Iodometric and Iodimetric titrations - Iodine titrations, their types, Chemical reactions, Practice problems & FAQs

Are you aware?

Ascorbic acid, generally known as vitamin C, is a crucial mineral for the immune system and has earned the epithet "wonder worker." Numerous illnesses, from basic disorders like the common cold to deadly conditions like cancer, have been reported to be healed by it.

Iodine titration was used to evaluate the ascorbic acid concentration of seven different fruits, including grapefruit, lime, banana, watermelon, strawberries and orange, in order to recognize which fruit would best meet the body's ascorbic acid requirements.

Let's go over the details of how these titrations function and what their main principles are!

Table of contents:

• Titrations
• Types of Iodine Titration
• Iodimetric titration
• Iodometric titrations
• Practice problems
• Frequently asked questions-FAQs

Titrations:

In the presence of an indicator, titration is a quantitative and volumetric method for determining the concentration of an unknown solution from the concentration of a known solution. The law of equivalence is applied in this technique. A solution with a known concentration known as the titrant is used to titrate the material whose concentration needs to be determined. The concentration of a chemical in a solution should be determined by progressively adding a certain extra component (typically with a burette) until the reaction is complete, which is shown by the indicator's colour changing.

Type of iodine titration:

Basically there are two types of iodine titrations which can be described as:

1. Iodimetric titration
2. Iodometric titration

Iodimetric titration:

In this titration I₂ is used as an oxidizing agent and it is used to determine the strength or the concentrations of a reducing agent which is directly titrated with I₂.

The single-step reaction is given as follows

$I_2+reducing agent \to 2I^{-}+ product$

Starch is employed in this titration as the indicator, which produces a dark blue color when I₂ is added. Usually, neutral or moderately alkaline to weakly acidic solutions are used for these titrations. I₂ will be disproportionate to hypoiodite and iodide ions if the pH is too alkaline.

$I_2+2{OH}^{-}\to I{O}^{-}+I^{-}+H_{2}O$

Strong acid has a tendency to hydrolyze or break down the starch employed for endpoint detection. In a neutral solution, the reducing power of various reducing agents is increased. I^- created during the reaction has a tendency to oxidize due to oxygen dissolved in the acidic solution.

${4I}^{-}+O_2+4H^{+}\to I_2+{2H}_{2}O$

Iodometric titrations:

To ascertain the potency or concentration of an oxidizing agent, iodometric titration is performed. In this, the presence of an oxidizing agent causes I^- to oxidize into I₂. In this case, starch is employed as an indication that turns dark blue when I₂ is present.

In this method indirect estimation of iodine takes place.

Step 1: An oxidizing agent reacts with an excess solid KI (potassium iodide).

$KI + Oxidizing agent \to I_2$

From the equation,

Equivalents of oxidizing agent = Equivalents of I^- oxidized = Equivalents of I₂ formed

Step 2: The iodine liberated is then titrated with a standard hypo solution (Na₂S₂O₃).

$I_2+{Na}_2{S}_2{O}_3\to NaI+ {Na}_2{S}_4{O_6}_{}$

Equivalents of I₂ = Equivalents of ${S_2{O}_3^{2-}}_{}^{}$

Therefore, from the amount of iodine consumed by the thiosulphate ions, the amount of oxidizing agent can be calculated. The combined schematic representation can be:

$O.A.+KI\to I_2+{Na}_2{S}_2{O}_3\to NaI+{Na}_2{S}_4{O}_6$

Iodimetric titration is a volumetric analysis involving either titration with a standardized solution of iodine or the quantitative examination of a solution of an oxidizing agent by adding an iodide that reacts to create iodine. The following table compares and contrasts Iodometric titrations and Iodimetric titrations.

Practice problems:

Q 1. Find the n-factor of Cr in ${\mathit{C}\mathit{r}}_2{\mathit{O}}_7^{2-}$ for the reaction${Cr}_2{O}_7^{2-}+6I^{-}+{14H}^{+}\to 2{Cr}^{3+}+{3I}_2+7H_{2}O$ ?

