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The Exhaustive Guide to Iodimetry & Iodometry
Welcome to the definitive masterclass on Iodine-based Redox Titrations. In the JEE Advanced syllabus, questions based on iodine chemistry are virtually guaranteed. The confusion between "Iodimetry" and "Iodometry", the exact n-factors in comproportionation reactions, and the specific timing of adding the starch indicator form the crux of the toughest numericals. We will dismantle all of these concepts mathematically and theoretically.
1. The Iodine Conundrum: A Mild but Mighty Element
Iodine ($I_2$) is a relatively mild oxidizing agent, positioned moderately in the electrochemical series with a standard reduction potential of $E^\circ_{I_2/I^-} = +0.54 \text{ V}$.
This intermediate position is exactly what makes it analytically brilliant.
- Because it is a mild oxidizing agent, it can quantitatively oxidize strong reducing agents (like Thiosulfate $S_2O_3^{2-}$, Sulfite $SO_3^{2-}$, Arsenite $AsO_3^{3-}$, and $Sn^{2+}$).
- Conversely, the Iodide ion ($I^-$) is a mild reducing agent. It can be quantitatively oxidized to free $I_2$ by strong oxidizing agents (like Permanganate $MnO_4^-$, Dichromate $Cr_2O_7^{2-}$, Cupric ions $Cu^{2+}$, Hydrogen Peroxide $H_2O_2$, and Ozone $O_3$).
The Solubility Issue: Molecular iodine ($I_2$) is highly non-polar and practically insoluble in water. To use it in aqueous titrations, it is dissolved in a concentrated solution of Potassium Iodide ($KI$). The $I_2$ reacts with $I^-$ to form the highly soluble, dark brown Triiodide complex ($I_3^-$).
*Note: For stoichiometric calculations, we treat $I_3^-$ simply as $I_2$ because as $I_2$ is consumed in a reaction, the equilibrium shifts completely to the left, releasing more $I_2$.
2. The Fundamental Divide: Iodimetry vs. Iodometry
These terms sound identical but describe completely distinct analytical procedures. Grasping this difference is the foundational step for JEE problems.
Iodimetry (Direct)
- Method: Direct titration.
- Titrant in Burette: Standard Iodine solution ($I_2$ in $KI$).
- Analyte in Flask: A reducing agent (e.g., Hypo, $H_2S$, Vitamin C).
- What happens: As $I_2$ drops into the flask, it gets reduced to colorless $I^-$. The analyte gets oxidized.
- Endpoint: The first drop of excess $I_2$ remains unreacted, turning the starch indicator deep blue.
Iodometry (Indirect / Back)
- Method: Indirect / Back titration (Two steps).
- Step 1: Unknown Oxidizing Agent + Excess $KI$ $\rightarrow$ Liberates $I_2$.
- Step 2: Liberated $I_2$ is titrated against standard Sodium Thiosulfate (Hypo) from the burette.
- Endpoint: Disappearance of the deep blue starch-iodine complex color.
3. The Magic and Timing of the Starch Indicator
Iodine in an aqueous $KI$ solution is brown. As it gets consumed (reduced to $I^-$), the color fades to a pale yellow. While you could technically use the disappearance of this pale yellow color as an endpoint, it is too faint for the human eye to detect accurately.
We use a freshly prepared, 1% aqueous Starch solution as a specific indicator. Starch (specifically the amylose fraction, which forms a helical structure) traps iodine molecules ($I_5^-$ chains specifically) within its helix, forming an intense, dark blue-black adsorption complex. This color is visible even at iodine concentrations as low as $10^{-5} \text{ M}$.
The Timing Trap: When to add Starch in Iodometry?
In an Iodometric titration (where the flask initially contains a massive amount of liberated brown $I_2$), you MUST NOT add starch at the beginning of the titration.
If added early, the very high concentration of $I_2$ causes the starch to coagulate, forming a highly stable, water-insoluble complex. The iodine gets "trapped" so securely that the thiosulfate titrant cannot reach it effectively. This leads to a falsely low titrant volume and a sluggish, dragging endpoint.
