the titration I can work out the concentration of it.
Moles = Concentration x Volume
Concentration = Moles
Volume
Concentration = 0.000207 mol dm-3
5
Concentration = 0.0000414 mol dm-3
This answer tells me how many mols are present in 1000cm3 but as I only a 100cm3 solution I need to multiply my results by 1000.
0.0000414 x 1000 = 0.0414 mols dm-3
These results as expected vary with temperature however I would have expected the results to get nearer to the actually concentration as the temperature increased. This is because as the reaction occurring works better at a higher temperature so I should have been able to monitor the reaction better and noted the end point accordingly.
Experiment Five - How much Iron (II) can be extracted from 15 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq)
Experiment A
5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O
The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II) is present in the spinach extract solution and then use this to work out how much Iron (II) is present in 100 grams of Spinach Oleracea.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 21.83 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.0002183 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II) used in the titration I need to divide the volume of moles by 3 and then multiply it by 5.
0.0002183 mol dm-3 x 5 = 0.000363833 mol dm-3
3
Now that I know the mols of Iron (II) present in 20 cm3 of spinach extract solution I can use this to work out the moles of Iron (II) present in 100 cm3 of spinach extract solution. To do this the moles present in 20 cm3 will have to be multiplied by 5.
0.000363833 mol dm-3 X 5 = 0.001819165 mol dm-3
Only 15 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II) present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be divided by 15 and multiplied by 100.
0.001819165 mol dm-3 X 100 = 0.012127766 mol dm-3
15
Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II) present in the 100 grams of Spinach Oleracea.
Mass = Moles X Ar
The Ar of Iron (II) is 55.9 and therefore:
Mass = 0.012127766 mol dm-3 X 55.9
Mass = 0.677942119 mg
This answer tells me that there are 0.678 mg’s of Iron (II) present in 100 grams of Spinach. This is not the expected result of 4 mg’s which is what the many textbooks suggest the answer should be.
Experiment B
5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O
The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II) is present in the spinach extract solution and then use this to work out how much Iron (II) is present in 100 grams of Spinach Oleracea.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 18.97 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.0001897 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II) used in the titration I need to divide the volume of moles by 3 and then multiply it by 5.
0.0001897 mol dm-3 x 5 = 0.000316166 mol dm-3
3
Now that I know the mols of Iron (II) present in 10 cm3 of spinach extract solution I can use this to work out the moles of Iron (II) present in 100 cm3 of spinach extract solution. To do this the moles present in 10 cm3 will have to be multiplied by 10.
0.000363833 mol dm-3 X 10 = 0.003161666 mol dm-3
Only 15 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II) present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be divided by 15 and multiplied by 100.
0.003161666 mol dm-3 X 100 = 0.021077777 mol dm-3
15
Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II) present in the 100 grams of Spinach Oleracea.
Mass = Moles X Ar
The Ar of Iron (II) is 55.9 and therefore:
Mass = 0.021077777 mol dm-3 X 55.9
Mass = 1.178247778 mg
This answer tells me that there are 1.178 mg’s of Iron (II) present in 100 grams of Spinach. Although this is not the expected answer of 4 mg’s, it is closer to the expected result than Experiment A. This could be due to the smaller volume of spinach extract solution that I used in Experiment B’s titrations.
Experiment Six - How much Iron (II) can be extracted from 20 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq)
Experiment A
5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O
The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II) is present in the spinach extract solution and then use this to work out how much Iron (II) is present in 100 grams of Spinach Oleracea.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 23.20 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.000232 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II) used in the titration I need to divide the volume of moles by 3 and then multiply it by 5.
0.000232 mol dm-3 x 5 = 0.000386666 mol dm-3
3
Now that I know the mols of Iron (II) present in 10 cm3 of spinach extract solution I can use this to work out the moles of Iron (II) present in 100 cm3 of spinach extract solution. To do this the moles present in 10 cm3 will have to be multiplied by 10.
0.000363833 mol dm-3 X 10= 0.003866666 mol dm-3
Only 20 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II) present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be multiplied by 5.
0.001819165 mol dm-3 X 5 = 0.019333333 mol dm-3
Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II) present in the 100 grams of Spinach Oleracea.
Mass = Moles X Ar
The Ar of Iron (II) is 55.9 and therefore:
Mass = 0.019333333 mol dm-3 X 55.9
Mass = 1.080733333 mg
This answer tells me that there are 1.081 mg’s of Iron (II) present in 100 grams of Spinach. This is not the expected result of 4 mg’s, which is what the many textbooks suggest the answer should be.
