Now, calculate the concentration of apotransferrin present in the original 4.00

Question

Now, calculate the concentration of apotransferrin present in the original 4.00 mL solution.

The protein apotransferrin binds two Fe(III) ions for cell transport. Consider when a 4.00 mL solution of apotransferrin is titrated with 113 mu L of a 1.37 M Fe(III) solution. The initially colorless solution turns red, and the color change can be measured with a spectrophotometer. First, explain why there is a difference in spectrophotometric response during titration before and after the equivalence point by placing the statements below into the appropriate categories. Now, calculate the concentration of apotransferrin present in the original 4.00 mL solution.

Explanation / Answer

Ans. Before the equivalence point: a. The color change is minimal- because color change takes when Fe3+ binds to apoprotein (APP) binding site.

            b. No binding sites are available for Fe(III)- Because there is no AAP with Fe3+ when both the reagents are NOT mixed yet, iron (III) does not have a binding site to bind.

Note: Such condition may also come after mixing the two reagents but only when there is EXCESS of Fe(III) in the solution. Since, the condition ‘Fe (III) is in excess’ or similar one is mentioned, it’s not safe to group this point in ‘after the equivalence point is reached’.   

After the equivalence point: b. All Fe(III) added will bind to the protein.

            c. The color change is dramatic.

Ans. 2. It’s assumed that reaction between the two the two reagents is exactly ‘complete’ i.e. no reagent is in excess or limiting.

            1 APP + 2 Fe(III) ----> Colored complex

For the reaction to be complete, the number of moles (here, micromoles- it won’t affect our result) of Fe(III) must be equal to 2 x number of moles of APP.

Now, using, M1V1 = M2V2        --- equation 1

Where, M1= molarity of initial solution 1, V1= volume of initial solution 1 (APP)

            M2= molarity of final solution 2, V2= volume of final solution 2 (Fe3+ solution)

Or, M1 x 4.0 mL = 1.37 M x 0.113 mL      [1 mL = 1000 uL ; both volumes shall have same unit]

Or, M1 = (1.37 M x 0.113 mL) / (4.0 mL) = 0.0387025 M

Thus, concentration of APP in 4.0 mL solution = 0.0387025 M

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