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Define 2 of the 4 titration types: Masking, Displacement, Indirect and Back titr

ID: 969662 • Letter: D

Question

Define 2 of the 4 titration types: Masking, Displacement, Indirect and Back titrations. Two different metals (X^n+ and Y^n+) with the same concentrations (buffered to a pH of 10) are titrated with a 0.100 M EDTA solution. Explain in detail how these titrations curve are similar for each metal and how they differ. A 75.00 mL 0.0530 M solution of C^2+ (buffered to a pH of 4.00, a= 3.0 times 10 9, log Kt = 16.5) was titrated with a 0.250 M EDTA solution. Based on this information answer the questions below. Determine the volume at the equivalence point and the Calculate the pCd^2+ at 1.00 mL of EDTA added after the equivalence point.

Explanation / Answer

A) definition of titrations

1. Masking : In this titration selective determination of the amount of one ion present in a mixture of two or more ions in solution is done by the process of masking the ion which we do not want to titrate by a masking or complexing reagent. The left ion of interest is then titrated with the right titrant and the amount of titrant used gives the concentration of ion present in solution.

2. Back titration : Here the solution with the ion is reacted with excess of a reagent and then the excess reagent added is titrated with another titrant. The amount of reagent initilly added - the amount of reagent reacted with the titrant gives the concentration of ion of interest in solution.

C) moles of Cd2+ present = 0.0530 M x 75 ml = 3.975 mmol

a. Volume at equivalence point for EDTA = 3.975 mmol/0.250 M = 15.9 ml

Total volume of solution would be = 75 + 15.9 = 90.9 ml

b. after 1.00 ml excess of 0.250 M EDTA added

[CdY2-] formed = 3.975 mmol/91.9 ml = 0.0432 M

remaining [EDTA] = 0.250 M x 1 ml/91.9 ml = 0.0027 M

Kf' = Kf x alpha[Y4-] = 3.16 x 10^16 x 3 x 10^-9 = 9.487 x 10^7

Kf' = [CdY2-]/[Cd2+][EDTA]

9.487 x 10^7 = 0.0432/[Cd2+](0.0027)

[Cd2+] = 1.686 x 10^-7 M

pCd2+ = -log[Cd2+] = 6.77

  

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