31 Analysis of Volhard’s Method’s Calculations of the experiment Step 1- Determining the average titre of Potassium Thiocyanate for the titrations The first step of the calculations will involve calculating the mean titre of the potassium thiocyanate used for every individual cheese titrations. Three concordant titres should be taken within 0.1cm³ of each other during the titration, these three titres will be used to calculate the average titre. We calculate the average titre by following the formula given below:
X₁ + X₂ + X₃ = Average titre n (Where “X” is the concordant titre and “n” is the number of values used.) The values are rounded to 1 decimal place in the average titre and are measured in ‘cm³’. Example- Philadelphia cheese spread 5.6 + 5.5 + 5.5 = 5.5cm³ (Rounded to 1 decimal place) 3 Average titres for all the different cheese experimented with in Volhard’s Method: Philadelphia Cheese spread = 5.5 cm³ Blue Stilton = No result, as no precipitation was formed during titration. Red Leicester titrated with original batch of solution = 5.5cm³ Red Leicester titrated with new batch of solution = 5.1 cm³ Lancashire, sample boiled in 20 minutes = 8.4 cm³ Lancashire, sample boiled in 55 minutes = 9.2 cm³ Edam = 8.6 cm³ Edam repeat = 9.6 cm³ Step 2- Calculating the moles of potassium thiocyanate used The next step is to calculate the moles of potassium thiocyanate we used in the titrations assuming the value we use is the average titre value. The concentration of potassium thiocyanate solution we used to titrate with the cheese solution was 0.1 mol. Therefore we multiply the average titre value with the concentration of the potassium thiocyanate solution, and then divide the value by 1000 as we need to convert the value of cm³ to dm³.
Average titre × 0.1 = mole of potassium thiocyanate used 1000 Example- Philadelphia cheese spread Kingsley Wong
32 5.5 × 0.1 = 5.5 ×10⁻⁴ mol (Rounded to 1 decimal place) 1000 Moles of potassium thiocyanate used in the different cheeses experimented with: Philadelphia cheese spread = 5.5 ×10⁻⁴ mol Blue Stilton = no results Red Leicester titrated with original batch of solution = 5.5 ×10⁻⁴ mol Red Leicester titrated with new batch of solution = 5.1 ×10⁻⁴ mol Lancashire, sample boiled in 20 minutes = 8.4 ×10⁻⁴ mol Lancashire, sample boiled in 55 minutes = 9.2 ×10⁻⁴ mol Edam = 8.6 ×10⁻⁴ mol Edam repeat = 9.6 ×10⁻⁴ mol Step 3- Working out the moles of unreacted silver nitrate in the cheese extract solution From the background information, we have found out the equation of the reaction between silver ions and thiocyanate ions are as follows: Ag +(aq) + SCN- (aq) → AgSCN(s) From the equation, we know the ratio of the reaction between silver ions and thiocyanate ions are 1:1. Therefore, the mole of potassium thiocyanate used in the reaction will be the same as the silver ions reacted in the cheese extract solution. Moles of silver ions reacted in the different cheeses experimented with (1 decimal place): Philadelphia cheese spread = 5.5 ×10⁻⁴ mol Blue Stilton = no results Red Leicester titrated with original batch of solution = 5.5 ×10⁻⁴ mol Red Leicester titrated with new batch of solution = 5.1 ×10⁻⁴ mol Lancashire, sample boiled in 20 minutes = 8.4 ×10⁻⁴ mol Lancashire, sample boiled in 55 minutes = 9.2 ×10⁻⁴ mol Edam = 8.6 ×10⁻⁴ mol Edam repeat = 9.6 ×10⁻⁴ mol Step 4- Working out the moles of unreacted silver nitrate in 500 cm3 of cheese extract solution
Kingsley Wong
33 In the method stated, we used 100 cm3 of cheese extract solution each time from the total of 500 cm3 of the solution for the titrations. As we have worked out the moles of unreacted silver ions in 100cm3 of cheese extract solution in the previous step, we multiple the values by 5 to work out the total moles of unreacted silver ions. Example- Philadelphia cheese spread 5.5 ×10⁻⁴ × 5 = 2.75×10⁻3 mol (Rounded to 2 decimal places) Moles of silver ions reacted in the different cheeses experimented with in 500 cm3: Philadelphia cheese spread = 2.75 ×10⁻3 mol Blue Stilton = no results Red Leicester titrated with original batch of solution = 2.75 ×10⁻3 mol Red Leicester titrated with new batch of solution = 2.55 ×10⁻3 mol Lancashire, sample boiled in 20 minutes = 4.20 ×10⁻3 mol Lancashire, sample boiled in 55 minutes = 4.60 ×10⁻3 mol Edam = 4.30 ×10⁻3 mol Edam repeat = 4.80 ×10⁻3 mol Step 5- Calculate the moles of silver nitrate in the 50cm3 of solution added during sample preparation During sample preparation, we added 50cm3 of 0.1 mol silver nitrate solution to the grated cheese. We now need to work out the mole in which we added to find out total amount used. Thus showing us the moles of silver ions involved in the whole procedure. To work out the mole, we convert the cm3 of the silver nitrate solution used to dm3 by dividing the value by 1000, then multiple it by the concentration which is 0.1 mol.
