The Turbidity Test for Pasteurized Milk

The Turbidity Test for Pasteurized Milk Milk and dairy products, such as cream and yoghurt, are an important food group in the food pyramid. This food group provides us with calcium, which is not only crucial in strengthening our bones, but also important in many biological processes, such as facilitating the release of neurotransmitters that transmit nerve impulses across a synapse. Since dairy products serve such importance in our diet, dairy products manufacturing industry takes extra precaution in ensuring that these products meet the guidelines set by statutory bodies, one of which is that the maximum lactic acid content allowed in milk is 0.15% w/w. Hence, the industry will employ various methods to determine the quality of milk. As such, in order to better understand these industrial methods, 2 groups of experiments relating to titratable acidity (TA) of selected foods and turbidity test for pasteurized, UHT and sterilized milk were carried out. The titratable acidity test allows us to determine the titratable acidi ty of a sample as lactic acid (for dairy products) or citric acid (for lemon curd) equivalent. Basically, TA, as an acid equivalent, of a food product measures the total amount of that particular reference acid in the selected food. This reference acid is the major acid component, amongst all types of acid present in the food, which we want to quantify. TA is different from pH as pH only measures the [H+] dissociated from the acid molecules. Hence, TA is a more accurate measure of the degree of spoilage of dairy products than pH. The turbidity test however, serves a different function in terms of quality control. It is usually used by the industry to test if sterilized milk products have been sufficiently sterilized. Titratable Acidity of Selected Foods Materials Phenolphthalein as indicator 50.00ml burette 10.0ml graduated pipette White porcelain basin Magnetic stirrer Experiment 1: Titratable Acidity of Milk Pasteurized milk (Farmhouse Fresh Milk), expires on 20/9/12 UHT milk (Marigold UHT Full Cream), expires on 15/6/13 0.01M sodium hydroxide (actual concentration is 0.0107M) Experiment 2: Titratable Acidity of Cream Sour cream (Bulla Sour Cream), expires on 14/9/12 Yoghurt (FN Alive Yoghurt), expires on 11/9/12 0.1M sodium hydroxide (actual concentration is 0.105M) pH meter Experiment 3: Titratable Acidity of Lemon Curd Lemon curd (Waitrose lemon curd), expired on Feb 12 0.1M sodium hydroxide (actual concentration is 0.105M) Methods Titration of selected food products against NaOH of known concentrations were carried out in order to determine the titratable acidity of these food products. The titratable acidity in lactic acid or citric acid equivalent was then determined by via stoichiometric ratio of the acid to the amount of NaOH, as seen in the stoichiometric calculations below. 3 sets of titrations for 3 different groups of food products, mainly pasteurized milk and UHT milk, sour cream and yoghurt, and lemon curd, were carried. Experiment 1: Titratable Acidity of Milk 10.0 ml of pasteurized milk was transferred to a white porcelain basin. 1.0 ml of phenolphthalein indicator was then added to this sample. The burette was filled up with 0.01M NaOH and then titrated against the pasteurized milk sample. End-point of titration was identified when a pale pink colouration persisted for at least 10 s. Initial and final burette readings were recorded in Table 1 below. The procedure was repeated thrice for both pasteurized and UHT milk. Experiment 2: Titratable Acidity of Cream 10.00 g of sour cream was transferred to a white porcelain basin. 10.0 ml of water was added to the sample and mixed and pH was then measured. 1.0 ml of phenolphthalein indicator was added to the diluted sample. The burette was filled up with 0.1M NaOH and then titrated against the sour cream sample. End-point of titration was identified when a pale pink colouration persisted for at least 10 s. Initial and final burette readings were recorded in Table 2 below. The procedure was repeated thrice for both sour cream and yoghurt. Experiment 3: Titratable Acidity of Lemon Curd 10.00 g of lemon curd was transferred to a white porcelain basin. 10.0 ml of water was added to the sample and mixed. 