Correct Procedure: Titrate the dark brown $I_2$ solution with Hypo until it fades to a pale straw-yellow color (indicating 95% of $I_2$ is consumed). Only then add the starch. The solution turns deep blue. Continue titrating drop-by-drop until the blue color vanishes sharply, leaving a colorless solution.
4. Iodimetry: Direct Titration Chemistry
In direct iodimetry, we use a standard $I_2$ solution to estimate reducing agents. Let's look at the most classic reaction: the oxidation of Sodium Thiosulfate ($Na_2S_2O_3$, commonly called 'Hypo').
Let's analyze the n-factors:
- For Iodine ($I_2$): $I_2^0 \rightarrow 2I^{-1}$. Total change in oxidation state = $2 \times |(-1) - 0| = 2$. Therefore, the n-factor of $I_2$ is 2. Equivalent weight = $M/2$.
- For Hypo ($Na_2S_2O_3$): The oxidation state of Sulfur in $S_2O_3^{2-}$ is $+2$ (average). In Tetrathionate ($S_4O_6^{2-}$), it is $+2.5$ (average).
Change per Sulfur atom = $|2.5 - 2| = 0.5$.
There are 2 Sulfur atoms in one molecule of Hypo. Therefore, total change = $2 \times 0.5 = 1$.
The n-factor of $Na_2S_2O_3$ is 1. Equivalent weight = $M/1 = M$.
5. Iodometry: The Art of Indirect Estimation
This is the heavy hitter for JEE Advanced. We use this to quantify strong Oxidizing Agents (let's call it $OA$). The process involves two stoichiometrically linked steps:
Step 1: Liberation of Iodine
A known volume of the unknown $OA$ is reacted with an unmeasured excess of Potassium Iodide ($KI$) in an acidic medium. The $OA$ reduces itself, and quantitatively oxidizes the $I^-$ ions into free molecular $I_2$.
Step 2: Titration of Liberated Iodine
The amount of $I_2$ produced is exactly proportional to the amount of the original $OA$. We find out how much $I_2$ was liberated by titrating it against a standard solution of Sodium Thiosulfate ($Na_2S_2O_3$).
The Master Law of Equivalence for Iodometry
Because these are sequential quantitative reactions, the equivalents cascade down the chain perfectly:
Equivalents of Oxidizing Agent ($OA$) = Equivalents of $I_2$ liberated = Equivalents of Hypo consumed.
$$ \left( \frac{\text{Mass}}{\text{Eq. Wt}} \right)_{OA} = (N \times V)_{Hypo} = (M \times n\text{-factor} \times V)_{Hypo} $$
*Remember, since the n-factor of Hypo is 1, its Normality is numerically equal to its Molarity ($N = M$).
6. Key Iodometric Reactions (The JEE Repertoire)
To solve numericals, you must know the exact chemical equations and the n-factors of the specific oxidizing agents used in Step 1.
1. Estimation of Cupric Ions ($Cu^{2+}$)
This is the most common iodometric estimation. $Cu^{2+}$ oxidizes $I^-$, but the resulting Cupric Iodide ($CuI_2$) is highly unstable and instantly decomposes into white, insoluble Cuprous Iodide ($Cu_2I_2$ or $CuI$) and free $I_2$.
Reaction: $2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 \downarrow (\text{white}) + I_2$
n-factor of $Cu^{2+}$ = 1 (since $Cu^{2+} \rightarrow Cu^+$). Equivalent weight = $M/1$.
2. Estimation of Dichromate ($Cr_2O_7^{2-}$)
Used to standardize Hypo solutions.
Reaction: $Cr_2O_7^{2-} + 6I^- + 14H^+ \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$
n-factor of $K_2Cr_2O_7$ = 6 (since $2Cr^{6+} \rightarrow 2Cr^{3+}$). Equivalent weight = $M/6$.
3. Estimation of Bleaching Powder ($CaOCl_2$)
Determines "Available Chlorine". Bleaching powder is treated with dilute acid to release $Cl_2$, which then acts as the oxidizing agent for $KI$.
Reaction 1: $CaOCl_2 + 2H^+ \rightarrow Ca^{2+} + H_2O + Cl_2 \uparrow$
Reaction 2: $Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$
n-factor of $Cl_2$ = 2. Equivalent weight of $Cl_2$ = $71/2 = 35.5$.
