quantitative analysis chemistry

Question

Calculate the concentration of Mn²⁺ at the following points in the titration of 25.0 mL of 0.0200M MnSO₄ solution with a solution of 0.0100 M EDTA in a solution buffered at pH 8. Sketch the titration curve.

a)     40.0 mL EDTA added

b)     50.0 mL EDTA added

60.0 mL EDTA added

Answer 1

Two constants this needs, both table values from Harris - substitute your own book's if they differ:

log Kf for MnY2- = 13.89, so Kf = 7.8 x 10^13
α(Y4-) at pH 8.00 = 5.6 x 10^-3

Because the pH is buffered, combine them once into a conditional formation constant and use it for the whole titration:

K'f = α(Y4-) x Kf = (5.6 x 10^-3)(7.8 x 10^13) = 4.3 x 10^11

First, find the equivalence volume

mmol Mn2+ = (25.0 mL)(0.0200 M) = 0.500 mmol. EDTA binds metals 1:1 regardless of charge, so

Ve = 0.500 mmol / 0.0100 M = 50.0 mL

That single number tells you which regime each of your three points is in.

a) 40.0 mL - before the equivalence point

Here there is leftover free Mn2+, and it swamps anything released by the complex. Just do the bookkeeping:

EDTA added = (40.0)(0.0100) = 0.400 mmol
excess Mn2+ = 0.500 - 0.400 = 0.100 mmol
total volume = 25.0 + 40.0 = 65.0 mL

[Mn2+] = 0.100 / 65.0 = 1.54 x 10^-3 M (pMn = 2.81)

b) 50.0 mL - at the equivalence point

Now the only Mn2+ present is what the complex gives back by dissociating. Formal concentration of the complex:

[MnY2-] = 0.500 mmol / 75.0 mL = 6.67 x 10^-3 M

MnY2- dissociates to give equal amounts, so with x = [Mn2+] = [EDTA]:

(6.67 x 10^-3 - x) / x^2 = K'f, and since x is tiny the numerator is just 6.67 x 10^-3

x^2 = 6.67 x 10^-3 / 4.3 x 10^11 = 1.53 x 10^-14

[Mn2+] = 1.24 x 10^-7 M (pMn = 6.91)

c) 60.0 mL - past the equivalence point

Excess EDTA now drives the metal down further. Both species are known, so solve K'f for [Mn2+]:

excess EDTA = (10.0 mL)(0.0100) = 0.100 mmol, total volume 85.0 mL, so [EDTA] = 1.18 x 10^-3 M
[MnY2-] = 0.500 / 85.0 = 5.88 x 10^-3 M

[Mn2+] = [MnY2-] / (K'f [EDTA]) = 5.88 x 10^-3 / ((4.3 x 10^11)(1.18 x 10^-3))

[Mn2+] = 1.15 x 10^-11 M (pMn = 10.94)

The titration curve

pMn
 12 +                                   .  60.0 mL, pMn 10.94
    |                                  /
 10 +                                 /
    |                                 |
  8 +                                 |   <-- steep break at the
    |                                 *       equivalence point
  6 +                                 |   50.0 mL, pMn 6.91
    |                                /
  4 +                        _____.-'
    |      _________________/     40.0 mL, pMn 2.81
  2 +_____/
    +----+----+----+----+----+----+----+
    0   10   20   30   40   50   60   70
            mL of 0.0100 M EDTA added

Plot pMn upward against volume. The curve is nearly flat while free metal remains, then jumps roughly four pMn units within a couple of milliliters of 50.0 mL, then flattens again as excess EDTA accumulates. That vertical break is the whole point of the titration - it is what an indicator or an ion-selective electrode detects.

One thing worth carrying forward: the size of that break is set by K'f, not by Kf. Run the same titration at pH 5 instead of 8 and α(Y4-) falls by roughly three orders of magnitude, K'f falls with it, and the break flattens until the endpoint is unusable. That is why EDTA titrations always specify a buffer, and why the pH is part of the method rather than an afterthought.