50 × 0.1 = 5 ×10⁻3 mol 1000 So the mole of silver nitrate in 50cm3 of solution is 5 ×10⁻3 mol. Step 6- Calculate the moles of silver nitrate reacted with the salt from the cheese From the previous steps we have worked out the silver nitrate added and the excess silver nitrate reacted during titrations. So in order to work out the moles of silver nitrate which reacted directly with the salt from the cheese, we subtract the excess silver nitrate reacted from the total moles of silver nitrate added. Example-Philadelphia cheese spread 5 ×10⁻3 mol - 2.75 ×10⁻3 mol = 2.25 ×10⁻3 mol
Kingsley Wong
34 Moles of silver nitrate reacted with the cheese in the different cheeses experimented with: Philadelphia cheese spread = 2.25 ×10⁻3 mol Blue Stilton = no results Red Leicester titrated with original batch of solution = 2.25 ×10⁻3 mol Red Leicester titrated with new batch of solution = 2.45 ×10⁻3 mol Lancashire, sample boiled in 20 minutes = 0.80 ×10⁻3 mol Lancashire, sample boiled in 55 minutes = 0.40 ×10⁻3 mol Edam = 0.70 ×10⁻3 mol Edam repeat = 0.20 ×10⁻3 mol Step 7- Working out the moles of chloride ion in the sample of cheese From the background research, we found the equation of the reaction between silver and chloride ions:
Ag+ (aq) + Cl- (aq) → AgCl(s) This equation shows the reaction between silver and chloride ions have a 1:1 ratio. This means the mole of silver ions reacted will be the same as the mole of chloride ions reacted. As we have worked out the silver ions reacted, therefore we know the mole of chloride ions reacted. Moles of chloride ions reacted with the cheese in the different cheeses experimented with: Philadelphia cheese spread = 2.25 ×10⁻3 mol Blue Stilton = no results Red Leicester titrated with original batch of solution = 2.25 ×10⁻3 mol Red Leicester titrated with new batch of solution = 2.45 ×10⁻3 mol Lancashire, sample boiled in 20 minutes = 0.80 ×10⁻3 mol Lancashire, sample boiled in 55 minutes = 0.40 ×10⁻3 mol Edam = 0.70 ×10⁻3 mol Edam repeat = 0.20 ×10⁻3 mol Step 8- Converting the moles of chloride ions to grams We have found out the moles of chloride ions extracted in all the different cheeses we experimented with, now we will need to convert the moles to grams in order to make an Kingsley Wong
35 easier comparison. To convert moles to grams, we must first find out the Atomic Mass of chlorine. From the periodic table, we can find out the Atomic Mass of chlorine is 35.5. We multiply the Atomic mass of chlorine with the moles of chloride ions found to make the conversion. Example- Philadelphia cheese spread 2.25 ×10⁻3 × 35.5 = 0.079875g Grams of chloride ion found in the different cheeses experimented with: Philadelphia cheese spread = 0.079875g Blue Stilton = no results Red Leicester titrated with original batch of solution = 0.079875g Red Leicester titrated with new batch of solution = 0.086975g Lancashire, sample boiled in 20 minutes = 0.0284g Lancashire, sample boiled in 55 minutes = 0.0142g Edam = 0.02486g Edam repeat = 0.0071g Step 9- Finding out the mass of chloride ion per 100 grams of cheese Now that the mass of chloride ions have been worked out the 500cm3 of cheese extract solution which contains around 6g of cheese in each sample varying due to accuracy difference during weighing. So to work out the gram of chloride ion per type of cheese, we divide it by the mass of cheese added during sample preparation. Then multiple it by 100 to find out the value of chloride ions per 100g of cheese. Example- Philadelphia cheese spread Mass of cheese added- 6.01g 0.079875 × 100 = 1.329g (Rounded to 3 decimal places) 6.01 Mass of chloride ion per 100 grams for the different cheeses experimented with: Philadelphia cheese spread = 1.329g Blue Stilton = no results Red Leicester titrated with original batch of solution = 1.333g Red Leicester titrated with new batch of solution = 1.450g Lancashire, sample boiled in 20 minutes = 0.473g Lancashire, sample boiled in 55 minutes = 0.237g Kingsley Wong
36
Edam = 0.414g Edam repeat= 0.118g Comparisons of results gathered from Volhard’s Method Type of cheese Philadelphia cheese spread
Chloride ion expected per 100g of cheese (g) 0.617
Chloride ion found per 100g of cheese (g) 1.329
Chloride ion differences (g) 0.712 more
Blue Stilton Lancashire (Fast boiling)
N/A 0.972
N/A 0.473
N/A 0.499 less
Lancashire (Slow boiling)
0.972
0.237
0.735 less
Red Leicester ( Old solution) Red Leicester (New solution)
1.111
1.333
0.222 more
1.111
1.450
0.339 more
0.414 0.118
1.129 less 1.410 less
Edam 1.543 Edam repeat 1.543 (Bold figures represent possible anomalies) Control experiment