4. Estimation of Ozone ($O_3$)
Used for environmental analysis of ozone depletion.
Reaction: $O_3 + 2I^- + H_2O \rightarrow O_2 \uparrow + I_2 + 2OH^-$
n-factor of $O_3$ = 2. (Notice that only one oxygen atom in $O_3$ changes its oxidation state from 0 to -2 to form $OH^-$; the other two form $O_2$). Equivalent weight = $M/2 = 48/2 = 24$.
7. 10 Advanced JEE Level Solved Problems
The theory must be tempered with mathematical rigor. Here are 10 highly complex, multi-step problems designed to test your mastery of iodometry and iodimetry. Do not reveal the solution until you have attempted the math!
Problem 1: Copper Ore Estimation
A $2.5 \text{ g}$ sample of a copper ore was dissolved in acid and all copper was converted to $Cu^{2+}$ ions. The solution was made up to $250 \text{ mL}$.
A $50 \text{ mL}$ aliquot of this solution was treated with excess $KI$. The liberated iodine required $20 \text{ mL}$ of $0.1 \text{ M}$ Sodium Thiosulfate solution for complete reduction using starch indicator.
Calculate the percentage of Copper ($Cu$) in the ore. (Atomic weight of $Cu = 63.5$)
Reveal Detailed Solution
Step 1: Apply the Law of Equivalence to the aliquot.
Equivalents of $Cu^{2+}$ in $50 \text{ mL}$ = Equivalents of Hypo consumed.
For Hypo, n-factor = 1. Therefore, Normality ($N$) = Molarity ($M$) = $0.1 \text{ N}$.
Eq of Hypo = $N \times V(\text{in L}) = 0.1 \times (20 \times 10^{-3}) = 2 \times 10^{-3} \text{ equivalents}$.
So, there are $2 \times 10^{-3} \text{ equivalents}$ of $Cu^{2+}$ in the $50 \text{ mL}$ aliquot.
Step 2: Scale up to the total solution volume.
The total solution is $250 \text{ mL}$. The multiplier is $250 / 50 = 5$.
Total equivalents of $Cu^{2+}$ in the sample = $5 \times (2 \times 10^{-3}) = 10 \times 10^{-3} = 0.01 \text{ equivalents}$.
Step 3: Convert equivalents to mass.
In the reaction $2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 + I_2$, the $Cu^{2+}$ reduces to $Cu^+$.
The n-factor of $Cu$ is 1.
Therefore, Equivalent Weight of $Cu$ = Molar Mass / 1 = $63.5 \text{ g/eq}$.
Mass of pure Copper = Equivalents $\times$ Equivalent Weight = $0.01 \text{ eq} \times 63.5 \text{ g/eq} = 0.635 \text{ grams}$.
Step 4: Calculate Percentage.
Percentage of Cu = $(\text{Mass of pure Cu} / \text{Total mass of ore}) \times 100$
$\% \text{ Cu} = (0.635 / 2.5) \times 100 = 25.4\%$
Answer: 25.4%
Problem 2: Bleaching Powder ($CaOCl_2$)
$3.55 \text{ g}$ of bleaching powder is suspended in water and made up to $500 \text{ mL}$. $50 \text{ mL}$ of this suspension is treated with excess $KI$ and dilute acetic acid. The iodine liberated requires $25 \text{ mL}$ of $0.1 \text{ N}$ $Na_2S_2O_3$ for complete neutralization.
Calculate the percentage of available chlorine in the bleaching powder.
Reveal Detailed Solution
Step 1: Aliquot Equivalents.
Equivalents of Hypo used = $N \times V = 0.1 \times 25 \times 10^{-3} = 2.5 \times 10^{-3} \text{ eq}$.
By Law of Equivalence: Eq of $Cl_2$ in $50 \text{ mL} = 2.5 \times 10^{-3} \text{ eq}$.
Step 2: Total Equivalents.
Total volume is $500 \text{ mL}$. Multiplier = $500 / 50 = 10$.
Total Eq of $Cl_2 = 10 \times (2.5 \times 10^{-3}) = 0.025 \text{ eq}$.
Step 3: Mass of Chlorine.