Experiment B
5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O
The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II) is present in the spinach extract solution and then use this to work out how much Iron (II) is present in 100 grams of Spinach Oleracea.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.01 mol dm –3 x 15.65 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.0001565 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II) used in the titration I need to divide the volume of moles by 3 and then multiply it by 5.
0.0001565 mol dm-3 x 5 = 0.000260833 mol dm-3
3
Now that I know the mols of Iron (II) present in 5 cm3 of spinach extract solution I can use this to work out the moles of Iron (II) present in 100 cm3 of spinach extract solution. To do this the moles present in 5 cm3 will have to be multiplied by 20.
0.000260833 mol dm-3 X 20 = 0.005216666 mol dm-3
Only 20 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II) present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be multiplied by 5.
0.005216666 mol dm-3 X 5 = 0.026083333 mol dm-3
Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II) present in the 100 grams of Spinach Oleracea.
Mass = Moles X Ar
The Ar of Iron (II) is 55.9 and therefore:
Mass = 0.026083333 mol dm-3 X 55.9
Mass = 1.458058333 mg
This answer tells me that there are 1.458 mg’s of Iron (II) present in 100 grams of Spinach. This is not the expected result of 4 mg’s, which is what the many textbooks suggest the answer should be, however it is again closer to this expected result. This could be due to the smaller volume of spinach extract solution that was used in the titration.
Experiment C
5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O
The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II) is present in the spinach extract solution and then use this to work out how much Iron (II) is present in 100 grams of Spinach Oleracea.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.025 mol dm –3 x 7.17 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.00017925 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II) used in the titration I need to divide the volume of moles by 3 and then multiply it by 5.
0.000017925 mol dm-3 x 5 = 0.000386666 mol dm-3
3
Now that I know the mols of Iron (II) present in 5 cm3 of spinach extract solution I can use this to work out the moles of Iron (II) present in 100 cm3 of spinach extract solution. To do this the moles present in 5 cm3 will have to be multiplied by 20.
0.00017925 mol dm-3 X 20 = 0.003585 mol dm-3
Only 20 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II) present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be multiplied by 5.
0.003585 mol dm-3 X 5 = 0.017925 mol dm-3
Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II) present in the 100 grams of Spinach Oleracea.
Mass = Moles X Ar
The Ar of Iron (II) is 55.9 and therefore:
Mass = 0.017925 mol dm-3 X 55.9
Mass = 1.0020075 mg
This answer tells me that there are 1.002 mg’s of Iron (II) present in 100 grams of Spinach. This is not the expected result of 4 mg’s, which is what the many textbooks suggest the answer should be. It is also lightly less than the previous experiment and this could be due to the higher concentration of Potassium Manganate (VII) (aq) that I used.
Experiment D
5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O
The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II) is present in the spinach extract solution and then use this to work out how much Iron (II) is present in 100 grams of Spinach Oleracea.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.025 mol dm –3 x 7.27 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.00018175 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II) used in the titration I need to divide the volume of moles by 3 and then multiply it by 5.
0.00018175 mol dm-3 x 5 = 0.000302916 mol dm-3
3
Now that I know the mols of Iron (II) present in 5 cm3 of spinach extract solution I can use this to work out the moles of Iron (II) present in 100 cm3 of spinach extract solution. To do this the moles present in 5 cm3 will have to be multiplied by 20.
0.000302916 mol dm-3 X 20 = 0.00605832 mol dm-3
Only 20 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II) present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be multiplied by 5.
0.00605832 mol dm-3 X 5 = 0.0302916 mol dm-3
Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II) present in the 100 grams of Spinach Oleracea.
Mass = Moles X Ar
The Ar of Iron (II) is 55.9 and therefore:
Mass = 0.0302916 mol dm-3 X 55.9
Mass = 1.69330044 mg
This answer tells me that there are 1.693 mg’s of Iron (II) present in 100 grams of Spinach. This is not the expected result of 4 mg’s, which is what the many textbooks suggest the answer should be, but the results of this experiment are getting closer to the expected value suggesting that my modifications are working.
I completed this experiment twice using two different spinach extract solutions, which were produced using the same method. This was done so that I could compare the two results and find an average.
