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 correct option is (c) 40 mL.

Step 1: Determine Moles of Oxalic Acid Dihydrate

Calculate the moles of oxalic acid dihydrate present in 6.3 g. The molar mass of oxalic acid dihydrate \(( ext{COOH})_2 ext{·} 2 ext{H}_2 ext{O}\) is approximately \(126 ext{ g/mol}\). The number of moles is given by: \[\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{6.3 ext{ g}}{126 ext{ g/mol}} = 0.05 ext{ mol}\]

Step 2: Determine Concentration of Oxalic Acid in Solution

The concentration of the oxalic acid solution is found by dividing the number of moles by the total volume in liters. The solution volume is \(250 \text{ mL} = 0.25 \text{ L}\):\[\text{Concentration (C)} = \frac{0.05 \text{ mol}}{0.25 \text{ L}} = 0.2 \text{ M}\]

Step 3: Volume of Solution in Reaction

The volume of oxalic acid solution used for reaction is \(10 \text{ mL} = 0.01 \text{ L}\). Use this volume to find moles in this reaction:\[\text{Moles in } 10 \text{ mL} = 0.2 \text{ M} \times 0.01 \text{ L} = 0.002 \text{ mol}\]

Step 4: Neutralization of Oxalic Acid

Oxalic acid is a diprotic acid (provides two protons per molecule). Thus, it requires 2 equivalents of base per molecule of oxalic acid to neutralize. Therefore, the equivalents required are:\[\text{Equivalents required} = 0.002 \text{ mol} \times 2 = 0.004 \text{ equivalents}\]

Step 5: Volume of NaOH Solution Required

The given normality of NaOH is \(0.1 \text{ N}\). Use the equivalents formula \(\text{Volume NaOH (L)} \times \text{Normality NaOH} = \text{Equivalents required}\):\[V_{\text{NaOH}} = \frac{0.004 \text{ eq}}{0.1 \text{ N}} = 0.04 \text{ L} = 40 \text{ mL}\]

Step 6: Conclude the Required Volume of NaOH

The volume of \(0.1 \text{ N}\) NaOH solution needed to completely neutralize \(10 \text{ mL}\) of the oxalic acid solution is \(40 \text{ mL}\). This corresponds to option (c).

Key Concepts

Neutralization Reaction

A neutralization reaction is an important concept in chemistry, especially when dealing with acid-base titrations. It occurs when an acid and a base react to form water and a salt, effectively cancelling out their respective acidic and basic properties.
In our scenario, oxalic acid, which is a diprotic acid (meaning it can donate two protons or hydrogen ions per molecule), reacts with sodium hydroxide (NaOH), a strong base. This reaction can be represented by the equation: \[\text{(COOH)}_2\text{2H}_2\text{O} + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 4\text{H}_2\text{O}\] This equation shows that each molecule of oxalic acid can react with two molecules of NaOH to achieve neutralization.
Understanding this stoichiometric relationship is crucial because it helps determine how much of the NaOH solution is needed to fully neutralize the given oxalic acid solution. It's all about balancing the number of equivalents, which brings us to equivalencies.

Oxalic Acid Dihydrate

Oxalic acid dihydrate is a specific form of oxalic acid, characterized by two water molecules associated with each oxalic acid molecule. Its chemical formula is \((\text{COOH})_2 \cdot 2\text{H}_2\text{O}\), and it has a molar mass of about 126 grams per mole.
This compound is commonly used in acid-base titration labs to explore basic principles of chemistry. Understanding its molar mass is crucial, as it affects how we calculate the number of moles present in a given mass, which is essential for accurate titration calculations.
To find the number of moles of oxalic acid dihydrate, you have to divide the mass (in grams) by its molar mass. This step was carried out in the exercise where 6.3 grams was used with a molar mass of 126 grams per mole, resulting in 0.05 moles of oxalic acid dihydrate.
Remember, the amount of oxalic acid in a solution is directly related to the degree of reaction it can undergo with a base during titration. Factors like molar mass help in calculating this yield.

Molarity and Normality Calculations

When dealing with acid-base titration, it's essential to understand the concepts of molarity and normality, as these measurements help in determining the concentration of solutions used.
**Molarity (M)** refers to the concentration of a solution expressed as moles of solute per liter of solution. For oxalic acid dihydrate, the molarity was calculated as 0.2 M using the formula: \[\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\] In the exercise, this calculation helped determine how concentrated the oxalic acid solution was after dissolution.
**Normality (N)**, on the other hand, accounts for how reactive a solution is, particularly involving acids and bases. It is defined as the number of equivalents of solute per liter of solution. For instance, in our titration, the normality of NaOH was given as 0.1 N. This means there are 0.1 equivalents of NaOH per liter.
For oxalic acid, since it releases two protons, its normality is twice its molarity when fully ionized. Calculating using normality is crucial because it ensures the correct amount of titrant is used to reach the equivalence point in a titration, making it a pivotal part of quantitative chemical analysis.