The n-factor for $Cl_2 \rightarrow 2Cl^-$ is 2.
Equivalent weight of $Cl_2 = M/2 = 71/2 = 35.5 \text{ g/eq}$.
Mass of available $Cl_2 = 0.025 \text{ eq} \times 35.5 \text{ g/eq} = 0.8875 \text{ grams}$.
Step 4: Percentage Calculation.
$\% \text{ Available Chlorine} = (0.8875 / 3.55) \times 100 = \mathbf{25.0\%}$
Answer: 25.0%
Problem 3: Standardizing Hypo with Dichromate
A primary standard solution is prepared by dissolving $1.47 \text{ g}$ of pure Potassium Dichromate ($K_2Cr_2O_7$) in water and making the volume up to $100 \text{ mL}$.
$20 \text{ mL}$ of this solution is pipetted into a flask, acidified with dilute $H_2SO_4$, and an excess of $KI$ is added. The liberated iodine requires $36 \text{ mL}$ of a Sodium Thiosulfate solution for titration.
Calculate the Molarity of the Sodium Thiosulfate solution. (Molar mass of $K_2Cr_2O_7 = 294 \text{ g/mol}$)
Reveal Detailed Solution
Step 1: Molarity and Normality of $K_2Cr_2O_7$.
Moles of $K_2Cr_2O_7$ = $1.47 \text{ g} / 294 \text{ g/mol} = 0.005 \text{ moles}$.
Molarity ($M$) = $0.005 \text{ moles} / 0.1 \text{ L} = 0.05 \text{ M}$.
In acidic medium, n-factor of $K_2Cr_2O_7 = 6$.
Normality ($N$) = $M \times 6 = 0.05 \times 6 = 0.3 \text{ N}$.
Step 2: Equivalents in Aliquot.
Volume used = $20 \text{ mL}$.
Equivalents of Dichromate = $N \times V = 0.3 \times 20 \times 10^{-3} = 6 \times 10^{-3} \text{ eq}$.
Step 3: Equate with Hypo.
Equivalents of Hypo = Equivalents of Dichromate.
$N_{\text{hypo}} \times V_{\text{hypo}} = 6 \times 10^{-3}$
$N_{\text{hypo}} \times (36 \times 10^{-3} \text{ L}) = 6 \times 10^{-3}$
$N_{\text{hypo}} = 6 / 36 = 1/6 \text{ N} \approx 0.1667 \text{ N}$.
Step 4: Molarity of Hypo.
For Sodium Thiosulfate, n-factor = 1.
Therefore, Molarity = Normality = $\mathbf{1/6 \text{ M} \text{ (or } 0.1667 \text{ M)}}$.
Answer: 1/6 M or 0.1667 M
Problem 4: Ozone ($O_3$) Depletion Analysis
A certain volume of air containing ozone was passed through an alkaline solution of $KI$. The ozone oxidizes the iodide. The solution is then carefully acidified and the liberated iodine was titrated with $15 \text{ mL}$ of $0.02 \text{ M}$ $Na_2S_2O_3$ solution.
Calculate the mass of ozone in the air sample. (Molar mass of $O_3 = 48$)
Reveal Detailed Solution
Step 1: The Chemistry.
Reaction: $O_3 + 2I^- + H_2O \rightarrow O_2 + I_2 + 2OH^-$
Only one oxygen atom in $O_3$ undergoes reduction (from 0 to -2).
So, n-factor of $O_3 = 2$. Equivalent weight = $48/2 = 24 \text{ g/eq}$.
Step 2: Equivalents of Hypo.
$N_{\text{hypo}} = M_{\text{hypo}} = 0.02 \text{ N}$.
Eq of Hypo = $0.02 \times 15 \times 10^{-3} = 3 \times 10^{-4} \text{ eq}$.
Step 3: Mass of Ozone.
Eq of $O_3$ = Eq of Hypo = $3 \times 10^{-4} \text{ eq}$.
Mass of $O_3$ = $3 \times 10^{-4} \text{ eq} \times 24 \text{ g/eq} = 72 \times 10^{-4} \text{ grams} = \mathbf{7.2 \text{ mg}}$.
