Question

pk1=0.0(CO2H)

PK2=1.5(CO2H)

PK3=2.0(CO2H)

PK4=2.66(CO2H)

PK5=6.16(NH+)

PK6=10.24(NH+)

1. The total protonated from of EDTA is H₆Y²⁺.
Based on the pK values, calculate the pH when the following form of EDTA exists dominantly in a 0.100 M aqueous solution.

(a) the H₆Y²⁺ form of EDTA. (Hint: treat it as the monoprotic weak acid using the K₁ as K_a to calculate H⁺ concentration for pH.)

(b) the zwitterion form (H₄Y) of EDTA.

(c) the Na₂H₂Y form (H₂Y^2-) of EDTA.

(d) the Y^4- form of EDTA. (Hint: treat it as the monobasic weak base using the K_b = K_w /K₆ to calculate the OH^- concentration, then get the pH from it.)

Answer 1

There are two many parts for s single questions.....

a) the H₆Y²⁺ form of EDTA.

     Its a case of Strong acid , assuming as monoprotic acid we can solve as follows

       Ka1 = Cα²/ (1-α)

solving quadratic equation we get α = 0.27 = 27 %

[ H⁺] = c α = 0.1 x 0.27 = 2.7x10^-2

pH = 1.56

b) the zwitterion form (H₄Y) of EDTA.

pH = (pka1+Pka2)/2 = (2+2.66)/2 = 2.33 its a zwitter ion

please post remaining question separately