I conducted several control experiments to prove that Volhard’s Method does work by using just salt, in the first trial; I followed the full procedure of the method using 0.65g of salt. But the solution was titrated with, no precipitation was formed. In the second trial of control experiment, I used the same amount of salt but I missed out the steps involving addition of nitric acid and potassium permanganate as the there were no cheese that was needed to be digested. However, when titrating, no precipitation was formed again. Instead, there was a strong scent of chlorine from the solution. This led me to believe that excess chloride ions from the salt had reacted with the water in chlorination to form chlorine water. This causes a problem as less chloride ion will react with the silver ions. Or all of the silver ions had reacted with the chloride ions, but there were still an excess of it. In the third trial, I reduced the silver nitrate concentration to 0.02mol and continued to skip the addition of nitric acid and potassium permanganate. Once again, no precipitation was formed. Possible reasons were that the salt was lost during filtration, and that the chloride had reacted with water again to form some chlorine water. In the fourth trial, I recognised that I was using far too much salt, as lower amount of salt means less will react with the silver ions, meaning there will be more silver ions available to react with the sodium thiocyanate during titration. I reduced the salt to 0.15g, and continued to use 0.1mol of silver nitrate. In the titration, precipitations formed. The results gathered had an average titre of 7.9cm³. And following the methods of calculations, the result was 1.05 ×10¯³ mol of chloride ions were found. Converting it into grams, it was 0.037g (3 decimal places). 0.15g of salt was added at the start of the control experiment, to find out the grams of chloride ion present in that. We simply divide the mass of the salt by the molecular mass of the compound, which is 58.5. This finds the mole of sodium chloride present, we then multiple that value with the atomic mass of chlorine which is 35.5 to find the mass of chlorine in 0.15g of sodium chloride. The result is 0.091g (3decimal places). The difference in chloride ion was 0.054g. This is a big difference, but Kingsley Wong
37 factors such as chlorination of water and lost of solution through filtration all contribute to the inaccuracy. The main thing gathered from the control experiment was that the method does work, although I was unable to completely follow the procedure, it was the best possible control experiment possible. Analysis of the results After calculating the chloride ion found in the different types of cheeses experimented with, I have formulated the results to a table for comparisons. In general, there were no perfect results. It would have been near impossible to gather a set of perfect measurements of the chloride ion content. The results found out are compared with theoretical chloride ion expected, it is not the actual value but of a rough estimation. Philadelphia cheese spread- the chloride ion found was the third highest in the list of cheeses tested. It had the greatest difference compared to the chloride ion expected, with 0.712g more. This result is extremely unusual when I expected Philadelphia cheese to only have around 0.617g of chloride ions, the lowest of all the cheeses tested on. Philadelphia cheese spread was a soft cheese, so I expected easier digestion of the cheese which would therefore release the maximum amount of sodium chloride, hence chloride ions. However, the results suggest a much higher amount of chloride ions was present in the cheese. Blue Stilton- there was no results gathered for the Blue Stilton cheese experiment. This is because when titrating the solution containing the cheese extract, there was no precipitation forming in the addition of potassium thiocyanate. After two repeats, I concluded there was no chloride ion present in the solution. This could be because the cheese was unable to be digested during sample preparation. The silver nitrate solution may be exposed to the light for too long, causing oxidation of the silver ions. It could be the reasons to why no precipitation was formed during the titrations. After conducting Volhard’s Method using Blue Stilton cheese and with no results shown, I thought of possible reasons to why it could have occurred. One of the possible reasons to which I thought might have contributed to the lack of results was the stock solutions. I questioned the stock solutions accuracy in concentration. So I made a complete new set of standard solution to conduct with in the next cheese experiment, along with the original solutions to see if there were going to be difference in results. The next cheese used was Red Leicester, in which two separate set of chemicals were used during sample preparation and titration in two separate experiments. Red Leicester (Original solutions)- The chloride ions found was 1.333g per 100 grams of cheese and the expected value was around 1.111g. So 0.222g more