Answer: 7.2 mg
Problem 5: Volume Strength of $H_2O_2$
$10 \text{ mL}$ of a given Hydrogen Peroxide ($H_2O_2$) solution is diluted to $100 \text{ mL}$. $20 \text{ mL}$ of this diluted solution is treated with excess $KI$ in acidic medium. The liberated iodine requires $30 \text{ mL}$ of $0.1 \text{ N}$ Hypo solution.
Calculate the "Volume Strength" of the original $H_2O_2$ solution. (Assume standard conditions for Volume strength calculation).
Reveal Detailed Solution
Step 1: Analyze the diluted solution.
Eq of Hypo used = $0.1 \times 30 \times 10^{-3} = 3 \times 10^{-3} \text{ eq}$.
So, Eq of $H_2O_2$ in $20 \text{ mL}$ of diluted solution = $3 \times 10^{-3} \text{ eq}$.
Normality of diluted $H_2O_2$ ($N_{\text{dil}}$) = Eq / Vol(L) = $(3 \times 10^{-3}) / (20 \times 10^{-3}) = 0.15 \text{ N}$.
Step 2: Find Normality of Original Solution.
Using dilution law: $N_1 V_1 = N_2 V_2$
$N_{\text{original}} \times 10 = 0.15 \times 100$
$N_{\text{original}} = 1.5 \text{ N}$.
Step 3: Calculate Volume Strength.
The standard formula connecting Normality and Volume Strength for $H_2O_2$ is:
Volume Strength = Normality $\times$ $5.6$
Volume Strength = $1.5 \times 5.6 = \mathbf{8.4 \text{ V}}$.
Answer: 8.4 V
Problem 6: The $KIO_3$ Comproportionation Trap
A solution containing $x \text{ grams}$ of Potassium Iodate ($KIO_3$) is treated with an excess of $KI$ in acidic medium. The liberated iodine completely reacts with $60 \text{ mL}$ of $0.1 \text{ M}$ Sodium Thiosulfate.
Find the value of $x$. (Molar mass of $KIO_3 = 214$)
Warning: Think carefully about the n-factor of $KIO_3$!
Reveal Detailed Solution
Step 1: The Chemical Reaction (The Trap).
Reaction: $IO_3^- + 5I^- + 6H^+ \rightarrow 3I_2 + 3H_2O$
This is a comproportionation reaction. Iodine in $+5$ ($IO_3^-$) and Iodine in $-1$ ($I^-$) both go to $0$ ($I_2$).
We want to estimate $KIO_3$. Look at the electron transfer for the Iodate ion alone:
$I^{5+} \rightarrow I^0$. It gains 5 electrons.
However, the standard equivalent relationship equates equivalents of analyte directly to equivalents of Hypo. The $3I_2$ produced requires 6 electrons from Hypo (since $3I_2$ acts as a 6 electron acceptor).
Wait, 1 mole of $KIO_3$ produces 3 moles of $I_2$.
3 moles of $I_2$ will consume 6 moles of Hypo ($Na_2S_2O_3$).
Therefore, 1 mole of $KIO_3 \equiv 6 \text{ moles of Hypo (1 electron transfer each)}$.
For the purpose of total titration equivalence, the effective n-factor of $KIO_3$ is 6.
Step 2: Calculations.
Equivalents of Hypo = $0.1 \times 60 \times 10^{-3} = 6 \times 10^{-3} \text{ eq}$.
Equivalents of $KIO_3 = 6 \times 10^{-3} \text{ eq}$.
Equivalent weight of $KIO_3 = M / 6 = 214 / 6 \text{ g/eq}$.
Mass ($x$) = $6 \times 10^{-3} \times (214 / 6) = \mathbf{0.214 \text{ grams}}$.
Answer: x = 0.214 g
Problem 7: Adsorbed Iodine & KSCN Effect
In a copper estimation experiment, if Potassium Thiocyanate (KSCN) is NOT added near the endpoint, a student notes the endpoint at $24.5 \text{ mL}$ of Hypo. When the experiment is repeated precisely, but KSCN is added near the endpoint, the correct endpoint is achieved.
Will the correct endpoint volume be greater than, less than, or equal to $24.5 \text{ mL}$? Explain briefly.
