Question

The M^+n (100.0 mL of 0.050 M metal ion buffered to pH 9.00) was titrated with 0.050 M EDTA.

a) Calculate the concentration of M^+n when one-half of the equivalence volume of EDTA (V=1/2Ve) is added.

b) What fraction of free EDTA is in the form Y^-4 at pH of 9.00?

c) If the formation constant Kf is 10^12.00, calculate the concentration of M^+n at V=Ve.

d) What is the concentration of M^+n at V=1.100 Ve?

Answer 1

1) concentration of m⁺n at 1/2 of equivalence volume

first we find equivalence volume

mole of 100.0 mL of 0.050 M metal ion = 0.05 M x 0.1 L = 0.005 moles

so equivalence mole of EDTA = 0.005 moles / 0.05 M = 0.1 L = 100 mL

concentration = 0.1(0.025)/0.150 L = 0.01667M .......(0.025 M = 0.05 M /2)  

2) fraction of free EDTA is in the form Y^-4 at pH of 9.00

0.05 x á

0.05 x 0.041 = .00205

3) formation constant Kf is 10^12.00

Kf = [EDTA . V²⁺ ] / [V²⁺ ][EDTA]

[EDTA . V²⁺ ] = 0.1(0.05)/ 0.2 = 0.025

[V²⁺ ] = [EDTA] = X

Kf = (0.025 - X ) / X² ......{0.025 - X ~ 0.025 }

X² = 0.025/Kf

X = sqrt(0.025/ 10¹² ) = 1.58x10^-7

4) concentration at V = 1.100 Ve

0.1(0.055)/0.210 L = 0.0262 M .... (0.05 x 1.10 = 0.055 M)