of chloride ion content is found, this was another unusual result as I was not expecting the cheese to have more chloride ions than expected. Red Leicester (New solutions)- the chloride ions found was 1.450g per 100 grams of cheese and the expected value was around 1.111g, Around 0.339g more of chloride ions was found, again, an unexpected result as more chloride ion is found in comparison to around the expected value. In both experiment, there were still undigested cheese particles remaining during sample preparation. There was most definitely a difference in the results of titration between the same cheeses by using different chemicals. The original solutions proved to give a slightly more accurate result in comparison to the new solutions. And both sets of solutions used provided some results, which eliminates the possibility of the stock solution in not gaining any results in the previous experiment. As the original set of solution provided a more accurate result, I assumed that very small variation in the concentration of stock solutions could affect the end result. From the experiment, I decided to use the original solutions for the remaining cheeses to be tested. But the fact that both cheeses had more chloride ions Kingsley Wong
38 than expected raised other questions such as problems in my method in sample preparation. So I decided to review the method, which could perhaps provide a more accurate salt extraction. In the next cheese I conducted with, Lancashire, I carried out two different experiments using the same stock solutions and same mass of cheese. The difference between the two experiments was that one would be heated for 20 minutes during sample preparation, and the other would be heated for 55 minutes. I chose two extreme values to observe the effect time under heating has on salt extraction and efficiency of which the cheeses are dissolved. Lancashire (20 minutes boiling time) – 0.473g of chloride ions was found per 100 grams of cheese and the expected value was around 0.972g. So there was 0.499g less chloride ion than expected, this result was completely the opposite of the previous results found. The results seemed more reliable as I expected less chloride ions to be found as they are lost through filtration and not all of which are released. Lancashire (55 minutes boiling time) – 0.237g of chloride ions was found per 100 grams of cheese and the expected value was around 0.927g. There is 0.735g less than the expected value. This result is far less accurate than the less boiling time experiment. I expected this experiment to be more accurate as I expected longer heating would release more salt from the cheese. However, there were far less chloride ions than expected. This set of results from the Lancashire cheese shows that the longer heating time provided a less accurate result which was opposite to my expectations. However, in the longer boiling time experiment, I discovered that I was conducting my method wrong. I did not allow the reaction between nitric acid and potassium permanganate to be completed fully in the previous experiments. So in order for the reactions to be complete, a long duration in heating is necessary, but the longer duration heating also made the results less accurate. At the end of both the experiments, there were still large amount of cheese particles left undigested. These results helped me to improve my method in conducting the experiments; however, it raised furthermore questions to the problems with the method. Prolonged heating of the solutions without digesting cheese and releasing salt could have promoted chlorination of the solution, forming some chloride water as free chloride ions have increased activation enthalpy in the solution from the heating. Formation of chlorine water or the release of any chlorine gas would of decreased concentration of chloride ions in solution, meaning less will react with the silver nitrate. Meaning more silver ions are left, thus increasing the titres during titration. The next cheese used to conduct Volhard’s Method with was Edam. I conducted the experiment with the original solutions and I heated it for the longer duration of time in order to try and dissolve more cheese. I recognised that perhaps the lowered concentration of nitric acid from 6mol to 5mol due to health and safety regulation in school could be a reason to why there was always a large amount of cheese particles left undigested. In order to try and digest more of the cheese, I increased the amount of nitric acid added from 20cm³ to 30cm³ and the amount of potassium permanganate added from 40cm³ to 60cm³. Edam- 0.414g of chloride ions was found per 100 grams of cheese and the expected value was around 1.543g. There is a difference of 1.129g. 1.129g of chloride ions less than expected was unusual, as I expected Edam cheese to contain