Reveal Detailed Solution
Logic:
During the reaction $2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 \downarrow + I_2$, the white precipitate of $Cu_2I_2$ heavily adsorbs some of the liberated $I_2$ onto its surface.
If KSCN is NOT added, this adsorbed $I_2$ is trapped and escapes titration by the Hypo solution. Because some $I_2$ is "hidden", less Hypo is consumed. Thus, $24.5 \text{ mL}$ is an underestimation.
When KSCN is added, $SCN^-$ displaces the $I_2$ from the precipitate surface (forming more stable $CuSCN$). This releases the hidden $I_2$ into the solution, allowing it to react with Hypo.
Therefore, more Hypo will be required to neutralize this newly released $I_2$.
Answer: The correct volume will be GREATER THAN 24.5 mL.
Problem 8: Mixture of Oxidants
A solution contains a mixture of $KMnO_4$ and $K_2Cr_2O_7$. $25 \text{ mL}$ of this mixture was treated with excess $KI$ in highly acidic medium. The total liberated iodine required $45 \text{ mL}$ of $0.1 \text{ M}$ Hypo solution.
In a separate experiment, $25 \text{ mL}$ of the exact same mixture was titrated directly with $0.1 \text{ M}$ Mohr's salt ($Fe^{2+}$) solution, requiring $45 \text{ mL}$.
Is it possible to determine the individual molarities of $KMnO_4$ and $K_2Cr_2O_7$ from this data? Prove mathematically.
Reveal Detailed Solution
Step 1: Analyze Experiment 1 (Iodometry).
Both $KMnO_4$ and $K_2Cr_2O_7$ oxidize $I^-$.
Let Molarity of $KMnO_4 = M_1$, n-factor = 5. Normality = $5M_1$.
Let Molarity of $K_2Cr_2O_7 = M_2$, n-factor = 6. Normality = $6M_2$.
Total Eq of Oxidants = Eq of Hypo
$(5M_1 \times 25) + (6M_2 \times 25) = 0.1 \times 45$
$25(5M_1 + 6M_2) = 4.5$ --- (Equation A)
Step 2: Analyze Experiment 2 (Direct Titration with Mohr's salt).
Both $KMnO_4$ and $K_2Cr_2O_7$ oxidize $Fe^{2+}$.
The n-factors for the oxidants remain exactly the same in acidic medium (5 and 6 respectively).
Total Eq of Oxidants = Eq of Mohr's salt
$(5M_1 \times 25) + (6M_2 \times 25) = 0.1 \times 45$ (Since n-factor of $Fe^{2+}$ is 1)
$25(5M_1 + 6M_2) = 4.5$ --- (Equation B)
Conclusion:
Equation A and Equation B are mathematically identical. We have two variables but only one unique equation. The data provided is redundant.
Answer: NO, it is impossible. The equations are linearly dependent.
Problem 9: Direct Iodimetry of Arsenic
A $0.198 \text{ g}$ sample containing $As_2O_3$ (Molar mass = 198) and inert impurities is dissolved in a mildly basic $NaHCO_3$ buffer. It is titrated directly with $0.05 \text{ M}$ Iodine ($I_2$) solution using starch indicator. The endpoint occurs at $30 \text{ mL}$.
Calculate the percentage purity of $As_2O_3$ in the sample.
Reveal Detailed Solution
Step 1: Understand the reaction and n-factor.
Reaction: $As_2O_3 + 2I_2 + 2H_2O \rightarrow As_2O_5 + 4HI$
Arsenic goes from $+3$ in $As_2O_3$ to $+5$ in $As_2O_5$.
Change per As atom = 2. There are 2 As atoms in $As_2O_3$.
Therefore, n-factor of $As_2O_3 = 2 \times 2 = \mathbf{4}$.
n-factor of $I_2 = 2$.
Step 2: Equivalents calculation.
Equivalents of $I_2$ used = $M \times n \times V(\text{L}) = 0.05 \times 2 \times (30 \times 10^{-3}) = 3 \times 10^{-3} \text{ eq}$.
Equivalents of $As_2O_3 = 3 \times 10^{-3} \text{ eq}$.
Step 3: Mass calculation.
Equivalent weight of $As_2O_3 = 198 / 4 = 49.5 \text{ g/eq}$.
