Question

Consider the titration of 25.0 mL of 0.0200 M MnSO4 with 0.0100 M EDTA in a solution buffered to pH=6.00. Calculate pMn2+ at the following volumes added EDTA, and use your data to sketch a titration curve for they hypothetical sample: 0, 20.0, 40.0, 49.0, 49.9, 50.0, 50.1, 55.0, and 66.0 mL.

Answer 1

1)The Mn+2 ion will react in 1:1 ratio with EDTA.

We know that the moles of two titrant becomes equal to titrand at equivalent point

M1V1 = M2V2

M1= molarity of EDTA

V1 = Volume of EDTA

M2 = molarity of Mn+2

V2 = Volume of Mn+2

0.01 X V1 = 0.02 X 25

V1 = 50mL of EDTA

a) at 0 mL

[Mn+2] = 0.02

So pMn+2 = -log[Mn+2] = 1.698 = 1.7 (approx)

b) at 20mL

moles of Mn+2 added = 25 X 0.02 = 0.5 millimoles

Moles of EDTA = 0.01 X 20 = 0.2 millimoles

so moles of Mn+2 remained unreacted = 0.5-0.2 = 0.3 millimoles

Concentration of Mn+2 = Millimoles / total volume of solution in mL = 0.3 / 25 + 20 = 0.0067

So pMn+2 = -log 0.0067 = 2.17

c) at 40mL of EDTA

moles of Mn+2 added = 25 X 0.02 = 0.5 millimoles

Moles of EDTA = 0.01 X 40 = 0.4 millimoles

so moles of Mn+2 remained unreacted = 0.5-0.4 = 0.1 millimoles

Concentration of Mn+2 = Millimoles / total volume of solution in mL = 0.1 / 25 + 40 = 0.0015

So pMn+2 = -log 0.0015 = 2.82

d) at 49 mL

moles of Mn+2 added = 25 X 0.02 = 0.5 millimoles

Moles of EDTA = 0.01 X 49 = 0.49 millimoles

so moles of Mn+2 remained unreacted = 0.5-0.49 = 0.01 millimoles

Concentration of Mn+2 = Millimoles / total volume of solution in mL = 0.01 / 25 + 49 = 1.35 X 10^-4

So pMn+2 = -log 1.35 X 10^-4 = 3.86

d) at 49.9 mL

moles of Mn+2 added = 25 X 0.02 = 0.5 millimoles

Moles of EDTA = 0.01 X 49.9 = 0.499 millimoles

so moles of Mn+2 remained unreacted = 0.5-0.49.9 = 0.001 millimoles

Concentration of Mn+2 = Millimoles / total volume of solution in mL = 0.001 / 25 + 49.9 = 1.34 X 10^-5

So pMn+2 = -log 1.34 X 10^-5 =4.87

d) at 50mL

Already calculated , this is the equivalence point so moles of MnSO4= moles of EDTA

So all ions are in the form of metal-EDTA complex

[MnY^-2] = moles / volume = 0.5 millimoles / 25 + 50 = 0.0067

Now here we need the formation constant of Mn+2-EDTA complex

Mn+2 + Y-4 --> MnY-2

Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4]

Let us apply ICE

                  Mn⁺² + Y^-4 --> MnY^-2

Initial          0           0          0.0067

Change    x x    -x

Equilibrium x    x    0.0067-x

Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4] = 0.0067 -x / x^2

We may ignore x in numerator as the Kf is very high

so

4.2 X 10^11 = 0.0067 / x^2

x^2 = 0.0067 / 4.2 X 10^11 = 0.0016 X 10^-11

x = 0.126 X 10^-6

pMn+2 = -logx = 6. 9

e) at 50.1 mL

The point just after equivalence point

Here the moles of EDTA will exceed the moles of Mn+2 ions

Excess of EDTA = Moles of EDTA added - Moles of Mn+2 present

Excess of EDTA = 50.1 X 0.01 - 25 X 0.02 = 0.001 millimoles

concentration of excess EDTA = 0.001 millimoles / 75.1 = 1.33 X 10^-5

Concentration of metal ion complex = 0.5 millimoles / 75.1 mL = 0.0066 Molar

Again we will apply the ICE table here

                  Mn⁺² + Y^-4 -->                  MnY^-2

Initial          0          1.33 X 10^-5           0.0066

Change    x x    -x

Equilibrium x    1.33 X 10^-5 +x    0.0066-x

Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4]

4.2 X 10^11 = 0.0066-x / x (1.33X 10^-5+x)

we may ignore x

4.2 X 10^11 = 0.0066 / x (1.33X 10^-5)

x = 0.0066 / (1.33X 10^-5)(4.2 X 10^11)

x = 0.00118 X 10^-6

pMn+2 = 8.92

f) at 55mL

We can calcualte similarly

Here the moles of EDTA will exceed the moles of Mn+2 ions

Excess of EDTA = Moles of EDTA added - Moles of Mn+2 present

Excess of EDTA = 55 X 0.01 - 25 X 0.02 = 0.05 millimoles

concentration of excess EDTA = 0.05 millimoles / 80 = 0.000625

Concentration of metal ion complex = 0.5 millimoles / 80 mL = 0.00625 Molar

Again we will apply the ICE table here

                  Mn⁺² + Y^-4 -->                  MnY^-2

Initial          0          0.000625       0.00625

Change    x x    -x

Equilibrium x 0.000625 +x    0.00625-x

Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4]

4.2 X 10^11 = 0.00625-x / x (0.000625+x)

we may ignore x

4.2 X 10^11 = 0.00625 / x (0.000625)

x = 10 / 4.2 X 10^11

pMn+2 = 10.62

g) at 66mL

Here the moles of EDTA will exceed the moles of Mn+2 ions

Excess of EDTA = Moles of EDTA added - Moles of Mn+2 present

Excess of EDTA = 66 X 0.01 - 25 X 0.02 = 0.16 millimoles

concentration of excess EDTA = 0.16 millimoles / 91 = 0.0017

Concentration of metal ion complex = 0.5 millimoles / 91 mL = 0.0055 Molar

Again we will apply the ICE table here

                  Mn⁺² + Y^-4 -->                  MnY^-2

Initial          0          0.0017       0.0055

Change    x x    -x

Equilibrium x 0.0017+x    0.0055-x

Kf = 4.2 X 10^11 = [MnY^-2] / [Mn+2] [ Y^-4]

4.2 X 10^11 = 0.0055-x / x (0.0017+x)

we may ignore x

4.2 X 10^11 = 0.0055 / x (0.0017)

x = 0.77 X 10^-11

pMn+2 = 11.11

Now we can plota a graph between the two

Volume of EDTA	pMn+2
0	1.7
20	2.17
40	2.82
49	3.86
49.9	4.87
50	6.9
50.1	8.92
55	10.62
66	11.11