Question

Consider titrations of the following two analytes, which each use a titre 0.01000 M EDTA: 50.00 mL of 0.00500 M Co2+ 50.00 mL of 0.00500 M CO3+ For each titration, which both take place at pH 1.00, find the percent present at the equivalence point. What do you conclude from this compariso about these two titrations? Explain briefly.

Answer 1

EDTA forms 1:1 complex with both Co²⁺ and Co³⁺ ions. The log(binding constants) of the two complexes are 16.45 and 41.4 respectively.

Co²⁺ + EDTA^4-$\rightleftharpoons$ [CoEDTA]^2- logK_f = 16.45, K_f = 2.8 x 10¹⁶

Co³⁺ + EDTA^4-$\rightleftharpoons$ [CoEDTA]^- logK_f = 41.4, K_f = 2.51 x 10⁴¹

1) For [CoEDTA]^2- complex formation,

Co²⁺ + EDTA^4-$\rightleftharpoons$ [CoEDTA]^2-

Initially at the equivalence point 0.005 0.005 0

At equilibrium 0.005-x 0.005-x x

K_f = [CoEDTA]^2-/([Co²⁺][EDTA^2-]) = 2.8 x 10¹⁶

=> x/((0.005-x)*(0.005-x)) = 2.8*10¹⁶

=> x = 2.8*10¹⁶ (2.5*10^-5 - 0.01x + x²)

=> 2.8 *10¹⁶*(2.5*10^-5 - 0.01x + x²) = 0 (since 2.8*10¹⁶*x² - x² = 2.8*10¹⁶*x²)

=> 2.5*10^-5 - 0.01x + x² = 0

=> x = 0.005

That means, all of the material is present as complex. There is no free Co²⁺ ion in the system.

2) Since K_f for [CoEDTA]^- is even higher, there no free Co³⁺ ions in the system at equilibrium. All Co³⁺ exists as [CoEDTA]^- complex.