Experiment 2: Determination of phosphate in water samples by the Ascorbic Acid Reduction Method
Introduction:
Phosphorus plays specific in many biochemical procedures as it is an essential nutrients for plants at a definite concentration; it is one of the most abundant elements on earth and is commonly found in the form of phosphate such as orthophosphate, hydrolysable phosphate and organic phosphorus compound. One of the key effects of phosphorus is the eutrophication of surface water. This is where the uncontrollable growth of algae occurs. Waterways should not exceed a concentration of phosphorus greater than 0.05mg/L. Increased levels of phosphates in water are attributed by fertilisers and sewage due to human factors will cause the water surface to glow green or red, depending on the type of algae. The most importantly consequence of this is that algae depletes the amount of dissolved oxygen in the water. The depletion of oxygen can be very dangerous to aquatic life. Therefore it is essential that waterways be carefully monitored for phosphorus content.
The analysis of phosphorus content is most commonly carried out using the ascorbic acid-molybdate blue method. The orthophosphate ion reacts with ammonium molybdate and antimony potassium tartrate under acidic conditions to form 12-molybdophosphoric acid. This complex is reduced with ascorbic acid to form a blue complex which absorbs light at 880 nm.The absorbance is proportional to the concentration of orthophosphate in the sample. The chemical equation can be seen below.

[P(Mo3O10)4]3- + 3H+ + Ascorbic acid → (MoO2.4MoO3)2.H3PO4+2MoO2+Dehydroascorbic acid+ 4H2O

Therefore, the intensity can be measured using a spectrophotometer. This instrument operates by passing a monochromatic beam of light through the sample. Then, the identification of the intensity of light reaching the detector will be performed.

The Beer-lambert law can then be used to calculate the concentration which states a relationship between the amount of light absorbed and the concentration. This law states that:

A = ε C l

Where, A= absorbance, C= Concentration, l = path length and ε = absorptivity constant.

The overall aim of this experiment is to determine the concentration of orthophosphate and total phosphorous in a water sample collected from Macquarie Lake Yerbury using the ascorbic acid reduction method followed by UV-visible spectrophotometer.

Experimental: Sampling and handling
The water sample was collected from Lake Yerbury, on a sunny day, in a clean plastic bottle, washed with a solution of HCl and rinsed with distilled water. To avoid suspended materials, the sample was filtered using a 0.45μm membrane filter.

Dihydrogen phosphate standard calibration:
Six standard solutions were prepared by accurately pipetting 0, 1, 2, 4, 7 and 10mL of 0.2197g/L KH2PO4 and 30mL of distilled water into 50mL volumetric flasks. Then 9mL of molybdate/ascorbic acid reagent was added to each sample. The solutions were allowed to develop a blue colour for 15 minutes .

Determination of orthophosphates:
From the filtered sample, 40 mL was pipetted into a 50mL volumetric flask followed by a drop of phenolphthalein indicator to obtain a red colour. then, three drop of 6M sulphuric acid were added to discharge the colour. Following the same procedure as before, we added 9ml of molybdate/ascorbic acid reagent. Then, The solutions were allowed to develop a blue colour for 15 minutes .

Determination of total phosphorous:
From the same filtered sample, 40 mL was pipetted into a 100mL conical flask followed by adding 0.4 ammoniumpersulfate and v=1 ml of sulphuric acid. Then some bloiling chips was added to the whole mixture. The sample was gently boiled on a heating plate for one hour fifteen minutes, with the volume of solution maintained between 25 and 50 mL. The end of the boiling was determined when the volume decreased to 10mL. To the cooled sample a drop of phenolphthalein indicator was added followed by 1 mol L-1 sodium hydroxide solution dropwise to adjust the colour to a faint pink and 6 mol L-1 sulfuric acid until became colourless again. After that, 9mL of molybdate reagent was added, subsequently dilution to 50 mL with distilled water. Then, the solution was allowed to develop a blue colour for 15 minutes.
At the end, all of our samples prepared before were analysed and the absorbances of the samples and standards were then measured at 880nm in 1cm cell using a Shimadzu UV- Visible mini 1240 spectrophotometer.
Results:
All the samples were analysed using UV spectrophotometer at 880 nm in a 1cm cell. Absorbance values are given in the tables I, II and III.
Table I: Absorbances of standards Sample | Concentration(mg/L) | Absorbance | Blank | 0.0000 | 0.000 | 1 | 0.0439 | 0.007 | 2 | 0.0879 | 0.016 | 3 | 0.1758 | 0.032 | 4 | 0.3076 | 0.053 | 5 | 0.4394 | 0.076 |

Table II: Absorbances for samples of orthophosphate | Absorbance | Blank | 0.000 | Sample solution | 0.027 |