the highest amount of chloride ions as it contains the highest salt level. The Edam cheese experiment’s results show that it contains even less chloride ions than the previous cheeses. I increased the nitric acid used during this experiment, but there was still undigested cheese particles left. This meant that the increased amount of nitric acid did not help to digest the Edam cheese. Edam is a very tough cheese with a chewy texture, this could be one of the reasons to why so little salt was extracted as it was harder to digest. Another reason why the chloride ion content is calculated to be low is the interpretation of the end point. The end point of the titration was difficult to judge, so I have noted down Kingsley Wong
39 several sets of possible values to which the titre could be in the Edam cheese experiment. I took the last set of possible values recorded and used it as my concordant titres. I repeated the Edam experiment again for a second set of results. Edam repeat- 0.118g of chloride ions were found per 100 grams of cheese and the expected value is around 1.543g. There is a massive difference of 1.410g. The repeat of the experiment seemed to have provided even less accurate results. The trends of the data found suggest several reasons to why the results values were unusual. The softer cheeses provided more chloride ions, as the type of cheese changed, to harder cheeses, the values of chloride ion content were smaller. This suggest that the softer cheese such as Philadelphia was digested more, providing better salt extraction in comparison to the harder cheeses such as Edam and Lancashire. The almost rubbery nature of Edam perhaps made it harder for the cheese to be digested and the reaction rate becomes slower. Another possible reason to the unusual trend of the data could be because additional salt is added to cheese after the manufacturing stage for better taste and prolonged preservation. This may not be included in the nutritional information provided, which therefore leads to inaccuracy when comparing the data. There was no trend suggesting that as the salt content of the cheese is higher there is a higher amount of chloride ion concentration, which would have been the logical expectation. I have highlighted several values that seem to be anomalies due to the extreme nature of the result. I have highlighted Philadelphia cheese spread result, one of the Lancashire result and both of the Edam results. Philadelphia showed a much higher chloride ion value and the other three results showed much lower chloride ion values, this could have arisen from all the reasons I suggested previously. The trend showed the opposite of what I expected the results to be; the logical expectation would be that the cheese with the highest salt content would contain the highest concentration of chloride ions. However, my experiment shows that the cheese with the least salt contained the highest concentration of chloride ions. This trend clearly highlights some form of mistake, as the chloride ion concentration in the Philadelphia cheese spread is almost at an impossible value. This led me to review my formatted data and my raw data again. After carefully analysing my rough notes and raw data again, I recognised a simple but major flaw in the experiment data which could be the reason to why the trend of the date is unusual. As stated earlier, the end point was down to interpretation and was hard to judge. For some of the experiments such as the Edam cheese experiments, I noted down more than 1 set of possible titre values. I took the highest set of values as concordant titres, which was a mistake. I used the values 8.50cm³, 8.60cm³ and 8.55cm³ which made the average titre 8.6cm³. However, I also noted down 7.90cm³, 7.80cm³ and 7.70cm³. If I took these 3 titres, the average titre would be 7.80cm³. Following the calculations procedures stated earlier with that starting value will provide me with 0.650g of chloride ion concentration per 100g of Edam cheese. In comparison to the value 0.414g in the original data, there is a notable difference. Although it would still make the value found around 0.893g less than the expected value, it proves to be more accurate. Therefore I can conclude that the data collected are less reliable than expected because of the judgement of the end points, it is hard to rely on human eyes to decide on the titres. This clearly highlights how human error is a major limitation and effect on the accuracy of the experiment from the analysis of the results. Although the values of expected chloride ion concentration for each type of cheese are estimations, it most certainly would have been impossible to get 100 percent accuracy in determining the chloride ion concentration. The data does show the method works, but it does not prove the reliability of the method.
Kingsley Wong