Average mg in 100 grams of Spinach Oleracea =1.0020075 + 1.69330044
2
= 1.34765397 mg
These results are similar but not close enough to suggest that they are very accurate results.
Experiment Six - How much Iron (II) can be extracted from 20 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq)
Experiment A
5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O
The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II) is present in the spinach extract solution and then use this to work out how much Iron (II) is present in 100 grams of Spinach Oleracea.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.025 mol dm –3 x 2.10 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.0000525 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II) used in the titration I need to divide the volume of moles by 3 and then multiply it by 5.
0.0000525 mol dm-3 x 5 = 0.0000875 mol dm-3
3
Now that I know the mols of Iron (II) present in 5 cm3 of spinach extract solution I can use this to work out the moles of Iron (II) present in 100 cm3 of spinach extract solution. To do this the moles present in 5 cm3 will have to be multiplied by 20.
0.0000875 mol dm-3 X 20 = 0.00175 mol dm-3
Only 20 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II) present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be multiplied by 5.
0.00175 mol dm-3 X 5 = 0.00875 mol dm-3
Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II) present in the 100 grams of Spinach Oleracea.
Mass = Moles X Ar
The Ar of Iron (II) is 55.9 and therefore:
Mass = 0.00875 mol dm-3 X 55.9
Mass = 0.489125 mg
This answer tells me that there are 0.489 mg’s of Iron (II) present in 100 grams of Spinach. This is not the expected result of 4 mg’s, which is what the many textbooks suggest the answer should be. The reason this experiment suggests there is such a low concentration of iron (II) present in 100 grams of Spinach Oleracea, could be due to the fact that the Spinach Oleracea was boiled in distilled water.
Experiment B
5Fe2+ + 5C2O42- + 3MnO4 + 24H+ 5Fe3+ + 10CO2 + 3Mn2+ + 12H2O
The equation above shows that 3 moles of Potassium Manganate (VII) (aq) react with 5 moles of Iron (II), and by using the equation below it is possible to work out how much Iron (II) is present in the spinach extract solution and then use this to work out how much Iron (II) is present in 100 grams of Spinach Oleracea.
Firstly work out the moles of Potassium Manganate (VII) (aq) in the titration.
Moles = Concentration x Volume
Moles = 0.025 mol dm –3 x 2.19 cm3
1000
The volume of the average titre is divided by 1000 to change the units from cm3 to dm3.
Moles = 0.00005475 mol dm-3
The ratio of Potassium Manganate (VII) (aq) to Iron (II) (aq) is 3:5 and therefore to work out the mols of Iron (II) used in the titration I need to divide the volume of moles by 3 and then multiply it by 5.
0.00005475 mol dm-3 x 5 = 0.00009125 mol dm-3
3
Now that I know the mols of Iron (II) present in 5 cm3 of spinach extract solution I can use this to work out the moles of Iron (II) present in 100 cm3 of spinach extract solution. To do this the moles present in 5 cm3 will have to be multiplied by 20.
0.00009125 mol dm-3 X 20 = 0.001825 mol dm-3
Only 20 grams of Spinach Oleracea were used to make up the 100cm3 spinach extract solution, so in order to work out the moles of Iron (II) present in 100 grams of Spinach Oleracea the volume of moles in the 100cm3 spinach extract solution will need to be multiplied by 5.
0.001825 mol dm-3 X 5 = 0.009125 mol dm-3
Now that I know the moles present in 100 grams of spinach I can use the equation below to work out the mass of Iron (II) present in the 100 grams of Spinach Oleracea.
Mass = Moles X Ar
The Ar of Iron (II) is 55.9 and therefore:
Mass = 0.009125 mol dm-3 X 55.9
Mass = 0.5100875 mg
This answer tells me that there are 0.510 mg’s of Iron (II) present in 100 grams of Spinach. This is not the expected result of 4 mg’s, which is what the many textbooks suggest the answer should be. The reason this experiment suggests there is such a low concentration of iron (II) present in 100 grams of Spinach Oleracea, could be due to the fact that the Spinach Oleracea was boiled in distilled water.
I completed this experiment twice using two different spinach extract solutions, which were produced using the same method. This was done so that I could compare the two results and find an average.
Average mg in 100 grams of Spinach Oleracea = 0.489125 + 0.5100875
2
= 0.49960625 mg
These two results were very near to each other suggesting that these are accurate result.