Table III: Absorbances for samples of total phosphorous | Absorbance | Blank | 0.000 | Sample solution | 0.037 |

Discussion: A calibration plot was constructed between the absorbances obtained for standard samples of dihydrogen phosphate using the regression sheet and their corresponding concentrations as the following:

From the obtained plot, Slope = 0.1728/mgL-1, Intercept = 0.0028833
Therefore the line of regression can be written as: Absorbance = 0.1728x [Phosphate] mg/L + 0.0028833.
T test will be performed to check for the significance of the correlation coefficient:
Step 1: Null Hypothesis: There is no statistically significant correlation between concentration of phosphate (mg/L) and absorbance.
Step2: Alternative Hypothesis: There is a statistically significant correlation between the concentration of phosphate (mg/L) and absorbance.
Step 3: Test Statistic = /r/n-2(1-r²) = 0.99966-2(1-(0.99962) = 70.68
Step 4: Critical Value from the two sided t table with 4 degrees of freedom is 2.78 at the 95% confidence level.
Step 5: As 70.68 >2.78, we reject null hypothesis
Step 6: We conclude that there is a statistically significant correlation between the concentrations of phosphate (mg/L) and the absorbance at 95% confidence level.

From the residual plot, all of the absorbances values lie horizontally around the horizontal line.
Thus, referring to the scatter plot and the residual plot, we can conclude that there is linear a relationship between the concentration of phosphate and absorbances which was showed by the T-test.
Determination of 95% confidence intervals of Slope and Intercept
1-Calculation of 95% confidence interval of slope:
(SD) slope = Sy/x(x-x)2 =0.000880.142886= 0.002
Confidence interval for slope (b) = ±t (n-2) (SD)slope = ±2.78 × 0.002= ± 0.0556
Calculation of 95% confidence interval of intercept:
(SD) intercept = S y/xx2n(x-x)2 = 0.00120.32836×0.142886 = ±0.000743
Confidence interval for intercept (a) = ±t (n-2) (SD)intercept= ± 2.78 × 0.0007 = ±0.002
The calibration plot equation can be given as:
Absorbance = 0.1728± 0.0556× [phosphate]mg/L+0.0028± 0.002
Determination of Orthophosphates concentration in water samples
Interpolation of the external calibration plot will be used to find the concentration of orthophosphate in the water sample.
Referring to table II:
The absorbance found for orthophosphate =0.027
Therefore the concentration of orthophosphate will be using the regression line:
Absorbance = 0.1728x [Phosphate] mg/L + 0.0028
Therefore, 0.027 = 0.1728x [Phosphate] mg/L + 0.0028
Orthophosphate concentration(mg/L) = (0.027-0.0028)/0.1728
Orthophosphate concentration = 0.1400mg/L
After considering dilution factor is: 0.1400 mg/L x (40/50) = 0.175mg/L
Calculation of 95% confidence interval for the orthophosphate concentration in the sample:
(SD)Orthophosphate = Sy/xb1+1n+Y-y2b²x-x2 = 0.00080.17281+16+(0.027-0.03)20.17282 (0.142886)=0.0500
Confidence interval for the Orthophosphate concentration in water sample:
±t (n-2) (SD) Orthophosphate= ±2.78 x 0.0500 = ± 0.1390
Therefore the orthophosphate concentration will be = 0.175± 0.1390 mg/L
Determination of Total phosphorous concentration in the water samples
Referring to table III:
The total phosphorous absorbance will be =0.037 The concentration of total phosphorous was calculated by substituting this measured absorbance of 0.038 in the regression line:
Absorbance = 0.1728 x [Phosphate] mg/L + 0.0028
Hence, 0.037= 0.1728x [Phosphate] mg/L + 0.0028
Total phosphorous concentration (mg/L) = (0.037-0.0028)/0.1482
Total phosphorous concentration = 0.2307 mg/L
Total phosphorous concentration after considering the dilution factor is calculated as the following: 0.2307 mg/L x (40/50) = 0.184 mg/L
Calculation of 95% confidence interval for the total phosphorous concentration in the sample:
(SD)Total phosphorous = Sy/xb1+1n+Y-y2b²x-x2 = 0.00080.17281+16+(0.037-0.0306)20.17282(0.14288)= 0.0050
Hence, confidence interval for the Total phosphorous concentration in water sample can be calculated as the following:
±t (n-2) (SD) Total phosphorous= ±2.78 x 0.0050 = ± 0.0139
Therefore the total phosphorous concentration= 0.184 ± 0.0139 mg/L.

