Complexometric Titration Reactions
Example 1:
Calculate the weight of Ni2+ in a solution if 15 ml of 0.11 M EDTA was
added to this solution and the excess EDTA required 3 ml of 0.101 M
ZnSO4 .
Solution:
VEDTA = 15 ml
MEDTA = 0.11 M
VZnSO4 = 3 ml
MZnSO4 = 0.101 M
(M x V)EDTA = (M x V)Ni2+ + (M x V)ZnSO4
(M x V)EDTA = (
wt
Awt
0.11 x 15 = (
wt
58.69
x 1000)Ni2+ +(M x V)ZnSO4
x 1000) + 0.101 x 3
wt Ni2+ = 0.096 g
g gg
Example 2:
A standard calcium ion solution was prepared by dissolving 0.4644 g of
calcium carbonate in HCl and diluted to 1 liter. 50 ml of this solution was
titrated with 31.4 ml EDTA. What is the molarity of the EDTA.
Solution:
Ca2+
MCa2+ =
MCa2+ =
0.4644 g in 1 L
wt
Mwt
x
0.4644
100
1000
Vml
x
1000
1000
MCa2+ = 4.644 x 10−3 M
(M x V)EDTA = (M x V)Ca2+
(?? x 31.4)
= 4.644 x 10−3 x 50
MEDTA = 7.39 x 10−3 M
Example 3:
A 24 hour urine specimen was diluted to exactly 2 liters. The Ca2+ and
Mg 2+ in a 100 ml portion of the diluted solution required 20.81 ml of 0.0083
M EDTA at pH=10. A second 100 ml portion was treated with oxalate to
precipitate Ca2+ as CaC2 O4 and the precipitate was filtered. The filtrate was
titrated with 5.98 ml of the EDTA solution. Calculate the milligrams of Ca2+
and Mg 2+ in the urine sample.
Solution:
1st Step
(M x V)EDTA = (
wt
Awt
(0.0083 x 5.98) = (
x 1000)Mg 2+
wt
24.3
x 1000)Mg 2+
wt Mg 2+ = 1.21 x 10−3 g
wt Mg 2+
100 ml
x
2000 ml
x = 0.0242 g = 24.2 mg
2nd Step
(M x V)EDTA = (
wt
Awt
x 1000)Ca2+ + (M x V)Mg 2+
(0.0083 x 20.81) = (
wt
40.1
x 1000)Ca2+ + (
wt Ca2+ = 3.44 x 10−4 g
wt Ca2+
100 ml
x
2000 ml
x = 6.98 x 10−3 g = 6.98 mg
1.21 x 10−3
24.3
x 1000)Mg 2+
Example 4:
1.05 g sample of Pd/Cd alloy was dissolved in an acid and diluted to
exactly 250 ml in a volumetric flask. A 50 ml aliquot of the diluted solution
was brought to a PH= 10 with ammonia buffer, the titration involved both
cations and required 28.89 ml of 0.0695 M EDTA. A second 50 ml aliquot
was brought to a PH=10 HCN-NaCN buffer which served to mask Cd2+ . the
titration is needed for Pb2+ 11.5 ml of EDTA. Calculate % Pb and Cd in the
sample (At.wt of Pb= 207.2, Cd= 112.4).
Solution:
Pd/Cd sample (1.05 g)
.
diluted to 250 ml
50 ml = EDTA
(28.89 ml , 0.0695 M)
50 ml (Pb2+ + masked Cd2+ = EDTA
(11.5 ml , 0.0695 M)
1st Step
wt
(M x V)EDTA = (
Awt
(0.0695 x 11.5) = (
x 1000)Pb2+
wt
207.2
𝑥 1000) Pb2+
wt of Pb2+= 0.165 g
50 ml
X
250 ml
wt of Pb2+=0.829 g
% Pb2+=0.829/1.059 =78.2%
2nd Step
(M x V)EDTA
= (
wt
Awt
x 1000)Pb2+ + (
( 0.0695 x 28.89 )EDTA = (
Wt of Cd2+ =0.136 g
X
Wt of Cd2+ = 0.679 g
wt
Awt
0.165
207.2
x 1000)Cd2+
x 1000)Pb2+ + (
wt
112.4
x 1000)Cd2+
50 ml
250 ml
% Cd2+=0.679/1.059 =64.1%