EDTA Titration Example [PDF]

EDTA Titration Example. Titration of 50.00 mL of 0.00500M Ca. 2+. (Analyte) with 0.0100M EDTA (Titrant). Titration React

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CHEM 311 Quantitative Analysis Spring 2013

EDTA Titration Example Titration of 50.00 mL of 0.00500M Ca2+ (Analyte) with 0.0100M EDTA (Titrant). Titration Reaction EDTA(aq) + Ca2+(aq)

CaEDTA2+(aq)

Equilibrium Reaction Ca2+(aq) + EDTA(aq)

CaEDTA2+(aq)

Kf = 1.75 x 1010 Note: For notational convenience I have represented H2EDTA2+ as simply EDTA.

0.00 mL Titrant Added Titration Reaction

end

EDTA

Ca2+

CaEDTA2+

=0M

= 0.00500 M

=0M

Equilibrium Reaction

Initial

EDTA

Ca2+

CaEDTA2+

0

0.0500

0

[Ca2+] = 0.00500 M

Concentration pCa = - log(0.00500) = 2.30

25.00 mL Titrant Added (Equivalence Point) Titration Reaction

start end

EDTA

Ca2+

CaEDTA2+

(0.0100M)(0.02500L) = 0.000250 mol

(0.0050M)(0.05000L) = 0.000250 mol

= 0 mol

0 mol

0 mol

0.000250 mol

=0M

=0M

(0.000250mol)/(0.07500L) = 0.00333M

Equilibrium Reaction EDTA

Ca2+

CaEDAT2+

Initial

0

0

0.00333

Change

+x

+x

-x

Equilibrium

x

x

~0.00333

1.75 x 1010 = (0.00333) / x2 x = [Ca+] = 4.36 x 10-7 M

Concentration pCa = - log(4.36 x 10-7) = 6.36

50.00 mL Titrant Added Titration Reaction

start end

EDTA

Ca2+

CaEDTA2+

(0.0100M)(0.05000L) = 0.000500 mol

(0.0050M)(0.05000L) = 0.000250 mol

= 0 mol

0.000250 mol

0 mol

0.000250 mol

(0.000250mol)/(0.1000L) = 0.00250 M =0M

(0.000250mol)/(0.1000L) = 0.00250M

Equilibrium Reaction EDTA

Ca2+

CaEDAT2+

Initial

0.00250

0

0.00250

Change

+x

+x

-x

Equilibrium

~0.00250

x

~0.00250

1.75 x 1010 = (0.00250) / (0.00250) x x = [Ca+] = 5.71 x 10-11 M

Concentration pCa = - log(5.71 x 10-11) = 10.24

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