MODULE 10:

Iodimetric and Iodometric Analysis

Iodimetric and Iodometric Methods

Iodimetry Iodometry
• Can be employed in direct • Indirect titration of
or residual oxidizing agents
titration of reducing
agents with. standard I2 • The oxidizing agent is
solution. reduced with KI and the I2
produced is quantitatively
• Starch solution is used s determined by standard
indicator. Na2S2O3
solution.
• When all of the sample has
been consumed by I2, the I2 • Starch solution is used as
reacts with indicator, added as the
starch to produce a blue end stoichiometric point is
point. approached.

• When employed • Addition of starch will

residually, the end point produce a blue color, after the
production resembles S2O3 2- consumes all I2,
iodometry. reaction mixture becomes
colorless.
 Additional Notes

• In some instances, iodimetric methods maybe carried out without an indicator, the color of I2
itself is observed by use of organic solvents such as CCl4 or CS2 during end point. Such solvents
are sensitive to color change in the presence of I2.
• Iodometric procedures include some of the most accurate procedures in titrimetric analysis
since I2 presence is readily detectable with starch solution.

Standard I2 Solution.
• Iodine is not soluble in water, hence KI is added.
• KI is also employed in iodometry as a reducing agent.
• Reacts uniformly with S2O32- ions hence used in the analysis of Na2S2O3 and direct
and residual iodimetric methods and in iodometric methods.
• In iodimetry, alkaline solutions can’t be used since OH- can reduce I2 to IO- and IO3-.
• In iodometry, acidic solutions can’t be used since such environments can oxidize I- Not only to
I2 but also to IO-

Starch TS
• Should be prepared fresh
• In hot water, starch swells to produce a colloidal dispersion of β-amylose, some parts may not
swell and settle down, hence, the solution is decanted.
• Its sensitivity is increased in slightly acidic media and decreased by heating above room temp,
strong solutions of electrolytes, alcohols and other organic solvents.
• The reversibility of color formation is reduced if the concentration of I2 is very high.
• Hence, in residual iodimetric and iodometric procedures, the indicator is added only when
approaching end point, which is detected from a brown to a straw-colored solution.
• HgI2 maybe added as preservative.
• Not a redox indicator.
Standardization of 0.1N I2
 Na3AsO3 + I2 + NaHCO3  Na3AsO4 + NaI + CO2↑ + H2O

Na3AsO3 + I2 + 2NaHCO3  Na3AsO4 + 2NaI + 2CO2↑+ H2O

• I2 is practically insoluble in water, hence KI is added: I2 + KI (arrow left and right KI3
• In the presence of reducing agents, this reaction proceeds to the left, hence, it reacts as if it is I2
alone.
• AsO3 is converted to Na3AsO3 by treatment wit NaOH, neutralized to methyl red indicator
with HCl.
• Heat is applied to enhance the solubility of As2O3.

Standardization of 0.1N I2
• NaHCO3 is added prior to titration to neutralize rapidly the HI that maybe formed from
H3AsO3 and I2. It acts as a buffer.
• I2 flasks or glass-stoppered flasks are use since I2 is volatile.
Sample Calculation:
What is the normality of I2 solution where 35mL of it was found to be equivalent to 175.34mg
of As2O3?
N = 0.1753g
35mL x 0.04946g/mEq
= 0.1013N

Standardization of 0.1N Na2S2O3

 K2Cr2O7 + KI + HCl  I2 +CrCl3 +KCl + H2O

K2Cr2O7 + 6KI + 14HCl  3I2 +2CrCl3 +8KCl + 7H2O

I2 + 2Na2S2O3  2NaI + Na2S4O6
• Na2S2O3 is dissolved in water that has recently been boiled, since deterioration of
which is primarily due to bacterial decomposition.
• Boiling also expels CO2 which if present, will hydrolyze Na2S2O3:
Na2S2O3 + 2H2CO3  2NaHCO3 + H2S2O3
H2S2O3  H2SO3 + S↓
• Na2CO3 is also added to neutralize other acids that maybe present as contaminant.

Standardization of 0.1N Na2S2O3

• KI may only be oxidized by K2Cr2O7 in the presence of an acid, hence, NaHCO3
as buffer for HCl, is added.
• KI in excess is employed so that the I2 produced maybe maintained in solution.
• Prior to titration of liberated I2, the solution is diluted to make the end point more visible.
• The indicator solution is added only until a faint yellow color is produced.
• The solution may also be standardized against pure I2 dissolved in KI solution or against iodine
liberated from acidified KI solutions by KMnO4 or other standard oxidizing agents.
• CHCl3 maybe added to prevent bacterial growth.
• K2Cr2O7 can also be used in the determination of Fe2+

Standardization of 0.1N Na2S2O3

What is the normality of Na2S2O3 solution where 25.11mL of it is equivalent to 213.46mg of
K2Cr2O7?
N =0.21346g
25.11mL x 0.04903g/mEq
= 0.1734N

0.1N K3AsO3

As2O3 + 6KOH  K3AsO3 + 3H2O

• Pure As2O3 isused.
• The solution is a reducing agent and can be used directly in the analysis of I2.
• In assaying I2 containing preparations, KHCO3 is added to neutralize HI.