1.0 ml of phenolphthalein indicator was added to the diluted sample. The burette was filled up with 0.1M NaOH and then titrated against the lemon curd sample. End-point of titration was identified when a pale pink colouration persisted for at least 10 s. Initial and final burette readings were recorded in Table 3 below. The procedure was repeated two more times. Results Experiment 1: Titratable Acidity of Milk Table 1: Titration of pasteurized and UHT milk against 0.01M NaOH Milk sample vol. of milk measured (ml) average vol. of milk (ml) initial burette reading (ml) final burette reading (ml) vol. of NaOH used (ml) average vol. of NaOH used* (ml) ÂÂ  Pasteurized Milk 10.0 10.0 50.00 37.65 12.35 12.35 10.0 37.65 25.15 12.50 10.0 25.15 12.80 12.35 ÂÂ  UHT Milk 10.0 10.0 50.00 37.70 12.30 12.30 10.0 37.70 25.30 12.40 10.0 25.30 13.00 12.30 Pasteurized Milk CH 3 CH OH C O- Na+ O CH 3 CH OH C OH O + NaOH Ã   + H2O (1) Amount of NaOH used = (Average vol. of NaOH used) x [NaOH] = (12.35/1000)(0.0107) = 1.32 x 10-4 mol From (1), lactic acid : NaOH is 1:1 amount of lactic acid in 10.0ml of pasteurized milk = 1.32 x 10-4 mol Concentration of lactic acid (in mol/100mL) equivalent in pasteurized milk = (1.32 x 10-4) / (10/100) = 1.32 x 10-3 mol/100mL Concentration of lactic acid equivalent in g/100mL in pasteurized milk = (molar concentration (in mol/100mL) of lactic acid equivalent) x (molar mass of lactic acid) = (1.32 x 10-3)(90.08) = 0.119 g/100mL UHT Milk Amount of NaOH used = (Average vol. of NaOH used) x [NaOH] = (12.30/1000)(0.0107) = 1.31 x 10-4 mol From (1), lactic acid : NaOH is 1:1 amount of lactic acid in 10.0ml of UHT milk = 1.31 x 10-4 mol Concentration of lactic acid (in mol/100mL) equivalent in UHT milk = (1.31 x 10-4) / (10/100) = 1.31 x 10-3 mol/100mL Concentration of lactic acid equivalent in g/100mL in UHT milk = (molar concentration (in mol/100mL) of lactic acid equivalent) x (molar mass of lactic acid) = (1.31 x 10-3)(90.08) = 0.118 g/100mL From the calculations, it can be seen that both the titratable acidities of pasteurized milk and UHT milk in lactic acid equivalent are below 0.15%, the maximum allowed titratable acidity of milk in lactic acid equivalent. As such, both samples are deemed safe for consumption. The titratable acidity of pasteurized milk is also observed to be slightly above that of UHT milk by a very minute concentration of 0.001 g/100mL. This suggests that pasteurized milk contains slightly more microbes than UHT milk, which goes in tandem with the properties of pasteurized milk. This is because pasteurized milk is heated to about 65oC for at least 30 minutes in order to preserve the flavor of milk, while UHT milk is heated at 135oC for about 2 seconds6. Hence, fewer microbes are killed in pasteurized milk than UHT milk. As such, pasteurized milk will have slightly higher lactic acid concentration which is produced from the fermentation of lactose by microbes. However, the magnitude of difference of 0.001 g/100mL obtained from the titration results is too small to make the above conclusive deduction. The average vol. of NaOH used is almost identical for both milk samples as there is only a difference of 0.05 ml, making the titration results somewhat anomalous. The main reason for this anomaly is the subjectivity of the end-point of titration. The colour change of phenolphthalein from colourless to pale pink is very difficult to ascertain by naked eye for the inexperienced, unlike workers in this industry who carry out large volumes of titrations every day. As such, the faint pink that I observed in pasteurized milk is most probably not the true end-point of titration or it could be that the faint pink I observed in UHT milk is over the end-point of titration for UHT milk. Experiment 2: Titratable Acidity of Cream Table 2: Titration of sour cream and yoghurt against 0.1M NaOH Cream sample pH of sample average pH mass of sample (g) average mass of sample (g) initial burette reading (ml) final burette reading (ml) vol. of NaOH used (ml) average vol. of NaOH used* (ml) ÂÂ  Sour Cream 4.48 4.49 10.00 10.00 50.00 44.60 5.40 5.40 4.50 10.01 44.60 39.20 5.40 4.50 9.99 39.20 33.80 5.40 ÂÂ  Yoghurt 4.43 4.38 10.01 9.99 50.00 37.90 12.10 12.30 4.34 10.00 37.90 25.60 12.30 4.38 9.98 25.60 13.30 12.30 Sour Cream Amount of NaOH used = (Average vol. of NaOH used) x [NaOH] = (5.40/1000)(0.105) = 5.67 x 10-4 mol From (1), lactic acid : NaOH is 1:1 amount of lactic acid in 10.00g of sour cream = 5.67 x 10-4 mol Mass of lactic acid in 10.00g of sour cream = (amount of lactic acid) x (molar mass of lactic acid) = (5.67 x 10-4)(90.08) = 0.0511g Concentration of lactic acid equivalent (in %w/w) in sour cream = (mass of lactic acid in 10.00g of sour cream) / (average mass of sour cream) x 100% = (0.0511) / (10.00) x 100% = 0.511% (w/w) Yoghurt Amount of NaOH used = (Average vol. of NaOH used) x [NaOH] = (12.30/1000)(0.105) = 1.29 x 10-3 mol From (1), lactic acid : NaOH is 1:1 amount of lactic acid in 9.99g of yoghurt = 1.29 x 10-3 mol Mass of lactic acid in 9.99g of yoghurt = (amount of lactic acid) x (molar mass of lactic acid) = (1.29 x 10-3)(90.08) = 0.116 g Concentration of lactic acid equivalent (in %w/w) in yoghurt = (mass of lactic acid in 9.99g of yoghurt) / (average mass of yoghurt) x 100% = (0.116) / (9.99) x 100% = 1.16% (w/w) From the results of this experiment in Table 2, we can see