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 \(0.1 \text{ N NaOH}\) required is \(40 \text{ mL}\).

Step 1: Calculate Moles of Oxalic Acid

Oxalic acid dehydrate has a molar mass of approximately \(126 \text{ g/mol}\). Since we have \(6.3 \text{ g}\) of oxalic acid, the number of moles is calculated by dividing the mass by the molar mass: \(\frac{6.3 \text{ g}}{126 \text{ g/mol}} = 0.05 \text{ moles}\).

Step 2: Determine Molarity of Oxalic Acid Solution

The solution is diluted to a total volume of \(250 \text{ ml}\). The molarity (\(M\)) is calculated by dividing the moles of solute by the liters of solution: \(M = \frac{0.05 \text{ moles}}{0.25 \text{ L}} = 0.2 \text{ M}\).

Step 3: Calculate Moles in 10 mL of Solution

We need to find the number of moles in \(10 \text{ mL}\) of the \(0.2 \text{ M}\) solution. \(10 \text{ mL}\) is \(0.01 \text{ L}\). So, moles = \(0.2 \text{ M} \times 0.01 \text{ L} = 0.002 \text{ moles}\).

Step 4: Write Neutralization Reaction

The balanced reaction between oxalic acid \((\text{H}_2\text{C}_2\text{O}_4)\) and NaOH is \(\text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O}\). It shows each mole of oxalic acid reacts with 2 moles of NaOH.

Step 5: Calculate Volume of NaOH Required

Since \(0.002 \text{ moles}\) of oxalic acid require \(0.004 \text{ moles}\) of NaOH (from Step 4). Given that the NaOH concentration is \(0.1 \text{ N}\), we need \( \frac{0.004 \text{ moles}}{0.1 \text{ N}} = 0.04 \text{ L} = 40 \text{ mL}\).

Key Concepts

Oxalic Acid

Oxalic acid is a simple organic compound with the formula \(\text{H}_2\text{C}_2\text{O}_4\). It often appears as a colorless crystalline solid that is soluble in water. As a dicarboxylic acid, it contains two carboxyl groups, which classify it under organic acids. One common form is oxalic acid dihydrate, meaning each molecule of oxalic acid is accompanied by two water molecules in its crystalline form.
In chemistry, oxalic acid is significant due to its role in various reactions. It's commonly used as a primary standard in titrations because its stable and pure form allows for accurate measurement. Its ability to form strong bonds with metals makes it instrumental in rust removal and the industrial preparation of metal complexes.
It is also important in biological systems and must be handled with caution, as it is toxic in high concentrations. It's found in many plants, including rhubarb and spinach, contributing tartness and influencing nutrient absorption.

Molarity Calculation

Molarity (\(M\)) is a key concept in chemistry that describes the concentration of a solution. It is defined as the number of moles of a solute divided by the volume of the solution in liters. Calculating molarity is crucial when preparing chemical solutions, as it indicates how much solute is present in a given solution volume.
To illustrate, let's consider a solution where 6.3 g of oxalic acid dihydrate is dissolved to form a 250 ml solution. First, calculate the moles of oxalic acid using its molar mass (126 g/mol): \[\text{Moles} = \frac{6.3 \text{ g}}{126 \text{ g/mol}} = 0.05 \text{ moles}\].
Next, the volume of the solution is converted to liters (250 ml = 0.25 L). The molarity is then: \[ M = \frac{0.05 \text{ moles}}{0.25 \text{ L}} = 0.2 \text{ M} \].
It is essential for various calculations, including determining reagent amounts needed for reactions or titrations.

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. It is based on the principle of the conservation of mass, where the total mass of reactants equals the total mass of products. This concept allows chemists to predict the outcomes of reactions by using balanced chemical equations.
Let's take the reaction between oxalic acid and sodium hydroxide (NaOH) as an example: \[\text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O}\].
This equation implies that one mole of oxalic acid reacts with two moles of NaOH. Hence, if you have 0.002 moles of oxalic acid, they will require 0.004 moles of NaOH for complete neutralization.
Understanding stoichiometry is critical for solving problems involving molarity, reactions, and conversions in chemistry. It forms the foundation for predicting how much of each reactant is needed or how much product will be formed.

Aqueous Solution Chemistry

Aqueous solution chemistry deals with substances dissolved in water, a universal solvent. Many reactions occur in aqueous solutions as water facilitates the movement and interaction of ions, enabling chemical transformations.
In the given exercise, oxalic acid is dissolved in water to form an aqueous solution, measured in molarity (0.2 M). Such solutions are common in titrations, where an acid reacts with a base. Here, oxalic acid acts as the solute, and water is the solvent forming the solution.
The significance of aqueous solutions in chemistry can't be overstated. Many biological, industrial, and environmental processes involve aqueous solutions. They allow for straightforward manipulation of concentrations, mixing, and conducting electrochemical reactions.
Understanding how solutes interact in water helps predict their behavior in reactions, such as the neutralization of an acid with a base, as shown in oxalic acid and NaOH's reaction in this exercise.