Pharmaceutical Analytical Chemistry
PHCM223
Lecture 12
Applications on different types of equilibria
Dr. Nesrine El Gohary
12th lecture
Revision lecture next week
Next Saturday (21-5-2016) in the 1st slot at H13.
Learning Outcomes
• Apply complexometric titrations to different samples.
• Apply redox titrations to different samples.
• Apply gravimetric analysis to different samples.
• Apply precipitation titrations to different samples.
Some important Relations
Number of moles 
Mass
Molec.wt
Number of moles  M V
No. of moles
Molarity 
Volume in Liter
For a given mass of a substance
For a solution of given conc. and
volume
Concentration of solution
Some important Relations Cont.
Wt (g)= M x V x Mwt
1000
Percent purity = Mass of pure compound
Total weight of the sample
Strength (g/l) = M x Molar mass
In EDTA Titrations
EDTA reacts 1:1 with all metal ions whatever their valencies
At equivalent point
(M V)EDTA = (M V)metal ion
x 100
Example 1
Chromel is an alloy composed of nickel, iron and chromium. A 0.6472 g sample was dissolved and
diluted to 250 ml. When a 50 ml aliquot of 0.05182 M EDTA was mixed with an equal volume of
the diluted sample and all the three ions were chelated and a 5.11 ml back titration with 0.06241
M copper (II) was required. The chromium in a second 50 ml aliquot was masked through the
addition of hexamethylenetetramine, titration of the Fe and Ni required 36.28 ml of 0.05182M
EDTA. Iron and chromium were masked with pyrophosphate in a third 50 ml aliquot and the nickel
was titrated with 25.91 ml of the EDTA solution. Calculate the percentage of nickel,
chromium and iron in the alloy.
5.11 ml
Cu2+
36.28 ml
0.06241 M
250 ml
0.6472 g
Ni, Fe, Cr
EDTA
0.05182 M
MVEDTA = MVNi
0.05182 x 36.28= M50Ni + M50Fe
MVCu = MVunreacted EDTA
0.06241 x 5.11 = 0.05182 x V
Vunreacted EDTA = 6.15ml
V of EDTA reacted with Ni, Fe and Cr = 50 – 6.15 = 43.85 ml
(MV)Ni + (MV)Fe + (MV)Cr= (MV)EDTA
(M50)Ni + (M50)Fe + (M50)Cr = (0.05182 x 43.85)EDTA
EDTA
0.05182 M
50 ml
Ni, Fe, Cr
50 ml
50 ml +50 ml of 0.05182 M EDTA
Ni, Fe, Cr
Ni, Fe, Cr
MVEDTA = MVNi+ MVFe
[Ni-EDTA] + unreacted EDTA
[Fe-EDTA]
[Cr-EDTA]
25.91 ml
0.05182 x 25.91 = M x 50
MNi=0.0268 M
MFe= 0.0108 M
MCr= 7.8 x 10-3 M
Example 1 Cont.
 After calculating the molarities, determine the weight of each metal in the sample.
Wt of Ni = (0.0268×250×58.693)/1000 = 0.393 g
Wt of Fe = (0.0108×250×55.845)/1000 = 0.151 g
Wt of Cr = (7.8×10-3×250×51.996)/1000 = 0.101 g
 Now calculate the percentage of each metal in the sample.
% Ni = (0.393/0.6472)x 100 = 60.72 %
% Fe = (0.151/0.6472)x 100 = 23.33 %
% Cr = (0.101/0.6472)x 100 = 15.61%
Example 2
Calcium in powdered milk is determined by taking 1.5 g sample and then titrating
calcium with EDTA solution, 12.1 mL being required.
The EDTA was standardized by titrating 10 mL of a Zn2+ solution prepared by
dissolving 0.632 g Zn metal in acid and diluting to 1.0 liter. 10.8 mL of EDTA is
required for the titration. What is the concentration of Ca in the powdered milk in
mg/kg. (Atomic masses: Ca = 40.08, Zn = 65.41)
12.1 ml
EDTA
10.8 ml
M???
(MV)EDTA= (MV) Zn
(M X 10.8)EDTA= (9.68 x 10-3 X 10) Zn
MEDTA= 8.96 x 10-3 M
Zn2+
10 ml
EDTA
0.632 g/l
Ca2+
1.5 g sample
MVEDTA = (Wt/at.wt) Ca
Strength Zn (g/l) = M x Molar mass
(8.96 x 10-3 x 12.1) EDTA = (Wt/40.078) Ca
0.632 = M x 65.3
1.5 g sample
(Wt) Ca = 4.34 mg
2+
-3
M of Zn = 9.68 x 10 M
1000 g sample
(Conc.) Ca = 2893.3 mg / Kg
Redox Titrations
Calculating the potential in a redox titration
Before the
endpoint
Apply
Nernst
equation
At the
endpoint
Apply the
following
equation
After the
endpoint
Apply
Nernst
equation
To the
sample half
reaction to
calculate the
potential
n 1 E 1  n 2 E 2
E
n1  n 2
To the titrant
half reaction
to calculate
the potential
Example 3
A 0.2 g sample containing copper is analyzed iodometrically. Copper (II) is
reduced to copper(I) by iodide:
2Cu2+ + 4I2CuI + I2
What is the percent of copper in the sample if 20 ml of 0.10 M Na2S2O3 is
required for the titration of the liberated iodine?