Evaluating Evidence and Procedures
Experiment One - How accurate is a Potassium Manganate titration in determining the concentration of an Iron (II) Ammonium Sulphate (aq)
Experiment A
There was one anomalies result in this experiment and that was a titre of 1.75 cm3. However this result would not have affected the final results, as it was not included in the equation that I used to work out the average titre.
Precision Errors
Precision Error (%) = Error of Equipment X 100
Measurement
Iron (II) Ammonium Sulphate on balance = 0.0005 X 100 = 0.012755102 %
3.92
Potassium Manganate (VII) on balance = 0.0005 X 100 = 0.012658227 %
3.95
100 cm3 Volumetric Flask = 0.2 X 100 = 0.2 %
100
250 cm3 Volumetric Flask = 0.2 X 100 = 0.08 %
250
Potassium Manganate (VII) in 50 cm3 burette = 0.005 X 100 = 0253807106 %
1.97
Iron (II) Ammonium Sulphate in a 10 cm3 pipette = 0.05 X 100 = 0.5 %
10
Sulphuric Acid in a 10cm3 pipette = 0.05 X 100 = 0.5 %
10
The 10 cm3 pipettes that were used in this experiment caused the greatest precision error, and thus would have had the greatest impact on the accuracy. However this is still a very small precision error and thus would not have any dramatic effect on the experiment. It would make the solutions seem slightly more or less concentrated.
Procedural Errors
It was difficult to judge the end point of the reaction even though a dramatic colour change took place. This was because a single extra drop of Potassium Manganate (VII) (aq) turned the solution a deeper shade of purple. Also the colour faded very quickly after the Potassium Manganate (VII) (aq) had been added. Even after the end-point had been reached (i.e. the solution remained coloured for 30 seconds) the colour continued to fade so that there was nothing to compare the solutions colour too, to help with determining the end point of the reaction. If the end point had been over shot, the average titre point would be higher; therefore the concentration of Iron (II) would appear higher suggesting that the solution was more concentrated. If the end point had not been reached, (i.e. the solution did not appear as dark in colour as the others at the end point) the average titre point would be lower; therefore the concentration of Iron (II) would appear lower suggesting that the solution was less concentrated. If either of these had occurred the redox titration would have appear less accurate.
The whole solid may not have dissolved fully in the acid/distilled water needed to create the correct concentration of solutions in both the Iron (II) Ammonium Sulphate (aq) and the Potassium Manganate (VII) (aq). If this had occurred the solutions produced would have been less concentrated resulting in any equations done using their concentration being inaccurate.
It was difficult to get all the solute and washings into the volumetric flask without it over shooting the graduation mark, this occurred as the solutions were being made into too concentrated solutions. This over shooting took up much time as every time this occurred a new solution had to be made up.
The Iron (II) Ammonium Sulphate (aq) appeared colourless at this concentration, and thus it was difficult to see the washings in the beaker. For this reason perhaps not all the solute was collected. This could result in a lower concentration of solution, which, would eventually result in inaccurate calculations later on.
Improvements
This experiment could be improved in a number of ways. The Iron (II) Ammonium Sulphate (aq) and the Potassium Manganate (VII) (aq) could have been made up into larger volumetric flasks or at a lower concentration, which would have made it easier to collect all the solute and washings without over shooting the graduation mark on the volumetric flasks.
As this was a preliminary experiment only a few results were collected. To obtain more accurate data for this experiment more titres within 0.1 of each other would have to be collected. This would produce similar titre results with which to calculate the average, and thus anomalous results would easily be identified.
More washing out of the beaker would mean any Iron (II) that has been left behind would be picked up; and as Iron (II) Ammonium Sulphate was appearing colourless it was difficult to tell if there was any solute left in the beaker. By re-washing the beaker this would ensure that all the Iron (II) was collected.
The conical flask should be washed out between each titration, so to avoid contamination of solutions.
Experiment B
There were a few anomalies that resulted in this experiment and they were 20.20 cm3 and 20.60 cm3 for the titres. However this result would not have affected the final results, as it was not included in the equation that I used to work out the average titre.
