Problem 56 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

(i) What is the weight of sodium bromate and molarity of solution necessary to prepare \(85.5 \mathrm{~mL}\) of \(0.672 \mathrm{~N}\) solution when the halfcell reaction is $$ \mathrm{BrO}_{3}^{-}+6 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow \mathrm{Br}^{-}+3 \mathrm{H}_{2} \mathrm{O} $$ (ii) What would be the weight as well as molarity if the half-cell reaction is : $$ 2 \mathrm{BrO}_{3}^{-}+12 \mathrm{H}^{+}+10 \mathrm{e}^{-} \rightarrow \mathrm{Br}_{2}+6 \mathrm{H}_{2} \mathrm{O} $$ [1987 - 5 Marks]

Step-by-Step Solution

Answer

(i) First reaction: 1.45 g and 0.672 M. (ii) Second reaction: 1.736 g and 0.336 M.

Step 1: Determining Equivalent Weight for First Reaction

For the first reaction \( \text{BrO}_3^- + 6\text{H}^+ + 6\text{e}^- \rightarrow \text{Br}^- + 3\text{H}_2\text{O} \), the equivalent weight (\(E.W.\)) is calculated by dividing the molar mass of \(\text{NaBrO}_3\) (which is 150.89 g/mol) by the number of electrons transferred (6 electrons), resulting in \( E.W. = \frac{150.89}{6} = 25.15 \text{ g/equiv} \).

Step 2: Calculating Weight of Sodium Bromate for First Reaction

Using the formula \( \text{Weight} = \text{Normality} \times \text{Volume} \times \text{Equivalent Weight} \), we substitute the values for the first reaction: \( \text{Weight} = 0.672 \times 0.0855 \times 25.15 = 1.45 \text{ g} \).

Step 3: Calculating Molarity for First Reaction

Since 1 mole of \(\text{NaBrO}_3\) provides one equivalent in this reaction, the normality is equal to molarity here, which is \(0.672 \text{ M}\).

Step 4: Determining Equivalent Weight for Second Reaction

For the second reaction \( 2 \text{BrO}_3^- + 12 \text{H}^+ + 10 \text{e}^- \rightarrow \text{Br}_2 + 6 \text{H}_2\text{O} \), the equivalent weight is calculated as follows: \( E.W. = \frac{2 \times 150.89}{10} = 30.178 \text{ g/equiv} \).

Step 5: Calculating Weight of Sodium Bromate for Second Reaction

Using the formula \( \text{Weight} = \text{Normality} \times \text{Volume} \times \text{Equivalent Weight} \), we substitute for the second reaction: \( \text{Weight} = 0.672 \times 0.0855 \times 30.178 = 1.736 \text{ g} \).

Step 6: Calculating Molarity for Second Reaction

For the second reaction, the normality is related to molarity by multiplying with the number of equivalents per mole of \(\text{NaBrO}_3\): \( \text{M} = \frac{\text{N}}{2} = \frac{0.672}{2} = 0.336 \text{ M} \).

Key Concepts

Equivalent Weight

Equivalent weight is a fundamental concept used to understand how much of a substance is required to react fully in a given chemical reaction. It tells us how much of a compound would provide one mole of charge in a reaction.
To calculate the equivalent weight, you need to know the molar mass of the substance involved and the number of electrons exchanged in the chemical equation. This highlights how chemical reactions involve the transfer of electrons.

In our example, for the first reaction, the equivalent weight is determined by dividing the molar mass of sodium bromate \(\text{NaBrO}_3\) by the total electrons transferred, i.e., 6 electrons. Hence, the equivalent weight for the first reaction turns out to be 25.15 g/equiv.

• Molar Mass of \(\text{NaBrO}_3\): 150.89 g/mol
• Electrons Transferred: 6
• Calculation: \(\frac{150.89}{6} = 25.15 \text{ g/equiv}\)

The process is similar for the second reaction, except we consider the combined effects of 2 moles, hence, the equivalent weight calculation is \(\frac{2 \times 150.89}{10} = 30.178 \text{ g/equiv}\).
Understanding equivalent weight is critical because it allows for precise calculations of the amounts needed in chemical reactions which ensures efficiency and accuracy.

Molarity Calculation

Molarity is a measure of the concentration of a solute within a solution. It is denoted as the number of moles of solute per liter of solution, giving us insight into how much solute is present compared to the volume of the solution.

For accurate molarity calculations, it's essential to know both the weight of the solute added and the volume of solution it is dissolved in. It's often used in labs to prepare specific concentrations of solutions for experiments.

• Formula: \(\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}\)

In the original exercise, after finding the equivalent weight, we see that for the first reaction, the normality and molarity happen to be equivalent, leading us to a molarity of 0.672 M.
This relationship exists because the reaction provides one equivalent for each mole of sodium bromate used. Whereas, for the second reaction, the molarity is adjusted because you only use half of the equivalents per mol of reactant, so the molarity is 0.336 M (as half of 0.672 N).
Molarity is crucial for understanding how solutions react and are often pivotal in titration and other chemical solution preparation protocols.

Normality Concept

Normality is similar to molarity but accounts for the number of equivalents of a solute in a solution. It provides a clearer picture when dealing with reactions that involve acid-base reactions or redox processes, where the transfer of particles is not straightforward.

Unlike molarity, normality varies depending on the reaction because it concerns the reactive capacity of the solute. It's the measure of concentration equivalent to the gram equivalent weight of solute per liter of solution.

• Formula: \(\text{Normality (N)} = \frac{\text{number of equivalents}}{\text{liters of solution}}\)

In the exercise given, normality is used to directly calculate the amount of sodium bromate needed given the number of electrons transferred and the desired concentration of the solution.
For instance, for the reaction involving \(6 \text{e}^-\) transfers, knowing the normality helps determine how many grams are needed to make a desired 0.672 N solution.
Normality becomes particularly useful in titration, where exact reactions require calculation of reactive equivalents, meaning it ensures the right amount of reactant product is always achieved.