Mapoob Sapling Learning You are titrating 1500 mL of 0060 M

Mapoob Sapling Learning You are titrating 150.0 mL of 0.060 M Ca2 with 0.060 M EDTA at pH 9.00. Log K for the Ca2 EDTA complex is 10.65, and the fraction of free EDTA in the Y4-form, , is 0.041 at pH 9.00. (a) What is Kt, the conditional formation constant, for Ca2 at pH 9.00 Number K, (pH 900)-O (b) What is the equivalence volume, Ve, in milliliters? Number mL (c) Calculate the concentration of Ca2 at V1/2 Ve Number

Solution

For the titration,

(a)

log Kf = 10.65

Kf = 4.46 x 10^10

Kf\' = Kf x alpha[Y4-]

= 4.46 x 10^10 x 0.041

= 1.83 x 10^9

(b)

Equivalence volume Ve (volume of EDTA) = 150 ml

(c)

When V = 1/2Ve = 75 ml

concentration of [Ca2+] = 0.06 M x 75 ml/225 ml

= 0.02 M

(d)

When V = Ve

concentration of [CaY2-] = 0.06 M x 150 ml/300 ml

= 0.03 M

Ca2+ + EDTA <==> CaY2-

let x be the amount of complex in solution

Kf\' = [CaY2-]/[Ca2+][EDTA]

1.83 x 10^9 = (0.03)/x^2

concentration of [Ca2+] = x = 4.05 x 10^-6 M

(e)

When V = 1.1Ve

[CaY2-] formed = 0.06 M x 150 ml/315 ml

= 0.0286 M

[EDTA] left = 0.06 m x 15 ml/315 ml

= 0.00286 M

Kf\' = [CaY2-]/[Ca2+][EDTA]

1.83 x 10^9 = (0.0286)/[Ca2+](0.00286)

concentration of [Ca2+] = 5.46 x 10^-9 M