Precision Errors
Precision Error (%) = Error of Equipment X 100
Measurement
Iron (II) Ammonium Sulphate on balance = 0.0005 X 100 = 0.012755102 %
3.92
Potassium Manganate (VII) on balance = 0.0005 X 100 = 0.012658227 %
3.95
100 cm3 Volumetric Flask = 0.2 X 100 = 0.2 %
100
250 cm3 Volumetric Flask = 0.2 X 100 = 0.08 %
250
Potassium Manganate (VII) in 50 cm3 burette = 0.005 X 100 = 0.025201612 %
19.84
Iron (II) Ammonium Sulphate in a 10 cm3 pipette = 0.05 X 100 = 0.5 %
10
Sulphuric Acid in a 10cm3 pipette = 0.05 X 100 = 0.5 %
10
The 10 cm3 pipettes that were used in this experiment caused the greatest precision error, and thus would have had the greatest impact on the accuracy. However this is still a very small precision error and thus would not have any dramatic effect on the experiment. It would make the solutions seem slightly more or less concentrated.
Procedural Errors
It was difficult to judge the end point of the reaction even though a dramatic colour change took place. This was because a single extra drop of Potassium Manganate (VII) (aq) turned the solution a deeper shade of purple. Also the colour faded very quickly after the Potassium Manganate (VII) (aq) had been added. Even after the end-point had been reached (i.e. the solution remained coloured for 30 seconds) the colour continued to fade so that there was nothing to compare the solutions colour too, to help with determining the end point of the reaction. If the end point had been over shot, the average titre point would be higher; therefore the concentration of Iron (II) would appear higher suggesting that the solution was more concentrated. If the end point had not been reached, (i.e. the solution did not appear as dark in colour as the others at the end point) the average titre point would be lower; therefore the concentration of Iron (II) would appear lower suggesting that the solution was less concentrated. If either of these had occurred the redox titration would have appear less accurate.
The whole solid may not have dissolved fully in the acid/distilled water needed to create the correct concentration of solutions in both the Iron (II) Ammonium Sulphate (aq) and the Potassium Manganate (VII) (aq). If this had occurred the solutions produced would have been less concentrated resulting in any equations done using their concentration being inaccurate.
It was difficult to get all the solute and washings into the volumetric flask without it over shooting the graduation mark, this occurred as the solutions were being made into too concentrated solutions. This over shooting took up much time as every time this occurred a new solution had to be made up.
The Iron (II) Ammonium Sulphate (aq) appeared colourless at this concentration, and thus it was difficult to see the washings in the beaker. For this reason perhaps not all the solute was collected. This could result in a lower concentration of solution, which, would eventually result in inaccurate calculations later on.
Improvements
This experiment could be improved in a number of ways. The Iron (II) Ammonium Sulphate (aq) and the Potassium Manganate (VII) (aq) could have been made up into larger volumetric flasks or at a lower concentration, which would have made it easier to collect all the solute and washings without over shooting the graduation mark on the volumetric flasks.
As this was a preliminary experiment only a few results were collected. To obtain more accurate data for this experiment more titres within 0.1 of each other would have to be collected. This would produce similar titre results with which to calculate the average, and thus anomalous results would easily be identified.
More washing out of the beaker would mean any Iron (II) that has been left behind would be picked up; and as Iron (II) Ammonium Sulphate was appearing colourless it was difficult to tell if there was any solute left in the beaker. By re-washing the beaker this would ensure that all the Iron (II) was collected.
The conical flask used in the titration should be washed out after each titration, so as to avoid contaminating the solutions.
Experiment Two - How accurate is a Calibration Graph in determining the concentration of an Iron (II) Ammonium Sulphate (aq)
This whole experiment resulted in anomalous results. I was hoping that the readings of the colorimeter would make an accurate Calibration Curve that I would be able to read the concentration of the spinach extract solution off but unfortunately the experiment did not work.
Precision Errors
Precision Error (%) = Error of Equipment X 100
Measurement
Iron (II) Ammonium Sulphate on balance = 0.0005 X 100 = 0.012755102 %
3.92
100 cm3 Volumetric Flask = 0.2 X 100 = 0.2 %
100
10cm3 Volumetric Flask = 0.2 X 100 = 2 %
10
Iron (II) Ammonium Sulphate in 10cm3 pipette = 0.05 X 100 = 0.625 %
8
Iron (II) Ammonium Sulphate in 10cm3 pipette = 0.05 X 100 = 0.83 %
6
Iron (II) Amonium Sulphate in 10cm3 pipette = 0.05 X 100 = 1.25 %
4
Iron (II) Ammonium Sulphate in 10cm3 pipette = 0.05 X 100 = 2.5 %
2
The largest precision error was of the 10cm3 pipette measuring out 2 cm3 of Iron (II) Ammonium Sulphate (aq). Another large precision error was created by the 10 cm3 volumetric flask when it was full of Iron (II) Ammonium Sulphate (aq). However overall these precision errors are still small and thus would not have any dramatic effect on the experiment. It would make the solutions seem slightly more or less concentrated.
