Problem 34 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

One gram of commercial \(\mathrm{AgNO}_{3}\) is dissolved in \(50 \mathrm{~mL}\). of water. It is treated with \(50 \mathrm{~mL}\). of a KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filterate is titrated with \((\mathrm{M} / 10) \mathrm{KIO}_{3}\) solution in presence of \(6 \mathrm{M} \mathrm{HCl}\) till all I- ions are converted into ICl. It requires \(50 \mathrm{~mL}\). of \((\mathrm{M} / 10) \mathrm{KIO}_{3}\) solution. \(20 \mathrm{~mL}\). of the same stock solution of KI requires \(30 \mathrm{~mL}\). of \((\mathrm{M} / 10) \mathrm{KlO}_{3}\) under similar conditions. Calculate the percentage of \(\mathrm{AgNO}_{3}\) in the sample. (Reaction: \(\left.\mathrm{KIO}_{3}+2 \mathrm{KI}+6 \mathrm{HCl} \rightarrow 3 \mathrm{ICl}+3 \mathrm{KCl}+3 \mathrm{H}_{2} \mathrm{O}\right)\)

Step-by-Step Solution

Answer

The percentage of \(\mathrm{AgNO}_{3}\) in the sample is approximately 42.47\%.

Step 1: Determine the Normality and Volume of KIO3

Given that 50 mL of \((M / 10) \mathrm{KIO}_{3}\) solution is used for titration, calculate its normality. The molarity is \(0.1 \, M\), and since it's a pentavalent iodine, its normality is equal to its molarity as \(1\) mole of \(\mathrm{KIO}_{3}\) reacts with \(2\) equivalents of \(KI\). Thus, \(N_{KIO_{3}} = 0.1\). The volume is \(50 \, \mathrm{mL}\), for both \(\mathrm{AgNO}_3\) solution excess titration and for \(\mathrm{KI}\) stock solution determination.

Step 2: Calculate Iodide Ion Concentration in Stock KI

For the \(20 \mathrm{~mL}\) of stock \(\mathrm{KI}\) solution, it requires \(30 \mathrm{~mL}\) of \((M/10)\cdot \mathrm{KIO}_{3}\). From the reaction, calculate \(\mathrm{KI}\) molarity:\[N(I^-) \times V(I^-) = N(KIO_3) \times V(KIO_3)\]\[N(I^-) = \frac{(0.1 \cdot 30)}{20}\]\[N(I^-) = 0.15\]Thus, the molarity \(M(I^-)\) equals \(0.15 \mathrm{~M}\).

Step 3: Determine Excess KI Used

For the excess KI in \(\mathrm{AgNO}_{3}\) solution: \[N(I_{excess}) \times V(I_{excess}) = N(KIO_3) \times V(KIO_3)\]\[N(I_{excess}) \times 50 = 0.1 \times 50\]\[N(I_{excess}) = 0.1\]Thus, the molarity of excess iodide ions \(M(I_{excess})\) is \(0.1 \mathrm{~M}\).

Step 4: Calculate Initial KI in AgNO3 Reaction

Since the total molarity available from initial KI is \(0.15 \mathrm{~M}\), and \(0.1 \mathrm{~M}\) was found as excess, determine the quantity initially mixed:\[M(I_{initial}) = M(I_{total}) - M(I_{excess})\]\[M(I_{initial}) = 0.15 - 0.1 = 0.05\]Using \(50 \mathrm{~mL}\):\[\text{Moles of } I^- = 0.05 \times 0.05 = 0.0025 \, \text{moles}\]

Step 5: Determine AgNO3 used in Reaction

The moles of \(\mathrm{AgNO}_{3}\) is equal to moles of initial \(I^-\):\[\text{Moles of } \mathrm{AgNO}_{3} = \text{Moles of precipitated } I^- = 0.0025 \, \text{moles}\]

Step 6: Calculate Mass and Percentage of AgNO3

Calculate the mass of \(\mathrm{AgNO}_{3}\) using its molar mass \((\approx 169.87 \mathrm{~g/mol})\):\[\text{Mass of } \mathrm{AgNO}_{3} = 0.0025 \times 169.87 = 0.424675 \, \text{grams}\]The percentage of \(\mathrm{AgNO}_{3}\) in the sample:\[\text{Percentage} = \left(\frac{0.424675}{1} \times 100\right) \approx 42.47\%\]

Step 7: Conclusion

The percentage of \(\mathrm{AgNO}_{3}\) in the commercial sample is approximately \(42.47\%\).

Key Concepts

Titration Calculation

Titration is a vital laboratory technique used to determine the concentration of a solute within a solution. In our exercise, we perform titration to find the quantity of ions present. After mixing a known solution with a reagent, we use a titrant of known concentration to determine how much was needed to react fully with the solute. This precise measurement allows us to calculate the necessary concentration.

For example, if you used 50 mL of a \((M / 10)\) potassium iodate (KIOe 3e) solution while titrating excess potassium iodide (KI), by knowing the normals of each, it becomes possible to determine the concentration of silver nitrate (AgNOe 3e) in your initial sample. Thus, titration calculations are integral for determining unknown concentrations based on the titrant used.

• The known titrant helps in quantifying the reactants in a given sample.
• Reactions like these help us understand the sample's composition and calculate amounts like the percentage of AgNOe 3e present.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction in which two solutions combine to form a solid, known as a precipitate. This is central in the described experiment with the formation of silver iodide (AgI) when combining silver nitrate with potassium iodide (KI). When these two react, silver (Ag+) ions react with iodide (I-) ions, resulting in the insoluble solid silver iodide, which appears as a precipitate.

The process steps include:

• Mixing equal volumes of AgNO e 3 e and KI solutions.
• Allowing silver iodide to form and subsequently filtering it off from the solution.

Once the solid is removed, the leftover or excess iodide ions can be quantified using another titration to ensure accuracy in measurements of the components initially present in the solution.
Precipitation reactions are essential in analytical chemistry for isolating compounds from complex mixtures and for purifying salts.

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate relative quantities of reactants and products in a chemical reaction. It allows chemists to understand both the theoretical and practical substance amounts required for a reaction.

In our exercise, stoichiometry plays a pivotal role as it helps in calculating how much KI and AgNO e 3 e reacts to form AgI. Based on the balanced equation provided, it outlines stoichiometric coefficients ensuring you know how many moles of one reactant are needed to react with another. Proper mole ratios derived from these coefficients are essential when calculating the moles of AgNO e 3 e reacting with KI to ensure complete precipitation of silver iodide.

Understanding stoichiometry is crucial because:

• It informs theoretical yields of product that should form under perfect conditions.
• Enables conversion between grams, moles, and particles, critical for quantitative analysis.
• justifies necessary steps to solve chemistry problems efficiently.

Normality and Molarity Concepts

Normality and molarity are both measures of concentration in a solution, key for chemical calculations. Molarity (M) refers to the number of moles of a solute per liter of solution, while normality (N) is the number of equivalents per liter. They're often synonymous; however, normality depends on the reaction involved.

In the problem, when dealing with KIO e 3 e, because of its specific reactiveness, although it's presented with a molarity of 0.1 M, it effectively equates to a normality of 0.1 N in iodide conversion processes. This simplification occurs due to equivalent factors and the reaction details.

• Normality is crucial in titration for directly establishing how much titrant is needed.
• Uses context distinctions, particularly when acids or bases are involved where more ions affect calculations.
• Molarity is utilized when calculating amounts of a specific chemical species within solutions.

Understanding these concepts ensures clarity in performing accurate chemical measurements, especially in titration and precipitation reactions.