1. 6
2. 2
3. 3
4. 4

Answer: (A)

Solution:

Let O.S of Cr in ${\mathit{C}\mathit{r}}_2{\mathit{O}}_7^{2-}$ = x

$$\begin{gathered} \Rightarrow 2x-14=-2 \\ \Rightarrow x=+6 \end{gathered}$$

Let O.S of Cr in ${\mathit{C}\mathit{r}}^{3+}$ = y

$\Rightarrow y=+3$

n-factor of Cr = 2 |6 - 3| = 6

Q 2. Find n-factor of S in ${\mathit{N}\mathit{a}}_2{\mathit{S}}_4{\mathit{O}}_6$, in reaction ${\mathit{I}}_2+{\mathit{N}\mathit{a}}_2{\mathit{S}}_2{\mathit{O}}_3\to \mathit{N}\mathit{a}\mathit{I}+\mathit{ }{\mathit{N}\mathit{a}}_2{\mathit{S}}_4{{\mathit{O}}_6}_{}$

1. 1
2. 2
3. 3
4. 4

Answer: (B)

Solution:

Let O.S of S in ${Na}_2{S}_2{O}_3$ = x

$$\begin{gathered} \Rightarrow 2+2x-6=0 \\ \Rightarrow x=+2 \end{gathered}$$

Let O.S of S in ${Na}_2{S}_4{O}_6$ = y

$$\begin{gathered} \Rightarrow 2+4y-12=0 \\ \Rightarrow y=+\frac{5}{2}=+2.5 \end{gathered}$$

n-factor of S= 4 |2.5 - 2| = 2

Q 3. For reaction, ${\mathit{I}}_2+2{\mathit{N}\mathit{a}}_2{\mathit{S}}_2{\mathit{O}}_3\to 2\mathit{N}\mathit{a}\mathit{I}+\mathit{ }{\mathit{N}\mathit{a}}_2{\mathit{S}}_4{{\mathit{O}}_6}_{}$ the endpoint is shown by

1. Appearance of blue colour
2. The disappearance of blue colour
3. Disappearance of red colour
4. Appearance of red colour

Answer: (B)

Solution: Endpoint is indicated by the disappearance of the blue colour. Starch is used as an indicator. At the endpoint with full consumption of iodine, the blue colour disappears.

Q 4. 12.5 mL of home bleach solution was treated with 100 mL of 0.20 M KI and 20 mL of 2 N acetic acid. 50 mL of 0.2 N Na₂S₂O₃ were used in the titration of the liberated iodine to achieve the endpoints. The molarity of the bleach is:

1. 0.1 M
2. 0.3 M
3. 0.4 M
4. 0.2 M

Answer: (C)

Solution: $\mathit{H}\mathit{o}\mathit{u}\mathit{s}\mathit{e}\mathit{h}\mathit{o}\mathit{l}\mathit{d}\mathit{ }\mathit{b}\mathit{l}\mathit{e}\mathit{a}\mathit{c}\mathit{h}\mathit{ }+2\mathit{K}\mathit{I}\to {\mathit{I}}_2+\mathit{P}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}$

$I_2+2{Na}_2{S}_2{O}_3\to 2NaI+ {Na}_2{S}_4{O_6}_{}$

n-factor of S in ${Na}_2{S}_2{O}_3$ = 1

Amount of moles of ${Na}_2{S}_2{O}_3$ used = $V\times M$ = $V\times N$ = $50\times 0.2$ = 10 millimoles

Amount of I2 generated = $\frac{Amount of moles of {Na}_2{S}_2{O}_3 used}{2}=\frac{10}{2}=5 millimoles$

Assuming 1 mol of household bleach produces 1 mole of I₂ , we will have

Amount of household bleach in 12.5 mL solution = 5 millimoles

Molarity of household bleach = $\frac{5\times {10}^{-3}moles}{12.5 mL}\times 1000$ = 0.4 M

Frequently asked questions-FAQs:

Q1. Do excessive amounts of indicator impact titrations?
Answer: A too-great indicator addition will modify the concentration of the solution to which the titrant is added, which will have an impact on the titration process.

Assume for the moment that the indicator is acidic in nature. It will increase the acid's acidity beyond what you predicted if you add it to the acid. Your entire experiment will be ruined since it will be inaccurate because more base will be needed to neutralize the acid.

Q2. Describe titrand.
Answer: Any solution to which the titrant is introduced and which includes the ion or species being determined is referred to as the titrand.

Q3. Why must iodometric titrations be completed in a bit of a rush?
Answer: Since an acid media is the ideal setting for air oxidation of the excess iodide ion, the titration of the liberated iodine in these situations must be completed rapidly to prevent unnecessary exposure to the atmosphere.

Q4. What function does titration serve?
Answer: Titration is a technique for determining a solution's concentration by reacting it with a reference solution of known concentration.