Procedural Errors
The solutions stood in their curvet for a few minutes and in this time the solute in the solution had begun to sink to the bottom causing the solution to be at different concentrations in the curvet. This would mean that when the light from the colorimeter was shun onto the curvet it would have been absorbed at different levels at different depths.
Improvements
Use even more concentrated Iron (II) Ammonium Sulphate (aq) to use in the colorimeter.
Experiment Three - How accurate is an Electrochemical Cell in determining the concentration of Iron (II) Ammonium Sulphate (aq)
I was unable to complete this experiment as I ran out of time, and thus I am unable to show whether or not anomalous results were produced, what the precision errors were, what procedural errors could have occurred and any improvements I could make to the experiment.
Experiment Four - How temperature effects a titration between Oxalic acid (aq) and Potassium Manganate (aq)
Precision Errors
Precision Error (%) = Error of Equipment X 100
Measurement
Oxalic Acid on balance = 0.0005 X 100 = 0.015873015 %
3.15
Potassium Manganate (VII) on balance = 0.0005 X 100 = 0.012658227 %
3.95
250 cm3 Volumetric Flask = 0.2 X 100 = 0.08 %
250
Potassium Manganate (VII) in 50 cm3 burette = 0.005 X 100 = 0.017349063 %
28.82
Potassium Manganate (VII) in 50 cm3 burette = 0.005 X 100 = 0.019762845 %
25.30
Potassium Manganate (VII) in 50 cm3 burette = 0.005 X 100 = 0.023169601 %
21.58
Potassium Manganate (VII) in 50 cm3 burette = 0.005 X 100 = 0.024154589 %
20.70
Oxalic Acid in a 25 cm3 pipette = 0.05 X 100 = 0.25 %
20
Oxalic Acid in a 10 cm3 pipette = 0.05 X 100 = 0.5 %
10
Thermometer = 0.05 X 100 = 0.071428571 %
70
The largest precision error was the 10cm3 pipette transferring Oxalic Acid (aq). However overall these precision errors are still small and thus would not have any dramatic effect on the experiment. It would make the solutions seem slightly more or less concentrated.
Procedural Errors
It was difficult to judge the end point of the reaction even though a dramatic colour change took place. This was because a single extra drop of Potassium Manganate (VII) (aq) turned the solution a deeper shade of purple. Also the colour faded very quickly after the Potassium Manganate (VII) (aq) had been added. Even after the end-point had been reached (i.e. the solution remained coloured for 30 seconds) the colour continued to fade so that there was nothing to compare the solutions colour too, to help with determining the end point of the reaction. If the end point had been over shot, the average titre point would be higher; therefore the concentration of Oxalic Acid would appear higher suggesting that the solution was more concentrated. If the end point had not been reached, (i.e. the solution did not appear as dark in colour as the others at the end point) the average titre point would be lower; therefore the concentration of Oxalic Acid would appear lower suggesting that the solution was less concentrated. If either of these had occurred the redox titration would have appear less accurate.
The whole solid may not have dissolved fully in the acid/distilled water needed to create the correct concentration of solutions in both the Oxalic Acid (aq) and the Potassium Manganate (VII) (aq). If this had occurred the solutions produced would have been less concentrated resulting in any equations done using their concentration being inaccurate.
It was difficult to get all the solute and washings into the volumetric flask without it over shooting the graduation mark, this occurred as the solutions were being made into too concentrated solutions. This over shooting took up much time as every time this occurred a new solution had to be made up.
The Oxalic Acid (aq) appeared colourless at this concentration, and thus it was difficult to see the washings in the beaker. For this reason perhaps not all the solute was collected. This could result in a lower concentration of solution, which, would eventually result in inaccurate calculations later on.
The solutions were only left for a few minutes to acclimatise, which could have resulted in the temperatures taken being inaccurate. Also as the Potassium Manganate (VII) (aq) was being added to the Oxalic Acid (aq) the temperature of it dropped or raised depending on what temperature it was already. This could have effected the results of this experiment as it was focusing on the effect temperature has on this particular redox reaction.