that titratable acidity is not equal to pH, and it shares an inverse relationship with pH, where pH = -lg[H+]. This is because lactic acid is an organic acid and hence it is a weak acid. As such, lactic acid only partially dissociates, giving a [H+] that is lower than the total lactic acid concentration. This is due to the low acid dissociation constant, Ka, of lactic acid. However, by proportionality, it is observed that higher concentrations of lactic acid molecules will give a higher deprotonated [H+]. This is observed in Table 2 where the lower pH of yoghurt corresponds to a higher average volume of NaOH required to neutralize the lactic acid present. In addition, another observation is that yoghurt requires more than twice the volume of 0.1M NaOH to neutralize the lactic acid present as compared to sour cream even though yoghurt is lower in pH by 0.11. This is mainly attributed to the presence of probiotics added into yoghurt. As such, this means that more lactose in yoghurt is converted into lactic acid, resulting in the marked difference in average vol. of NaOH required for neutralization. This second observation also proves that pH is not a true measure of total lactic acid content in dairy products as this small difference in pH is accompanied by a larger than proportionate difference in volume of NaOH required for neutralization. Experiment 3: Titratable Acidity of Lemon Curd Table 3: Titration of lemon curd against 0.105M NaOH Sample mass of sample (g) average mass of sample (g) initial burette reading (ml) final burette reading (ml) vol. of NaOH used (ml) average vol. of NaOH used* (ml) Lemon Curd 10.00 10.00 50.00 29.20 20.80 20.35 10.00 29.20 8.90 20.30 10.00 50.00 29.60 20.40 *As 3 sets of titration were conducted for each sample in order to improve the precision and reproducibility of the titration results, the average volume of NaOH was taken as the average of the 2 closest values of vol. of NaOH used in titration so as to be more precise. C Na+O- O CH 2 C OH C O- Na+ O CH 2 C O- Na+ O C HO O CH 2 C OH C OH O CH 2 C OH O + 3NaOH Ã   + 3H2O (2) Amount of NaOH used = (Average vol. of NaOH used) x [NaOH] = (20.35/1000)(0.105) = 2.14 x 10-3 mol From (2), citric acid : NaOH is 1:3 amount of citric acid in 10.00g of lemon curd = (amount of NaOH used) / 3 = 7.13 x 10-4 mol Molar mass of citric acid = 6(12) + 8(1) + 7(16) = 192 g mol-1 mass of citric acid in 10.00g of lemon curd = (amount of citric acid) x (molar mass of citric acid) = (7.13 x 10-4)(192) = 0.137 g Concentration of citric acid equivalent (in % w/w) in lemon curd = (mass of citric acid in 10.00g of lemon curd) / (average mass of lemon curd) x 100% = (0.137) / (10.00) x 100% = 1.37% (w/w) As calculated above, the concentration of citric acid equivalent in lemon curd is 1.37% (w/w), which is well above the minimum standard of 0.33% (w/w) set by legislation in some parts of the world. Hence, it can be deduced that this sample of lemon curd has passed the quality control measure. Citric acid is used as the reference for quality control of lemon curd mainly because citric acid is present in the largest quantity in lemons. Hence, measuring citric acid concentration present will be a good measure of the quality of the lemon curd. As such, this is a quality lemon curd sample. Even though this lemon curd product expired on February 2012, the citric acid content should not be significantly affected by microbial decomposition mainly because the acidic environment due to citric acid is not suitable for most bacteria to thrive. Discussion There are a few experimental procedures which can be improved on. Firstly, as mentioned in the results of experiment 1, the faint pink observed to mark the end-point of titration is subject to a large margin of human error. As such, a better method to solve the issue of colour subjectivity is to use a colorimeter to determine an intensity of pink as the end-point of titration, thus eliminating any inaccuracies that result from human error. Secondly, it was observed that the dilution of products of a more viscous consistency, such as sour cream and lemon curd, did not yield a homogenous consistency as compared with the milk samples and yoghurt. As such, the titrated NaOH may not have actually reacted with all the acid molecules as some acid molecules may be trapped inside the granular particles. This can be overcome by vortexing the cream and water mixture in a sealed round-bottom conical flask to ensure a homogenous solution is obtained, allowing us to obtain more accurate titration results. Thirdly, for runny liquid samples such as milk, there is a risk of spillage due to splashing when the magnetic stirrer operates probably due to the large exposed opening of the porcelain basin. Splashing can be overcome by using a conical flask to contain the samples and place a