Calculating the pZn value

Question

Calculate the pZn value after the addition of 25 mL of titrant in the titration of 50 mL of 0.01 M Zn2+ solution buffered to pH 11 with 0.02 M EDTA. (Kf = 3.2 x 1016 α4 = 0.85)

Answer 1

Zn²⁺+Y^4- ↔ ZnY^2-

50 mL of Zn²⁺ with 0.01 M and 25 mL of Y^4- with 0.02 M is at equivalency point

pH = 11 K_f = 3.2 × 10¹⁶ α_Y4- = 0.85

First calculate [ZnY^2-] with n_Zn2+ and V_tot

0.050 L Zn²⁺ • 0.01M Zn²⁺= 0.0005 moles Zn²⁺

0.0005 moles Zn²⁺ ≅ 0.0005 moles ZnY^2-

[ZnY^2-] ≅ 0.0005 moles ZnY^2- / 0.075 L= 0.0067 M

This is an assumption that all of the moles of Zn are reacted.

Unreacted concentrations of Zn and EDTA are equal

[Zn²⁺] = [Y^4-]

Use the formula for K_f ' (Since K_f^'' requires a buffer with an auxiliary complexing agent like ammonia or citrate)

K_f ^' = Ξ [products] / Ξ [reactants]

K_f ^' = [ZnY^2-] / ( [Zn²⁺] [Y^4-] )

Substitute in K_f ^' = K_f α _Y4-

K_f α _Y4- = [ZnY^2-] / ( [Zn²⁺] [Y^4-])

Use the equivalency relation for Zn and Y

K_f α _Y4- = [ZnY^2-] / [Zn²⁺] ²

Solve for [Zn²⁺]

[Zn²⁺] = √ [ZnY^2-] / (K_f α _Y4-)

Apply -log

pZn = -log √ [ZnY^2-] / ( α_Y4-K_f )

pZn = -log √ 0.0067 / (0.85 × 3.2 × 10¹⁶)

pZn = 9.3

Hope that helps!