The following solutions were prepared: 3.0 mL of Ferric ammonium sulfate dodecah

Question

The following solutions were prepared: 3.0 mL of Ferric ammonium sulfate dodecahydrate aqueous solution is added to a 25 ml. volumetric flask. 40 mg of ferrozine and 1.25 mL of ammonium acetate buffer is added before diluting to volume. 3.0 mL of Ferric ammonium sulfate dodecahydrate aqueous solution is added to a 25 mL volumetric flask. 3.0 mL of hydroxylamine hydrochloride (a reducing agent) was added and the flask was swirled for one minute. 40 mg of ferrozine and 1.25 mL of ammonium acetate buffer is added before diluting to volume. The absorbance of solution 1 was found to be 0.107 and the absorbance of solution 2 was found to be 0.837. What is the concentration of converted Fe(lll) rightarrow Fe(ll) in solution 2? (Report to 1 sig fig) What is the concentration of Fe(lll) in the 3.0 mL of undiluted stock solution of iron ammonium sulfate dodecahydrate aqueous solution? (Report to 1 sig fig)

Explanation / Answer

Ans. The calibration curve, the trendline equation or linear regression equation is in form of “Y = mX + C”.

In the graph, Y-axis indicates absorbance and X-axis depicts concentration. That is, according to the trendline (linear regression) equation y = 18.55x - 0.0886, one absorbance unit (1 Y = Y) is equal to 18.55 units on X-axis (concentration) minus 0.0886. Note that the unit of concertation (or, X-axis) is mM as shown in graph.

#A. Solution 1 gives absorbance of 0.107 because of Fe(II) already present in it. There is no reduction of Fe(III) into Fe(II) in tube 1. That is, even if there is no reduction of Fe(III), the solution gives some absorbance. Thus, solution 1 is called “reagent blank”.

Solution 2 gives absorbance due to Fe(II) initially present in it plus Fe(II) produced from Fe(III). Note that even if there were no reduction of Fe(III), the solution also gives absorbance of 0.107 like the reagent blank.

So,

Absorbance of solution 2 due of reduction of Fe(III) =

Total Abs of solution 2- Abs of solution 1 (reagent blank)

= 0.837 – 0.107 = 0.730

Now, putting y = 0.730 in trendline equation-

            0.730 = 18.55x - 0.0886

            Or, 18.55x = 0.730 – 0.0886 = 0.6414

            Or, x = 0.6414 / 18.55 = 0.0346

Therefore, concentration of Fe(III) converted to Fe(II) = 0.0346 mM

#B. Assuming complete reduction, the initial [Fe(III)] is equal to [Fe(II)] produced from reduction.

So, initial [Fe(III)] in 25.0 mL solution = 0.0346 mL

Original solution = 3.0 mL

Now, Using    C1V1 = C2V2                        - equation 2

C1= Concentration, and V1= volume of initial solution 1         ; Original solution

C2= Concentration, and V2 = Volume of final solution 2         ; Reaction mixture

Putting the values in above equation-

            C1 x 3.0 mL = 0.0346 mM x 25.0 mL

            Or, C1 = (0.0346 mM x 25.0 mL) / 3.0 mL = 0.288 mM

Thus, [Fe(III)] in 3.0 mL original solution = 0.288 mM

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