white tile under the conical flask so that the change in colour of milk can be made more obvious. This is because a conical flask has a much narrower neck and therefore a significantly narrower opening, thus minimizing spillage from splashing. In this way, more accurate titration results can be obtained. For lemon curd, simply measuring the citric acid concentration is insufficient to conclude a quality product. This is mainly due to the possibility of adulteration of lemon curd by adding more citric acid chemical, just like how milk was adulterated by the adding melamine. As such, additional qualitative methods can be employed, such as measuring the concentration of certain chemical substances more unique to lemon, such as limonene. Turbidity test for pasteurized, UHT and sterilized milk Materials Ammonium sulphate powder Pasteurized milk UHT milk Sterilized milk Method 4.0g of ammonium sulphate, (NH4)2SO4, was dissolved in 20.0 ml of pasteurized milk. The mixture was allowed to stand for at least 5 min and subsequently filtered. 5 ml of the filtrate was collected in a test-tube and then placed in boiling water bath for at least 5 min. The test-tube containing filtrate was then cooled in cold water and the contents were examined for presence of turbidity. Results Discussion Table 4: Turbidity test results Sample Observation Pasteurized milk A cloudy pale yellow solution with precipitation was observed. UHT milk A cloudy pale yellow solution was observed. Sterilized milk A clear pale yellow solution was observed. The turbidity test is useful in telling us if a sample of milk is sufficiently sterilized, whereby a clear solution will be observed. The turbidity test is first carried out by adding a denaturing agent, usually ammonium sulphate, (NH4)2SO4, to the milk sample. As NH4+ exhibits acidic properties, as shown in the following equation, NH4+ + H2O Ã   NH3 + H3O+ this addition of ammonium ions will bring about an increase in [H+], resulting in the disruption of casein micelle structure. This causes casein proteins to precipitate and coagulate as they interact with the ammonium and sulphate ions. For those casein and whey proteins that are already denatured by heat treatment during processing, ammonium and sulphate ions will form interactions with the charged R-groups of the acidic and basic amino acid residues, causing them to precipitate out of the solution. These precipitate are obtained as the residue from filtration. The filtrate obtained contains mostly undenatured whey proteins and probably some unprecipitated protein molecules encapsulated in the casein micelle structure amidst a solution of ammonium sulphate and other soluble milk products such as lactose. Upon heat treatment in a 100oC water bath, the milk proteins denature and are thus exposed to ammonium sulphate. They undergo the same mode of action with ammonium sulphate as described above, resulting in the observed precipitation. Referring to AVA regulations, pasteurised milk is defined to be milk that has been subjected to a single heat-treatment of 62.8 65.6oC for at least 30 min or 72 73.5oC for at least 15 s; UHT milk is defined as milk that has been heated at a temperature of at least 135oC for at least 2 s; sterilized milk is milk heated to 100oC long enough to sufficiently kill all microbes. As such, sterilized milk will have all the casein and whey proteins fully denatured and free in the milk due to prolonged heating. Whereas UHT milk will have a slight concentration of undenatured proteins present due to a short high heat treatment. For pasteurized milk however, it will contain the highest concentration of undenatured proteins due to lowest heat treatment temperature. Hence, the experimental observation in Table 4 clearly fits the hypothesis. On a side note, the yellow pale solution observed is most likely due to the Maillard reaction between lactose in and amino compounds in milk. Conclusion Titratable acidity and the turbidity test for milk is but only 2 out of the many methods that the milk processing industry employs to ensure that the heat treatments have produced milk that are safe for human consumption. The main disadvantage that lies with milk treated with higher heat processes is the loss of flavour. The nutrients that are lost during heat are usually replaced (e.g. enriched milk) and hence this is less of a concern for milk. As such, it is the companys decision on whether to process milk with an emphasis on taste or shelf-life. Nevertheless, it is of utmost importance that the milk products remain well within the margin of safety as stated by regulations. At the same time, regulatory bodies need to stay alert and play a part in ensuring that companies follow the safety guidelines, less an incident like the adulteration of dairy products through melamine addition may occur again.

