Calculate pCd^2+ at each of the following points in the titration of 55.00 mL of

ID: 496947 • Letter: C

Question

Calculate pCd^2+ at each of the following points in the titration of 55.00 mL of 0.0020 M Cd^2+ with 0.0020 M EDTA in the presence of the auxiliary complexing agent NH_3. The solution is buffered at a pH of 11.00 and the NH_3 concentration is fixed at 0.100 M. The fraction of EDTA in the form Y^4- as a function of pH can be found 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. (a) 0 mL. 10.00 mL (c) 50.00 mL (d) 55.00 mL (e) 60.00 mL

Explanation / Answer

Kf for Cd2+-EDTA = 2.88 x 10^16

alpha[Y4-] for EDTA at pH 11 = 0.85

So,

Kf' = Kf x alpha[Y4-] = 2.45 x 10^16

with,

[NH3] = 0.1 M

we have,

alpha[Cd2+] = 1/(1+ b1(0.1) + b2(0.1)^2 + b3(0.1)^3 + b4(0.1)^4)

b1, b2, b3 abd b4 are cumulative formation constants

So,

alpha[Cd2+] = 1/(1 + 35.5 + 1.02 + 0.022 + 0.0007) = 0.0266

Kf'' = Kf' x alpha[Cd2+] = 6.52 x 10^14

(a) no EDTA added

[Cd2+] = 0.002 M x 0.0266 = 5.32 x 10^-5 M

pCd2+ = -log[Cd2+] = 4.27

(b) 10 ml EDTA added

initial moles of Cd2+ = 0.002 M x 55 ml = 0.110 mmol

moles of EDTA added = 0.002 M x 10 ml = 0.02 mmol

[Cd2+] remained = 0.09/65 = 0.0014 M

[Cd2+] = 0.0014 x 0.0266 = 3.68 x 10^-5 M

pCd2+ = -log[Cd2+] = 4.43

[Cd2+] remained = (0.002 M x 55 ml - 0.002 M x 50 ml)/105 ml = 9.52 x 10^-5 M

[Cd2+] = 9.52 x 10^-5 x 0.0266 = 2.53 x 10^-6 M

pCd2+ = -log[Cd2+] = 5.60

(d) 55 ml EDTA added

equivalence point

[CdY2-] formed = 0.002 M x 55 ml/110 ml = 0.001 M

Cd2+ + EDTA <==> CdY2-

let x amount of complex had redissolved

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

6.52 x 10^14 = 0.001/x^2

x = [Cd2+] = 1.24 x 10^-9 M

[Cd2+] final = 1.24 x 10^-9 x 0.0266 = 3.29 x 10^-11 M

pCd2+ = 10.48

(e) 60 ml EDTA added

[CdY2-] formed = 0.11 mmol/115 ml = 9.56 x 10^-4 M

[EDTA] remained = 0.002 m x 5 ml/115 ml = 8.70 x 10^-5 M

6.52 x 10^14 = 9.56 x 10^-4/[Cd2+](8.70 x 10^-5)

x = [Cd2+] = 1.70 x 10^-14 M

[Cd2+] final = 1.70 x 10^-14 x 0.0266 = 4.48 x 10^-16 M

pCd2+ = 15.35

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