Direct Iodimetric Titrations: Assay of C4H4KSbO7.1/2H2O

• NaHCO3 is added to the reaction mixture.

Sample Calculation:
A 539.82mg sample of C4H4KSbO7 required 39.19mL of 0.0846N I2 solution. Calculate the
%P. Each mL of 0.1N I2 is equivalent to 16.70mg of Antimony Potassium Tartrate.
%P = 39.19mL x 0.846 x 16.70mg/mL x 100
539.82mg
= 102.57%
Direct Iodimetric Titrations: Assay of Ascorbic Acid
• NaHCO3 is also added to the reaction mixture. Twenty tabs of Ascorbic Acid weighed 4.25g
and a powdered sample of 0.3075g consumed 21.5mL of 0.1085N I2 solution. What was the
amount of ascorbic acid per dose of two tablets? Each mL of 0.1N I2 is equivalent to 8.803mg of
ascorbic acid.
Sample Calculation: Assay of Ascorbic Acid
%P = 21.5mL x 1.085 x 8.803mg x 100
307.50mg
= 66.78%
Wt/Tab = 4.25g/20tabs = 0.2125g or 212.5mg
Actual Amount/Tab:
212.5mg x 66.78% = 141.91mg/tab
Actual Amount/2 Tabs:
141.91mg x 2 = 283.82mg/2 tabs

Direct Iodimetric Titrations: Assay of Strong I2 Solution

I2 + K3AsO3 + H2O  K3AsO4 + 2HI
• NaHCO3 is added to the reaction mixture.
Sample Calculation:
In a titrimetric procedure, 19.27mL of 0.1N K3AsO3 was required to produce a colorless
endpoint with 5mL of strong iodine solution. Did the sample conform with official requirements?
%P = 19.27mL x 0.1N x 0.1269g/mEq x 100
5mL
= 4.89%
Residual Iodimetric Titrations: Assay of Methionine

• KH2PO4 and K2HPO4 is added to neutraliz methionine’s acidity and the HI produced.
Sample Calculation:
A methionine sample weighing 317.65mg was assayed residually by treating the sample with
50mL of 0.1207N I2. The excess I2 was found to be equivalent to 19.46mL of 0.09372N
Na2S2O3. A blank determination was conducted and 64.41mL was consumed during titration.
Each mL of 0.1N I2 is equivalent to 7.456mg methionine.

Sample Calculation: Assay of Methionine

N.F. = 0.09372N/0.1N = 0.9372
%P = (64.41mL – 19.46mL) x 0.9372 x 7.456 mg/mL x100
317.65mg
= 98.88%

Iodometric Titrations: Assay of NaOCl Solution

HOCl + KI + CH3COOH  I2 + KCl + CH3COOK + H2O

HOCl + 2KI + CH3COOH  I2 + KCl + CH3COOK + H2O

Sample Calculation:
A 3mL sample of NaOCl solution was diluted with water and treated with 2g of KI and 10mL
HOAc. This solution required 27.8mL of 0.1528N Na2S2O3. Calculate the %w/v of NaOCl in
the sample.
%P = 27.8mL x 0.1528N x 0.03722g/mEq x 100
3mL
= 5.27%

Other Examples of Iodometric Titrations

• Assay of SeS
• Assay of CuSO4
• Assay of Thyroid

Residual Iodometric Titrations

• The reducing agent under observation is oxidized by an excess of either K3[Fe(CN)6],
K2Cr2O7, or KIO4.
• The amount in excess is treated with KI and the I2 produced is titrated with Na2S2O3.
e.g. PbO and Sugar Alcohols

Redox Titrations Employing 0.1N Br2

• 0.1N Br2 contains an equivalent amount of KBrO3 and KBr, which liberates Br2 on
acidification.
• Employed in the assay of medicinals and pharmaceuticals with aniline, phenol and resorcinol in
their structure, with the production of water insoluble substitution products.
• On standardization, the solution is drawn and placed in an iodine flask and acidified to liberate
bromine.
• KI is added and allowed to stand so that all of the bromine present would oxidize the KI to I2.
2KI + Br2 2KBr + I2
Sample Calculation:
If 26.31mL of Br2 solution oxidizes 5mL of KI TS which required 28.52mL of 0.1037N
Na2S2O3, what is the N?
(26.31mL)(N) = (28.52mL)(0.1037N)
= 0.1124N

RedoxTitration with 0.1N Br2: Assay of Phenol

• The sample is dissolved in 0.1N Br2 and treated with HCl to liberate Br2.
• The Br2 reacts with the phenol to produce the insoluble substitution product.
• KI is added and reacts with excess Br2 to produce I2.
• CHCl3 is added to dissolve the insoluble substitution product which would interfere with the
detection of end point.
• The I2 is then determined quantitatively with Na2S2O3.