Improvements
This experiment could be improved in a number of ways. The Oxalic Acid (aq) and the Potassium Manganate (VII) (aq) could have been made up into larger volumetric flasks or at a lower concentration, which would have made it easier to collect all the solute and washings without over shooting the graduation mark on the volumetric flasks.
As this was a preliminary experiment only a few results were collected. To obtain more accurate data for this experiment more titres within 0.1 of each other would have to be collected. This would produce similar titre results with which to calculate the average, and thus anomalous results would easily be identified.
More washing out of the beaker would mean any Oxalic Acid (aq) that has been left behind would be picked up; and as Oxalic Acid (aq) is colourless it was difficult to tell if there was any solute left in the beaker. By re-washing the beaker this would ensure that all the Oxalate was collected.
The conical flask used in the titration should be washed out after each titration, so as to avoid contaminating the solutions.
The solutions should be allowed to acclimatise to their new temperature for 30 minutes before any titrations took place. This would ensure that the solutions were at the required temperature for the titrations.
Experiment Five - How much Iron (II) can be extracted from 15 grams of Spinach Oleracea when boiled in Sulphuric Acid (aq)
As this was a preliminary experiment to work out the volume of spinach extract solution to use in the final experiments, only a few readings were taken, and for this reason none of the results are within 0.1 of each other which is what we would expect for accurate results. Thus, all of the results could be anomalous.
Precision Errors
Precision Error (%) = Error of Equipment X 100
Measurement
Spinach Oleracea on balance = 0.0005 X 100 = 0.003333333 %
15
100cm3 Volumetric Flask = 0.2 X 100 = 0.2 %
100
Potassium Manganate (VII) on balance = 0.0005 X 100 = 0.012658227 %
3.95
250 cm3 Volumetric Flask = 0.2 X 100 = 0.08 %
250
Potassium Manganate (VII) in 50 cm3 burette = 0.005 X 100 = 0.02290426 %
21.83
Spinach Extract solution in a 20 cm3 pipette = 0.05 X 100 = 0.25 %
20
Spinach Extract solution in a 10 cm3 pipette = 0.05 X 100 = 0.5 %
10
Thermometer = 0.05 X 100 = 0.071428571 %
70
Procedural Errors
The spinach may not have been cut into equally sized pieces and the surface area to volume ratio was not equal which could have resulted in some pieces of the leave not being able to release all the Iron (II) that was stored inside it.
The spinach may have boiled for over or less than 5 minutes. If it was under 5 minutes then all the Iron (II) locked up in the leaves may not have had time to be released, which would have resulted in the Spinach Oleracea looking as though it contained less Iron (II).
The spinach solution was almost colourless and thus it was difficult to determine whether all the spinach extract solution had been collected. If washings were not done then maybe some of the Iron (II) remained in the beaker.
This experiment required me to heat the spinach extract solution to 70oc before using it in the titration. Once the solution was being used in the titration, its temperature quickly dropped. For this reason errors may have occurred.
It was difficult to judge the end point of the reaction even though a dramatic colour change took place. This was because a single extra drop of Potassium Manganate (VII) (aq) turned the solution a deeper shade of purple. Also the colour faded very quickly after the Potassium Manganate (VII) (aq) had been added. Even after the end-point had been reached (i.e. the solution remained coloured for 30 seconds) the colour continued to fade so that there was nothing to compare the solutions colour too, to help with determining the end point of the reaction. If the end point had been over shot, the average titre point would be higher; therefore the concentration of Oxalic Acid would appear higher suggesting that the solution was more concentrated. If the end point had not been reached, (i.e. the solution did not appear as dark in colour as the others at the end point) the average titre point would be lower; therefore the concentration of Oxalic Acid would appear lower suggesting that the solution was less concentrated. If either of these had occurred the redox titration would have appear less accurate.
The whole solid may not have dissolved fully in the acid/distilled water needed to create the correct concentration of solutions in both the Oxalic Acid (aq) and the Potassium Manganate (VII) (aq). If this had occurred the solutions produced would have been less concentrated resulting in any equations done using their concentration being inaccurate.
It was difficult to get all the solute and washings into the volumetric flask without it over shooting the graduation mark, this occurred as the solutions were being made into too concentrated solutions. This over shooting took up much time as every time this occurred a new solution had to be made up.
Improvements