Ohm’s Law Series-Parallel Circuits Calculation Essay

To end up the discussion of Series-Parallel Circuits, I would like to post this last one remaining topic which is about Ohm’s Law of Series-Parallel Circuits for currents and voltages. I did not even mentioned in my previous topics on how to deal with its currents and voltages regarding this type of circuit connection. Ohms Law in Series-Parallel Circuits Ohm’s Law in Series-Parallel Circuits – Current The total current of the series-parallel circuits depends on the total resistance offered by the circuit when connected across the voltage source. The current flow in the entire circuit and it will divide to flow through parallel branches. In case of parallel branch, the current is inversely proportional to the resistance of the branch – that is the greater current flows through the least resistance and vice-versa. Then, the current will then sum up again after flowing in different circuit branch which is the same as the current source or total current. The total circuit current is the same at each end of a series-parallel circuit, and is equal to the current flow through the voltage source. Ohm’s Law in Series-Parallel Circuits – Voltage The voltage drop across a series-parallel circuits also occur the same way as in series and parallel circuits. In series parts of the circuit, the voltage drop depends on the individual values of the resistors. In parallel parts of the circuit, the voltage across each branch are the same and carries a current depends on the individual values of the resistors. If in case of circuit below, the voltage of the series resistance forming a branch of the parallel circuit will divide the voltage across the parallel circuit. If in case of the single resistance in a parallel branch, the voltage across is the same as the sum of the voltages of the series  resistances. The sum of the voltage across R3 and R4 is the same   as the voltage across R2. Finally, the sum of the voltage drop across each paths between the two terminal of the series-parallel circuit is the same as the total voltage applied to the circuit. Let’s have a very simple example of this calculation for this topic. Considering the circuit below with its given values, lets calculate the total current, current and voltage drop across each resistances. What is the total current, current and voltage across each resistancesHere is the simple calculation of the circuit above: a. Calculate first the total resistance of the circuit: The equivalent resistance for R2 and R3 is: R2-3 = 25X50/ 25+50 = 16.67 ohms R total = 30 ohms + 16.67 ohms = 46.67 ohms b. Calculate the Total Current using Ohm’s Law: I1 = 120V / 46.67 Ohms = 2.57 Amp. Since R1 is in series connection, the total current is the same for that path. c. Calculating the voltage drop for R1: VR1 = 2.57 Amp x 30 ohms = 77.1 volts d. Calculate the voltage drop across R2 and R3. Since the equivalent resistance for R2 and R3 as calculated above is 16.67 ohms, we can now calculate the voltage across each branch. VR2 = VR3 = 2.57 Amp x 16.67 ohms = 42. 84 volts e. Finally, we can now calculate the individual current for R2 and R3: I2 = VR2 / R2 = 42.84 volts / 25 ohms = 1.71 Amp. I3 = VR3 / R3 = 42.84 volts / 50 ohms = 0.86 Amp. You may also check if the current in each path of the parallel branch are correct by adding its currents: I1 = I2 + I3 = 1.71 Amp + 0.86 Amp = 2.57 Amp. which is the same as calculated above. Therefore, we can say that our answer is correct.

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