Problem 75 - A Complete Resource Book in Chemistry for JEE Main

Question

An aqueous solution of \(6.3 \mathrm{~g}\) oxalic acid dehydrate is made up to \(250 \mathrm{ml}\). The volume of \(0.1 \mathrm{~N} \mathrm{NaOH}\) required to completely neutralize \(10 \mathrm{~mL}\) of this solution is (a) \(4 \mathrm{~mL}\) (b) \(20 \mathrm{~mL}\) (c) \(40 \mathrm{~mL}\) (d) \(60 \mathrm{~mL}\)

Step-by-Step Solution

Answer

The volume of NaOH required is \(40 \, \text{mL}\) (option c).

Step 1: Calculate Moles of Oxalic Acid

Oxalic acid has the formula \(\text{C}_2\text{H}_2\text{O}_4 \, \cdot \, 2\text{H}_2\text{O}\). The molar mass is \(126 \, \text{g/mol}\) (calculated as \(2 \times 12 + 2 \times 1 + 4 \times 16 + 2 \times (2 \times 1 + 16)\)). Given \(6.3 \, \text{g}\) of oxalic acid, we find moles by the equation: \[\text{Moles} = \frac{6.3}{126} = 0.05 \, \text{mol}\]

Step 2: Determine Molarity of the Solution

The solution containing 0.05 moles of oxalic acid is made up to a total volume of \(250 \, \text{mL}\) or \(0.25 \, \text{L}\). The molarity \(M\) of the solution is calculated as:\[M = \frac{\text{moles}}{\text{volume in L}} = \frac{0.05}{0.25} = 0.2 \, \text{M}\]

Step 3: Calculate Moles of Oxalic Acid in 10 mL

For \(10 \, \text{mL}\) of the solution, \[\text{Moles in } 10 \, \text{mL} = 0.2 \, \text{M} \times \frac{10}{1000} = 0.002 \, \text{mol}\]

Step 4: Use Reaction Stoichiometry

Oxalic acid reacts with NaOH following the equation: \(\text{C}_2\text{H}_2\text{O}_4 \, + \, 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 \, + \, 2\text{H}_2\text{O}\). Thus, \(2\) moles of NaOH are required per mole of oxalic acid. Therefore, for \(0.002 \, \text{mol}\) of oxalic acid:\[\text{Moles of } \text{NaOH} = 2 \times 0.002 = 0.004 \, \text{mol}\]

Step 5: Calculate Volume of NaOH Solution Required

The normality of the NaOH solution is given as \(0.1 \, \text{N}\), which means \(0.1 \, \text{mol/L}\). Hence, \[\text{Volume of } \text{NaOH} = \frac{0.004}{0.1} \times 1000 = 40 \, \text{mL}\]

Key Concepts

Oxalic Acid

Oxalic acid, known scientifically as \( \text{C}_2\text{H}_2\text{O}_4 \), and recognized for its role in many naturally occurring processes, is a simple dicarboxylic acid found in various plants. Unique for its ability to form strong complexes with most metal ions, oxalic acid is often used in applications like metal cleaning and bleaching. This compound typically appears in its dihydrate form (\( \text{C}_2\text{H}_2\text{O}_4 \cdot 2\text{H}_2\text{O} \)), meaning each molecule is associated with two molecules of water of crystallization. Understanding oxalic acid's physical and chemical properties is essential because these attributes influence how it reacts in chemical processes, such as neutralization reactions. It's also crucial to accurately determine the molar mass of the dihydrate form, which is crucial for calculations involving moles.

Molarity Calculation

In chemistry, accurately calculating the molarity (\( M \)) of a solution is fundamental to understanding its concentration. Molarity is defined as the number of moles of solute per liter of solution. Here's how one can calculate the molarity for a solution that uses oxalic acid, for instance:

• Calculate the moles of solute: Knowing the mass and the molar mass of the solute allows you to find moles. For oxalic acid, the formula used is: \( \text{Moles} = \frac{\text{mass of solute in g}}{\text{molar mass in g/mol}} \).
• Convert volume to liters if necessary: Since molarity is expressed in moles per liter, ensure the solution's volume is in liters; convert if initially provided in milliliters.
• Compute Molarity: Use the formula \( M = \frac{\text{moles of solute}}{\text{volume of solution in L}} \). This gives you the concentration of the solute within the solution.

Molarity is key to predicting how chemical reactions will proceed in solutions, making it invaluable for laboratory settings.

Stoichiometry

Stoichiometry involves calculations that relate quantities of reactants and products in chemical reactions. It is a fundamental concept used to predict the outcomes of reactions and the quantities needed or produced. In our oxalic acid example, stoichiometry helps determine the amount of NaOH needed for neutralization. The balanced chemical equation:\[\text{C}_2\text{H}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \]indicates that 2 moles of NaOH react with 1 mole of oxalic acid. This relationship helps to:

• Determine required quantities: Use the mole ratio from the balanced equation to calculate the required amount of a reactant or the expected amount of a product.
• Predict reaction outcomes: Knowing these ratios allows chemists to plan and scale industrial processes efficiently and safely.

Stoichiometry is not only foundational in laboratory work but also crucial in fields like pharmacology, where precise calculations ensure drug efficacy.

Normality

Normality (\( N \)) is a measure of concentration equivalent to molarity but multiplied by the equivalence factor, which depends on the reactive capacity of the solute. For acids and bases, the normality often equates to the number of moles of reactive species. Using normality provides a straightforward understanding of the specifics of a solution's reactive capacity. In the example provided:

• Relation to molarity: For oxalic acid, a dibasic acid, normality is twice the molarity because the acid can donate two protons (\( 2 \text{H}^+ \)) in a reaction.
• Neutralization calculations: Knowing that the NaOH solution is \( 0.1 \text{N} \) allows chemists to calculate precisely how much NaOH is needed to fully react with a known amount of oxalic acid.

Considering normality alongside molarity helps to navigate complex titrations and neutralization reactions more effectively. Understanding this equivalence enhances one's ability to predict and measure chemical reactions accurately.