The summary of the orthophosphate and the total phosphorous concentration collected from lake Macquarie will be showed in table VI: Sample | Concentration(mg/L) | 95% Confidence Interval | Orthophosphate | 0.175 | 0.1390 | Total phosphorous | 0.184 | 0.0139 |

From these results, we can demonstrate that there is a higher concentration of phosphorous than the recommended limits of 0.05 mg/l in Macquarie lake Yerbury. Therefore, this suggests that might be an influence of pollution carried out by humans leading to eutrophication and depletion of oxygen in our aquatic environment. There is a quite disturbance of this equilibrium of phosphorous in the ground and the surface water.
It has also to be considered possible errors during the analysis which might have been occurred due to temperature deviation and reaction time fluctuation. Some loss of the phosphorous compounds on the wall of containers can be another issue for variation of results. Additionally two people were involved during the analysis which increases the level of trueness.
Conclusion:
This experiment was based on the formation of the blue coloured phosphorus molybdenum from the reduction of phosphomolybdate with ascorbic acid. The intensity of the ‘blueness’ was proportional to the concentration of phosphate in the solution. Therefore a UV spectrometer was used to determine the absorbances of several solutions including our calibration standards. Using the standards we were able to determine a calibration plot with a correlation coefficient of 0.996, which is not great but can be worked with. We tested its significance with the t test and the unknown samples of orthophosphate and total phosphate was interpolated from the calibration plot. The confidence intervals of these concentrations were also calculated and reported.

We also calculated the total amount of phosphate in mg/L which was found to be
( 0.184mg/L). In our introduction we stated that normal waterways should have a phosphorus concentration of 0.05mg/L. Lake Yerbury has a phosphate concentration that increases the chance of eutrophication. We believe we have satisfactorily achieved our aim to determine the concentration of orthophosphate and total phosphate in a water sample collected from Lake Yerbury using the ascorbic acid reduction method. Therefore, to improve the precision of the result, more number of samples from Macquarie Lake should be collected and analysed. In addition, samples should be maintained in constant temperature before analysing time. Moreover, maintain of a constant reaction time for all samples should be considered. Finally, the current results warn us that Macquarie Lake is affected by human influence and further precautions and care can be taken, to minimize this effect and return the phosphate levels into normal.
Learning outcomes:
I agree that I do understand that increased phosphate levels will put the aquatic life through the risk of eutrophication and depletion of oxygen ,which can have detrimental effects in aquatic environments.

We also have learnt the practical use of UV spectrophotometers to get absorbance readings of samples that have a change in colour intensity or turbidity. As well the equations involved to find the concentration of total phosphorous.

In this experiment, adequate preparation was required as there were many factors especially time that could affect our end results. I agree that I have come to appreciate that there may be possible difficulties in analysis of real life samples.

Additional exercises:
Question 1:
It is a fertilizer produced by the action of concentrated sulphuric acid on powdered phosphate rock. It is found in naturally occurring ores, plant ,animal wastes and fertilisers.
Question2:
Eutrophication is the process where water is enriched by excessive concentrations of nutrients such as phosphate or nitrate causing the growth of algae on the surface of water. With more human activities and pollution , the rate of decomposition increases which lead to the depletion of oxygen causing the aquatic plants and animal to die.

Question3:
In solution, the equilibrium will be as follow:
H3PO4 + H2O ⇋ H2PO4- + H3O+
H2PO-4 + H2O ⇋ HPO4-2 + H3O+
HPO4-2 + H2O ⇋ PO4-3 + H3O+
[H3PO4] Ka = 7.5 x 10-3
[H2PO4-] Ka = 6.8 x 10-8
[HPO4-2] Ka = 1.7 x 10-12
If we let cT be the sum of the molar concentration of the phosphate – containing solution species in the solution:

cT =[H 3PO 4] + [H 2PO4] + [HPO42-] + [PO4-3] → (1) the alpha value for the free acid α0 is defined as α0= H3PO4cT → (2) And alpha value for H2PO4-, HPO42-;[ PO43-] are given by similar equation: α1= H2PO4-cT; α2= HPO42-cT; α3= PO43-cT →(3)
The sum of the alpha values for system must equal unity: α0 + α1 + α2 + α3 = 1 alpha values for the maleic acid system are expressed in terms of [H3O+], Ka1.,Ka2and Ka3 to obtain such expressions from (1), (2) and (3) α0= [H 3O+]3[H 3O+]3+ Ka1 [H 3O+]2+ Ka1.Ka2[H 3O+] + Ka1.,Ka2Ka3 α1=Ka1[H 3O+]2[H 3O+]3+ Ka1 [H 3O+]2+ Ka1.Ka2[H 3O+] +Ka1Ka2Ka3 α2= Ka2[H 3O+][H 3O+]3+ Ka1 [H 3O+]2+ Ka1.Ka2[H 3O+] + Ka1Ka2Ka3 α3= Ka1Ka2Ka3[H 3O+]3+ Ka1 [H 3O+]2+ Ka1.Ka2[H 3O+] + Ka1Ka2Ka3

