Calculate pCd^2+ at each of the following points in the titration of 50.00 mL of
ID: 1068454 • Letter: C
Question
Calculate pCd^2+ at each of the following points in the titration of 50.00 mL of 0.0020 M Cd^2+ with 0.0020 M EDTA in the presence of the auxillary completing ahent NH_3. The solution is buffered at a PH 11.00 and the NH_3 concentration is fixed at 0.100 M. The fraction of EDTA the form Y^4- as a function of pH can be found is here. The formation constant for the Cd^2+ -EDTA complex can be found here. The cumulative formation constants for the Cd(NH_3)_n+2+ complexes can be found here. 0 mL 50.00 mL 20.00 mL 55.00 mL 45.00 mLExplanation / Answer
The EP (equiv. point, completion degree 100%) is at
V = 50.00 mL x0.0020 M Cd2+ /0.0020 M EDTA = 50 mL EDTA
For 3 mL added the completion of titration is
100x3mL/50ml = 6%
Conc. of uncomplexed Cd2+ by EDTA is
0.94 x 0.004 M x 50 mL/53 mL = 0.00355 M
For 45 mL added the completion of titration is
100x45mL/50ml = 90%
Conc. of uncomplexed Cd2+ by EDTA is
0.10 x 0.004 M x 50 mL/95 mL = 0.000210 M
The formation of [Cd(EDTA)]2+ can be assumed to be complete (high Kf value), during titration.
Cd2+ + Y4- = CdY2- Kf = 1016.5 = 3x1016
Cd2+ uncomplexed by EDTA is available for the formation of [Cd(NH3)4]2+, having Kf2 = 1.3×107
1.[Cd2+] = [Cd(NH3)4]2+ / (Kf2 . [NH3]4) = 0.004/(1.3×107x1x10-4) = 3.07x10-3 M
pCd2+ = -log3.07x10-3 = 2.5
2. [Cd2+] = [Cd(NH3)4]2+ / (Kf2 . [NH3]4) = 0.00355 /(1.3×107x1x10-4) =
= 2.7x10-6 M
pCd2+ = 5.57
3. pCa2+ = = [Cd(NH3)4]2+ / (Kf2 . [NH3]4) = 0.000210 M / (1.3×107x1x10-4) =
= 1.6x10-7 M
pCd2+ = 6.8
4. At EP , [Cd2+] is controled by the dissociation at equilibrium of Cd(EDTA)2+
[Cd2+] = [EDTA] = ([Cd(EDTA)]2+ / Kf)1/2 = 4.5x10-10
pCd2+ = 9.35
5. [Cd2+] is controled by the excess of Cd(EDTA)2+
EDTA in excess is
10/100 x 0.004 x 50 mL/105 mL = 0.0002 M
[Cd2+] = [Cd(EDTA)]2+ / ([EDTA]Kf)
= 0.002/(0.0002x3x1016)
= 3.3x10-16
pCd2+ = 15.48
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.