at pH 11 AY4- is .81 Kf of Cd2+ is 3.162 x 10^16 Cd2+ formation constants: B1= 3

Question

at pH 11 AY4- is .81

Kf of Cd2+ is 3.162 x 10^16

Cd2+ formation constants: B1= 323.594 ; B2= 29512.0922 ; B3= 588843.6554 ; B4= 3630780.548

Calculate pCd2 at each of the following points in the titration of 55.00 mL of 0.0080 M Cd2 with 0.0080 M EDTA in the presence of the auxiliary complexing agent NH3. The solution is buffered at a pH of 11.00 and the NH3 concentration is fixed at 0.100 M. The fraction of EDTA in the form Y4 as a function of pH can be found here. The formation constant for the Cd2+-EDTA complex can be found here. The cumulative formation constants for the Cd(NH3)n2complexes can be found here. (a) 0 mL (d) 55.00 ml Number Number (b) 13.00 mL (e) 60.00 mL Number Number (c) 50.00 mL Number

Explanation / Answer

a. 0mL

Designating the total concentration of free EDTA as [EDTA] we can write

Kf’ = [CuY2-]/([Cu2+][EDTA] = Y4-•Kf = (.81)(3.162* 10^16) = 2.561 * 10^16

[EDTA] = (0.0080)(50.0) = 0.00016 M

[CuY2-] = 0

[Cu2+] = [CuY2-]/(Kf’[EDTA]) = 0

b.) 13 mL

Formal concentration of CuY2- = (13.0/63.0)(0.0080)= 0.00165 M

initial conc. of CuY2-=  0.00165 M

final conc. of CuY2-= (0.00165 -x )M

(0.00165 -x )/ x2 = 2.561 * 10^16

x= 2.53 X 10^-10

pCu2+ = 9.596

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