At the Ph= 4.5
The mole- percentage of [H 3PO 4]: α0 = [10-4.5]3[10-4.5]3+ 7.5×10-310-4.52+7.5×10-3×6.8×10-810-4.5+7.5×10-3×6.8×10-8×1.7×10-12

α0=4.19×10-3

The mole- percentage of [H 2PO 4-]: α1 = 7.5×10-3[10-4.5]2[10-4.5]3+ 7.5×10-310-4.52+7.5×10-3×6.8×10-810-4.5+7.5×10-3×6.8×10-8×1.7×10-12

α1=0.99

The mole- percentage of [HPO 4-2]: α2 = 6.8×10-8[10-4.5][10-4.5]3+ 7.5×10-310-4.52+7.5×10-3×6.8×10-810-4.5+7.5×10-3×6.8×10-8×1.7×10-12 α2=2.14×10-4 The mole- percentage of [PO 4-3]: α3 = 7.5×10-3×6.8×10-8[10-4.5][10-4.5]3+ 7.5×10-310-4.52+7.5×10-3×6.8×10-810-4.5+7.5×10-3×6.8×10-8×1.7×10-12 α3=1.15×10-10 By using the same way, we can calculate all the mole percentage in each of the Ph values: pH | H2PO4- | HPO4-2 | PO4-3 | H3PO4 | 4.5 | 0.99 | 2.14×10-3 | 1.154*10-10 | 4.19×10-3 | 6 | 0.94 | 0.06 | 1.08*10-7 | 1.25*10-4 | 7 | 0.60 | 0.41 | 6.88*10-6 | 7.94*10-6 | 8 | 0.19 | 0.87 | 1.48*10-4 | 1.71*10-7 | 9.5 | 0.0046 | 0.99 | 5.32*10-3 | 1.94*10-10 |

pH | H2PO4- | 4.5 | 0.99 | 6 | 0.94 | 7 | 0.6 | 8 | 0.19 | 9.5 | 0.0046 |

Figure 3 Plot of mole- percentage of H2Po-4 versus Ph value

In the graph, there is a decrease of the mole percentage of H2PO4 between PH=6 and Ph=8.
Thus, for the determination of the concentration of phosphates the pH should be less than 6.

Question4: we can use less sensitive maxima (a test that include solutions under low wavelengths) to reduce background signal.

Question5: Absorbance (A) | A365 | A700 | Solution 1 | 0.0235 | 0.617 | Solution 2 | 0.0714 | 0.755 | Solution 3 | 0.0945 | 0.920 | Solution 4 | 0.0147 | 0.592 | Solution 5 | 0.0540 | 0.685 | Molar absorptivities are: | Cobalt (Co) | 3529 | 428.9 | Nickel (Ni) | 3228 | 0 |

From equation : A= εLc

A365 = εCo365 L CCo + εNi365 L CNi A700 = εCo700 L CCo + εNi700 L CNi
Absorbance at 700nm: for solution 1 at 700 nm
0.617 = 428.9 M-1 x 1.00 cm x CCo + 0 M-1 x 1.00 cm x CNi
CCo=0.617/428.9*mol/L = 1.44 x 10-3 mol/L the concentration for the other solutions will be calculated using the same method : Solution | A700 | CCoin mol/L | CNi in mol/L | 1 | 0.617 | 1.44 x 10-3 | 0 | 2 | 0.755 | 1.76 x 10-3 | 0 | 3 | 0.920 | 2.15 x 10-3 | 0 | 4 | 0.592 | 1.38 x 10-3 | 0 | 5 | 0.685 | 1.60 x 10-3 | 0 |
Absorbance at 365nm: for solution 1 at 365nm .0235 = 3529 M-1 x 1.00 cm x 1.44 x 10-3 + 3228 M-1 x 1.00 cm x CNi
CNi =0.0235-3529 M-1 x 1.00 cm x 1.44 x 10-3 /3228
CNi=- 1.57 x 10-3

Using the same calculation for other solution:
Concentration of Cobalt and Nickel from sample solutions

Solution | A365 | CCoin mol/L | CNi in mol/L | 1 | 0.0235 | 1.44 x 10-3 | 1.57 x 10-3 | 2 | 0.0714 | 1.76 x 10-3 | 1.90 x 10-3 | 3 | 0.0945 | 2.15 x 10-3 | 2.32 x 10-3 | 4 | 0.0147 | 1.38 x 10-3 | 1.50 x 10-3 | 5 | 0.0540 | 1.60 x 10-3 | 1.73 x 10-3 |