Question

Calculate [Mn²⁺] when 50.00 mL of 0.1000 M Mn²⁺ is titrated with 25.00 mL of 0.2000 M EDTA. The titration is buffered to pH 11 for which = 0.81. K_f = 7.76 × 10¹³

Answer 1

Kf = 7.76 × 10¹³

α = 0.81

Kf¹ = 7.76 × 10¹³ x 0.81

      = 6.29 x 10^13

millimoles of Mn+2 = 50 x 0.1 = 5

millimoles of EDTA = 25 x 0.2 = 5

molarity of [MnY^2-] = 5 /75 = 0.0667 M

So, we have only MnY^2- at this point

Mn+2   + EDTA -------------> MnY^2-

0                0                         0.067

x                x                          0.067-x

Kf.= [MnY^2-]/[Mn2+][EDTA]

6.29 x 10^13 = 0.0667-x / x^2

x = 3.258 x 10^-8

x = [Mn2+] = 3.258 x 10^-8 M

[Mn2+] = 3.258 x 10^-8 M