3. The complexometric (EDTA) analysis of iron in two ore samples was performed.

Question

3. The complexometric (EDTA) analysis of iron in two ore samples was performed. The ore samples were crushed and digested in 25 mL nitric acid/hydrogen peroxide using microwave heating. Separate 5 mL aliquots of the extractate were transferred to 25 mL volumetric flasks, buffered to pH 5, and Variamine blue b base (VB) indicator was added before the flasks were filled to the line and mixed well. The replicate trials produced from the aliquots are reported below after titration to the VB endpoint with 0.009732 M EDTA (previously standardized using primary standard calcium carbonate):

Sample 1 - 0.1490 g ore, Trial 1 required 27.51 mL of EDTA

Sample 1 - 0.1490 g ore, Trial 2 required 27.38 mL of EDTA

Sample 1 - 0.1490 g ore, Trial 3 required 27.62 mL of EDTA

Sample 2 - 0.1671 g ore, Trial 1 required 31.47 mL of EDTA

Sample 2 - 0.1671 g ore, Trial 2 required 31.14 mL of EDTA

Sample 2 - 0.1671 g ore, Trial 3 required 31.07 mL of EDTA

Sample 2 - 0.1671 g ore, Trial 4 required 31.10 mL of EDTA

Calculate the percent iron in the ore samples. (AW Fe = 55.845)

Calculate the percent iron in the ore samples. (AW Fe = 55.845)

Does it appear that the analyst was wise to conduct a fourth trial for sample 2? Can any of the results in that set be rejected on statistical grounds? (The standard deviation for the four trials in Sample 2 is 0.3017 % Fe.)

From extensive previous experience, the relative standard deviation of this method is known to be 5 ppth – use this to calculate a value for the relevant standard deviation for this analysis and quote this value and the 95% confidence intervals for the two ore samples.

Is there a statistically significant difference between the results for the two ore samples? Support your answer using appropriate tests and state the result in proper statistical language.

Explanation / Answer

The balanced equation for the reaction of Fe2+ / Fe3+ with EDTA4- is

[Fe(H2O)6]2+(aq) + EDTA4–(aq) ----- > [Fe(EDTA)]2–(aq) + 6H2O(l)

[Fe(H2O)6]3+(aq) + EDTA4–(aq) ------> [Fe(EDTA)]–(aq) + 6H2O(l).

Hence 1 mol of Fe2+ / Fe3+ reacts with 1 mol of EDTA4-.

Sample-1:

Trail-1: moles of ETDA used = MxV(L) = 0.009732 mol/L x 27.51 mL x (1L / 1000 mL) = 2.6773x10-4 mol

Hence moles of iron (Fe2+ / Fe3+) in the ore = moles of EDTA = 2.6773x10-4 mol

=> mass of iron (Fe2+ / Fe3+) in the ore = 2.6773x10-4 mol x 55.845 g/mol = 0.01495 g

=> percentage of  iron (Fe2+ / Fe3+) in the ore = (0.01495 g / 0.1490 g) x100 = 10.0344 %

Trail-2: moles of ETDA used = MxV(L) = 0.009732 mol/L x 27.38 mL x (1L / 1000 mL) = 2.6646x10-4 mol

Hence moles of iron (Fe2+ / Fe3+) in the ore = moles of EDTA = 2.6646x10-4 mol

=> mass of iron (Fe2+ / Fe3+) in the ore = 2.6646x10-4 mol x 55.845 g/mol = 0.01488 g

=> percentage of  iron (Fe2+ / Fe3+) in the ore = (0.01488 g / 0.1490 g) x100 = 9.9870 %

Trail-3: moles of ETDA used = MxV(L) = 0.009732 mol/L x 27.62 mL x (1L / 1000 mL) = 2.6880x10-4 mol

Hence moles of iron (Fe2+ / Fe3+) in the ore = moles of EDTA = 2.6880x10-4 mol

=> mass of iron (Fe2+ / Fe3+) in the ore = 2.6880x10-4 mol x 55.845 g/mol = 0.01501 g

=> percentage of  iron (Fe2+ / Fe3+) in the ore = (0.01501 g / 0.1490 g) x100 = 10.0745 %

Hence average of percentage of iron in Sample-1 = ( 10.0344 % + 9.9870 % + 10.0745 %) / 3 = 10.0320 %

Standard deviation = 0.0438 %

Similarly for Sample - 2:

Trail-1: percentage of  iron (Fe2+ / Fe3+) in the ore = (0.017103 g / 0.1671 g) x100 = 10.2354 %

Trail-2: percentage of  iron (Fe2+ / Fe3+) in the ore = (0.016924 g / 0.1671 g) x100 = 10.1281 %

Trail-3: percentage of  iron (Fe2+ / Fe3+) in the ore = (0.016886 g / 0.1671 g) x100 = 10.1053 %

Trail-4: percentage of  iron (Fe2+ / Fe3+) in the ore = (0.016902 g / 0.1671 g) x100 = 10.1151 %

Average = 10.1460 %

Standard deviation = 0.06034 %

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