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Experiment Measuring Calcium Ions in Milk

Paper Type: Free Essay Subject: Chemistry
Wordcount: 1680 words Published: 18th Oct 2021

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Intro:

Minerals found in molds and vegetables are important for proper dental and calcium (Ca) development. Eating calcium during childhood is important in determining bone density in adults, and also affects rates. Dairy products provide the best sources of calcium in bioavail and can reduce osteoporosis through increased use, a disease that affects many people around the world when bones undergo brittle (bone dilution), leading to bone fractures over time.

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In this experiment, the calcium content in milk is measured by EDTA back titration. It means that one acid and one amine nitrogen can donate to one electron. EDTA is a hexaprotic ion (chair agent): it acts by binding metal ions (e.g. a complex of calcium ion). The chemical equation reveals that EDTA in the reaction of Ca is 1: 1 molar. Using a solution that includes a free metal ion (e.g. calcium) and a chelating agent solution (EDTA) this titanium transfer is measured against a normal Ca2 + ion solution. The final point is usually determined using a color-coded reference ligator with a free metal color indicating the difference in green / brown color. Personally, I want to see this as I want to know how much dairy products give your life so much calcium.

Figure 1: The EDTA Molecule

 

 

Image taken from: http://www.chm.bris.ac.uk/motm/edta/EDTA.gif [Accessed 11/28/19]

Figure 2: How it bonds with the calcium ions

Image taken from: http://openlearn.open.ac.uk/file.php/2986/S_ 1_002i.jpg[Accessed 11/28/19]

Experimental:


Results:

Showing the mass of milk used.

Sample #

1

2

3

Mass of empty glass in grams

115

90

144

Mass of powder and glass in grams

116

91

135

Mass of milk used in grams

1

1

1

Showing the masses of EDTA and CaCO3 used.

 

EDTA

CaCO3

Mass of empty beaker /g

32.1

32.1

Mass of beaker and salt /g

35.8

32.7

Mass of salt used /g

3.70

0.615

EDTA mass calculation needed to make the solution

EDTA: 372 g/mol-1

Number of moles = 0.03 mol

Therefore mass of salt required for 25mL = (372 g/mol-1×0.03 mol)/4

The Mass of EDTA = 2.79g

 CaCO3 Regular Solution Volumes versus 25mL 0.03 Molarity EDTA and Dairy Solutions

Sample #

1

2

3

Initial Analysis in cm3

1.0

0.0

0.0

Final Analysis in cm3

39

38

37

Volume of CaCO3 in cm3

38

38

37

 CaCO3 Regular Solution Volumes volumes versus 25mL 0.03 Molarity EDTA.

Sample #

1

2

3

Initial Analysis in cm3

0.0

0.0

0.0

Final Analysis in cm3

40

41

41

Volume of CaCO3 in cm3

40

41

41

Treatment of Results:

Concentration of EDTA

Mass of EDTA = 2.79g | Mole Raito of EDTA: 372 g/mol-1

Number of moles EDTA = 0.03 mol | Number of moles of EDTA present in 250mL

Moles = m/mole raito = 2.79g/372 g/mol-1 = 7.5 x10-3 mol

So there are 7.5 x10-3 moles in the 250mL solution.

Concentration = (7.5 x10-3 moles/250mL) x 1000 = 0.03 Molarity

Concentration of CaCO3

Mass of CaCO3 = 0.615g | Mole Raito CaCO3: 100 g/mol-1

Number of moles of CaCO3 present in 250mL

= 0.615g/100g/mol-1= 6.15 x10-3 mol

So there is 6.15 x10-3 mol present in 250mL of solution

Concentration = (6.15 x10-3 mol/250mL) x 1000 = 0.0246 Molarity

Finding the number of Ca moles that reacted to EDTA with 25 mL

Sample Blank 1 used 40mL of CaCO3

So in 40mL there is (0.0246M/1000) x 40mL = 9.84x10-4 mol used

9.84x10-4 mol of Ca reacted with 25mL of EDTA

Sample 1 used 38mL of CaCO3

So in 38mL there is (0.0246M/1000) x 38mL = 9.35x10-4 mol used

9.35x10-4 mol of Ca reacted with 25mL of EDTA

Therefore the number of moles of calcium in sample 1

= Total calcium reacting with 25 mL of EDTA – moles of calcium reacting with residual moles of calcium = 9.84x10-4 mol   - 9.35x10-4 mol = 4.9 x10-5 mol

To find the % of Ca in milk sample 1

Mole Raito Ca = 40.1 g/mol-1

= 4.9 x10-5 moles x 40.1 g/mol-1 = 1.96 x10-3 g in 1 g of milk

So the amount of Ca in 100g of milk = 196mg

Average calcium concentration = (196mg + 373mg+5\324)/3 = 190mg

So the standard deviation = 13.8. So the calcium concentration in milk samples = 190 ±13.8 mg

Conclusion: This lab has shown how calcium can be present in milk. Calcium, theoretically, reaches 900-1190 mg per 100 g of milk powder. The mass was 190 ± 13.8 mg in this experiment. The scale shows a high error rate and cannot be precise because the sample used did not show the exact value.

Discussion: The empty tithes in which Solochrome Black is used as a guide show the color difference with the introduction of calcium-EDTA. At the center, the solutions of Royal Blue (Ca2 + ion with a reference) were very ahead of the endpoint. Eventually, the center turned purple (all Ca2 plus EDTA ions are complex, an indicator that has not been completely changed). In the experiment, a color change was also observed when the blue sample solutions started well before the end, with clouds (opaque) gray, then before the end point, followed by a violet color. Overall, milk is probably the safest way to get your required amount of calcium to keep your teeth and bones strong and healthy.

References:

  1. CHEM 2460 Principles of Chemical Analysis Laboratory Manual.pg 17-21
  2. James N Miller & Jane C Miller, “Statistics and  Chemometrics for Analytical” Chemistry, 5th Ed(2005) Publ. Pearson Education Limited  pg 114
  3. http://www.csudh.edu/oliver/che230/labmanual/calcium.htm [Accessed 11/28/19]
  4. WO SAY DISCUSSION HELP http://www.ajcn.org/cgi/reprint/83/2/310.pdf [Accessed 10/04/10]
  5. http://www.cerlabs.com/experiments/10875404367.pdf[Accessed 11/28/19]
  6. Cassie. “The Determination of Calcium in Milk by EDTA Titration.” Scribd, Scribd, 27 Apr. 2010, www.scribd.com/doc/30580869/The-Determination-of-Calcium-in-Milk-by-EDTA-Titration.

 

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