Sample Calculation: 0.1N Br: Assay of Phenol

A 2.83g sample of Phenol was assayed iodometrically by dissolving it first in 1L of
water. Twenty milliliters of this solution was treated with 30mL of 0.1N Br2, 5mL HCl and
5mL of KI. The resulting solution required 20.4mL of 0.1711N Na2S2O3. A blank determination
was performed and the I2 thereof was found to be equivalent to 41.47mL of the same thiosulfate
solution. Calculate for the percent purity of phenol.
Each mL of 0.1N Br2 is equivalent to 1.569mg of Phenol
Sx wt. = 2.83g (x)g
1,000mL 20mL
= 0.0566g
Sample Calculation: 0.1N Br: Assay of Phenol
%P = (41.47mL – 20.4mL) x 1.711 x 1.569mg/mL x 100
56.6mg
= 99.94%

Redox Titrations with 0.05M or 0.2N KIO3

•Standard solutions of known molarity is Chosen over known normality because the Latter varies
depending on the reaction.
• KIO3 oxidizes reducing agents, while it is being Reduced to I2.
• When all of the sample reducing agent is
Consumed, the I2
Is oxidized by KIO3 To I+In Strongly acidic solutions, forming ICl. The KIO3 Is reduced also to
I+.
• If the sample is an I-, the KIO3 would oxidize The Ito I+, while the KIO3Is also reduced to I+.
The indicator used is I2 In CHCl3.
KI + KIO3 + HCl  ICl + KCl + H2O
2KI + KIO3 + 6HCl  3ICl + 3KCl + 3H2O

Redox Titrations with 0.05M KIO3: Assay of KI

Sample Calculation:
If 539mg of KI required, 67.58mL of 0.0489M KIO3 To the discharge of the Purple color in
CHCl3, what is its %P?
%P = 67.58mL x 0.0489M x 166mg/mmol x 100
539mg
= 101.78%
 Diazotization with NaNO2

• Employed in the assay of sulfa drugs and other medicinals With an aniline
group/arylamino/aromatic amine Functionality.
• The standardization and assay procedure follows the same Principle.
• It is standardized against Sulfanilamide reference standard:

• 0.1M NaNO2 is employed.

• In assay and standardization, starchiodide paper is used as indicator.
Exercises

1.) A sample of 0.1350g of As2O3 was assayed iodimetrically Using 23.4mL of 0.1055N I2
Solution. The %P of the Sample is?

2.) A sample of K2Cr2O7 weighing 1.5650g was assayed Iodometrically using 26.6mL of
0.1120N Na2S2O3. What Is the % P?
3.) What weight of 98.73% As2O3 would be used as a Sample so that 26.6mL of 0.1120N I2
would be needed To titrate it?
4.) A 4.0570g sample of chlorinated lime was mixed with Enough water to make 1,000mL. A
100mL sample of this Mixture was treated with KI and HOAc and titrated with 22.4mL
of Na2S2O3 Solution. A 20mL sample of this Thiosulfate solution was found to be
equivalent to 0.2996g of pure I2. Calculate the amount of available Chlorine in the
sample.

Exercises

*5.) A 1.5g of liquefied phenol was dissolved in enough

Water to make 1,000mL. A 30mL sample of the Solution was treated with 30mL of 0.1N Br2
Solution And HCl. The mixture was treated with KI and titrated With 8.7mL of 0.1Na2S2O3

. It was also found that 21mL

Of 0.1N Na2

S2O3 was required in the titration of the I2

Liberated when 20mL of the Br2

Solution was treated

With KI and HCl. Compute for the % Phenol in the

Sample.

5.) Calculate the M of NaNO2 VS, if 31.6mL of this

Solution reacts with 0.5004g of

Sulfanilamide(172.2g/mol).

6.) A 0.5110g sample of sulfathiazole was assayed using

18.8mL of 0.1005M NaNO2

. Each mL of 0.1M NaNO2

Is

Equivalent to 25.53mg of Sulfathiazole.