Mass of pure $As_2O_3 = 3 \times 10^{-3} \times 49.5 = 0.1485 \text{ grams}$.
Step 4: Percentage purity.
$\% \text{ Purity} = (0.1485 / 0.198) \times 100 = \mathbf{75.0\%}$
Answer: 75.0%
Problem 10: The Ultimate Multi-Step Iodometry
A $1.0 \text{ g}$ sample of $KIO_3$ and inert material is dissolved in water and made up to $100 \text{ mL}$. $20 \text{ mL}$ of this solution is treated with excess $KI$ in acidic medium. The liberated iodine is titrated with a $0.1 \text{ M}$ Hypo solution, but the student overshoots the endpoint!
To correct this, he adds $5.0 \text{ mL}$ of standard $0.02 \text{ M}$ $I_2$ solution to back-titrate the excess Hypo, reaching the precise starch endpoint. The total volume of Hypo run from the burette initially was $25.0 \text{ mL}$.
Calculate the mass percentage of pure $KIO_3$ in the original sample. ($M = 214$)
Reveal Detailed Solution
Step 1: Untangle the overshot titration.
Total Hypo added = $25.0 \text{ mL}$ of $0.1 \text{ M}$ (Normality = 0.1).
Total meq of Hypo added = $25 \times 0.1 = 2.5 \text{ meq}$.
This Hypo reacted with two things: the $I_2$ liberated from $KIO_3$, AND the standard $I_2$ added later to fix the mistake.
meq of standard $I_2$ added = $M \times n \times V = 0.02 \times 2 \times 5.0 = 0.2 \text{ meq}$.
Therefore, meq of Hypo that actually reacted with the $I_2$ from $KIO_3$ = $2.5 - 0.2 = \mathbf{2.3 \text{ meq}}$.
Step 2: Calculate $KIO_3$ in aliquot.
From Problem 6, we know the effective n-factor of $KIO_3$ in this entire sequence is 6.
meq of $KIO_3$ in $20 \text{ mL}$ aliquot = $2.3 \text{ meq}$.
Step 3: Scale to original solution and calculate mass.
Total volume = $100 \text{ mL}$. Multiplier = $100 / 20 = 5$.
Total meq of $KIO_3$ = $2.3 \times 5 = 11.5 \text{ meq} = 0.0115 \text{ eq}$.
Equivalent weight of $KIO_3 = 214 / 6 \text{ g/eq}$.
Mass of $KIO_3 = 0.0115 \times (214 / 6) = 0.4101 \text{ grams}$.
Step 4: Percentage.
$\% \text{ Purity} = (0.4101 / 1.0) \times 100 = \mathbf{41.01\%}$
Answer: 41.01%
8. Frequently Asked Questions (FAQ)
Why is $KI$ added in excess during Iodometry?
Two critical reasons: First, an excess ensures that the unknown oxidizing agent is completely reduced without limiting the reaction. Second, molecular iodine ($I_2$) is almost completely insoluble in water. The excess $I^-$ ions react with the newly formed $I_2$ to form the highly soluble $I_3^-$ (triiodide) complex, preventing iodine from evaporating or precipitating out of the solution before it can be titrated.
Can we use Nitric acid ($HNO_3$) instead of Sulfuric acid ($H_2SO_4$) to acidify the solution in Iodometry?
Absolutely not. Nitric acid is a strong oxidizing agent on its own. If added, it will oxidize the $KI$ to $I_2$ independently of the actual analyte you are trying to measure. This will create artificially high amounts of iodine, leading to massive overestimation errors. Dilute $H_2SO_4$ or $HCl$ (in specific non-oxidizing scenarios) are preferred.
Why does the blue color sometimes return after reaching the endpoint in an Iodometric titration?
If the solution turns blue again after a few minutes of standing colorless, it means atmospheric oxygen has dissolved into the acidic solution and oxidized some of the $I^-$ back into $I_2$. ($4I^- + O_2 + 4H^+ \rightarrow 2I_2 + 2H_2O$). This is why iodometric titrations should be completed swiftly and not left standing. (If it returns immediately, the reaction with the original oxidant might have been kinetically slow and incomplete).
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