Page 345 - Modern Analytical Chemistry

1400-CH09 9/9/99 2:13 PM Page 328

328 Modern Analytical Chemistry

Chloride is determined by titrating with Hg(NO 3 ) 2 , forming soluble HgCl 2 .
The sample is acidified to within the pH range of 2.3–3.8 where diphenylcarbazone,
2+
which forms a colored complex with excess Hg , serves as the visual indicator. Xy-
lene cyanol FF is added as a pH indicator to ensure that the pH is within the desired
range. The initial solution is a greenish blue, and the titration is carried out to a
purple end point.

Quantitative Calculations The stoichiometry of complexation reactions is given by
the conservation of electron pairs between the ligand, which is an electron-pair
donor, and the metal, which is an electron-pair acceptor (see Section 2C); thus

moles of electron pairs donated
´ moles ligand =
mole ligand
moles of electron pairs accepted
´ moles metal
mole metal
This is simplified for titrations involving EDTA where the stoichiometry is always
1:1 regardless of how many electron pairs are involved in the formation of the
metal–ligand complex.

9
EXAMPLE .8
The concentration of a solution of EDTA was determined by standardizing
against a solution of Ca 2+ prepared from the primary standard CaCO 3 . A
0.4071-g sample of CaCO 3 was transferred to a 500-mL volumetric flask,
dissolved using a minimum of 6 M HCl, and diluted to volume. A 50.00-mL
portion of this solution was transferred into a 250-mL Erlenmeyer flask and the
pH adjusted by adding 5 mL of a pH 10 NH 3 –NH 4 Cl buffer containing a small
2+
amount of Mg –EDTA. After adding calmagite as a visual indicator, the
solution was titrated with the EDTA, requiring 42.63 mL to reach the end
point. Report the molar concentration of the titrant.

SOLUTION
Conservation of electron pairs for the titration reaction requires that
Moles EDTA = moles Ca 2+
2+
Making appropriate substitution for the moles of EDTA and Ca gives
M EDTA ´V EDTA = M Ca ´V Ca
2+
The molarity of the Ca solution is
.
0 4071 g
moles CaCO 3 g CaCO 3
= =
.
V flask FW CaCO 3 ´ V flask 100.09 g/mol ´0 5000 L
.
= 8 135 ´10 -3 M Ca 2+
Substituting known values and solving for M EDTA gives
50
8
Ca
MV Ca (.135 ´10 -3 M)( .00 mL) -3
= = .541 ´10 M EDTA
9
V EDTA 42 .63 mL