Question

Calculate the pCo2+ after 13.00 mL of 0.03846 M EDTA in the titration of 25.00 mL of 0.020 M Co2+ (Kf = 2.04 x 1016) at pH = 6.00.

Answer 1

At pH=6 the fraction α_Y4- = [Y^4-]/C_EDTA = 2.4x 10^-5

[Y^4-] = α_Y4-C_EDTA

Co²⁺   + Y^4-     = CoY^2-

Kf = [CoY^2-] / ([Co²⁺][Y^4-] =

      = [CoY^2-] / ([Co²⁺] α_Y4-C_EDTA)

[Co²⁺] = [CoY^2-] / (Kf α_Y4- C_EDTA)

                                        

0.025 L x 0.020 mol/L = 0.00050 mol Co²⁺, initial quantity

0.013L x 0.03846 mol/L = 0.00050 mol EDTA added

The titration is at its end point EP.

At EP :

[Co²⁺] = C_EDTA   ( from dissociation of the complex)

[Co²⁺] = ([CoY^2-] / (Kf α_Y4-))^1/2

[CoY^2-] = 0.0005 mol / (0.025 +0.013) L = 0.01316 M

Put values in this equation:

[Co²⁺] = ([CoY^2-] / (Kf α_Y4-))^1/2

[Co²⁺] = (0.01316 / (2.04 x 10¹⁶ x 2.4x 10^-5) )^1/2

            = (2.688x10^-14 )^1/2

= 1.639x10^-7 M

pCo²⁺ = -log1.639x10^-7 = 6.78