20 ml
Na2S2O3
0.1 M
2Cu2+ + 4II2 + 2S2O32-
Cu2+
2I- + S4O62-
+ Excess of I-
0.2 g sample
% Cu???
2CuI + I2
1 mmol Cu2+ ≡ 1 mmol S2O32(Wt/ at.wt)Cu2+ = (MV) S2O32-
I2
(Wt/ 63.546)Cu2+ = (0.1 x 20) S2O32Wt of Cu = 127.092 mg = 0.127 g
% of Cu = 63.5 %
Example 4
A 25 ml aliquot of a sample containing Tl + ion was treated with K2CrO4. The
Tl2CrO4 was filtered, washed free of excess precipitating agent, and dissolved
in dilute H2SO4.. The Cr2O72- produced was titrated with 40.6 ml of 0.1004 M
Fe2+ solution. What was the mass of Tl in the sample? The reactions are:
2Tl+ + CrO422Tl2CrO4 (s) + 2H+
Tl2CrO4 (s)
4Tl+ + Cr2O72- + H2O
Cr2O72- + 6Fe2+ + 14 H+
6Fe3+ + 2Cr3+ + 7H2O
2 mmol of Tl+ ≡ 3 mmol of Fe2+
?
(40.6 x 0.1004)
mmol of Fe2+
mmol of Tl+ = 2.717 = Wt
At.wt
Wt= 555.4 mg = 0.555 g
Gravimetric Analysis
 Gravimetric calculations relate moles of the product finally weighed to moles of
analyte.
Precipitating agent (B)
Analyte (A)
Filtered
Washed
Dried
Weighed (P)
20 mls
nA + B
n Mwt
•Where W2 is the weight of
the analyte ion only
dissolved in 20 ml of
solution and W1 is the
weight of precipitate (ppt).
W2
mP + S
m Mwt
Convert moles into weights by
multiplying by the molecular
weight.
W1
weight of the analyte W2  weight of the precipitate W1 x
(nMwtanalyte/mMwtppt) is called the gravimetric factor.
n MWanalyte
m MWppt
Example 5
A sample containing only Al2O3 (FM 101.8) and Na2Cr2O7.2H2O (FM 298) was
brought into the laboratory for analysis. The analyst heated a 2.2846 g sample of
this material to 100oC for one hour, whereby only Al2O3 and Na2Cr2O7 remained.
This mixture was found to weigh 2.2312 g. Calculate the percent chromium in the
sample. (atomic mass of Cr = 52)
The loss in mass is due to loss of water present in Na2Cr2O7.2H2O
This is equal to
Loss in mass = 2.2846 – 2.2312 = 0.0534 g
1 mole of Na2Cr2O7.2H2O contains 36 g of H2O
0.0534 g of H2O
??
mass of H2O that
was present in the
mixture.
Moles of Na2Cr2O7.2H2O = (0.0534 x 1) / 36 = 1.48 x 10-3
Wt of Cr = 0.154 g
-3
-3
=
Wt
Moles of Cr = 1.48 x 10 x 2 = 2.96 x 10
% of Cr = 6.75 %
At.wt
Example 6
A 0.6407 g sample containing chloride and iodide ions gave a silver halide
precipitate weighing 0.443 g. The precipitate was then strongly heated in a stream
of Cl2 gas to convert the AgI to AgCl, on completion of this treatment, the
precipitate weighed 0.3181 g. Calculate the percentage of chloride and iodide in
the sample.
The decrease in mass of precipitate is because chloride is lighter than iodide.
Molar mass of Cl = 35.45 g/mol
Molar mass of l = 126.9 g/mol
If 1 mole of I is converted into Cl, the loss in mass would be = 91.45 g
In this problem, the loss in mass is = 0.443-0.3181 = 0.1249 g
So by a cross multiplication, the no. of moles of I present is = 1.36 x 10-3
Wt of I = 0.1733 g
% of I = 27.05 %
Now we can calculate wt of Agl present = 0.319 g
So wt of AgCl = 0.443 – 0.319 = 0.124 g
So can calculate wt of Cl = (0.124 x 35.45)/143.32= 0.0306 g
% of Cl = 4.78 %
Example 7
Precipitation Titrations
A 2.3527 g of an unknown sample containing chloride is dissolved and diluted to
volume in a 250 mL volumetric flask. A 50.00 mL aliquot is placed into a conical flask,
50 mL of deionized water is added, and 25.55 mL of 0.0333 M AgNO3 is required to
fluorescein indicator endpoint. Calculate percentage chloride in the original sample.
25.55 mL
AgNO3
0.0333 M
50.00 mL aliquot
250 ml
2.3527 g
sample
(MVn) titrant = (MVn) sample
(0.0333 x 25.55 x 1) AgNO3 = (M x 50 x 1) Chloride
M of Chloride = 0.017 M
Wt = (0.017 x 250 x 35.45) / 1000 = 0.15 g
% of Cl = 6.37 %
References
• D. A. Skoog, D.A. West, F.J. Holler, S.R. Crouch, Analytical
Chemistry, an introduction, 7th Edition, ISBN 0-03-020293-0.
• Revision lecture by Prof. Rasha Elnashar, GUC, SS 2015.
• Revision lecture by Dr. Raafat Faraghly, GUC, SS 2007.