Problem 112 - A Complete Resource Book in Chemistry for JEE Main

Question

An aqueous solution of \(6.3 \mathrm{~g}\) oxalic acid dihydrate 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) \(40 \mathrm{~mL}\) (b) \(20 \mathrm{~mL}\) (c) \(10 \mathrm{~mL}\) (d) \(4 \mathrm{~mL}\)

Step-by-Step Solution

Answer

40 mL of 0.1 N NaOH is needed to neutralize 10 mL of the oxalic acid solution.

Step 1: Determine Moles of Oxalic Acid Dihydrate

First, we need to calculate the moles of oxalic acid dihydrate present in the solution. The molar mass of oxalic acid dihydrate, \((\text{C}_2\text{H}_2\text{O}_4\cdot 2\text{H}_2\text{O})\), is computed as follows: \(2(12.01) + 2(1.01) + 4(16.00) + 2(18.02)\). This equals \(126.07\) g/mol. Thus, the moles of oxalic acid dihydrate are given by: \(\frac{6.3}{126.07} = 0.05\) moles.

Step 2: Determine Molarity of Oxalic Acid Solution

Now calculate the molarity of the oxalic acid in the solution. Since we have 0.05 moles in 250 mL, the molarity is \(\frac{0.05}{0.250} = 0.2\) M. This indicates that the concentration of oxalic acid is 0.2 mol/L.

Step 3: Calculate Moles in 10 mL of Solution

Next, find out how many moles of oxalic acid are contained in 10 mL of the 0.2 M solution. This is calculated as follows: \(0.2 \times \frac{10}{1000} = 0.002\) moles of oxalic acid in 10 mL.

Step 4: Determine Equivalent Moles of NaOH Required

Oxalic acid reacts with NaOH in a 1:2 molar ratio: \(\text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O}\). So, for 0.002 moles of oxalic acid, we need \(0.002 \times 2 = 0.004\) equivalents of NaOH.

Step 5: Calculate Volume of NaOH Solution Required

Given the normality of NaOH is 0.1 N, use the equation \(\text{Number of equivalents} = \text{Normality} \times \text{Volume (L)}\). Rearrange to find the volume: \(0.004 = 0.1 \times \text{Volume (L)}\), resulting in \(\text{Volume (L)} = \frac{0.004}{0.1} = 0.04\) L, or 40 mL.

Key Concepts

Oxalic Acid Dihydrate

Oxalic acid dihydrate, with the chemical formula \( ext{C}_2 ext{H}_2 ext{O}_4\cdot 2 ext{H}_2 ext{O}\), is a hydrate form of oxalic acid. It means that in this compound, two water molecules are coordinated with oxalic acid. Hydrates like this often appear as crystalline solids.

To understand its role in neutralization reactions, it's essential to look at its molar mass. Oxalic acid dihydrate has a molar mass of 126.07 g/mol. This is calculated by adding the atomic masses of all atoms present:

• Carbon (C): 12.01 g/mol, 2 atoms
• Hydrogen (H): 1.01 g/mol, 2 atoms
• Oxygen (O): 16.00 g/mol, 4 atoms
• 2 Water molecules (\(2\times\text{H}_2\text{O}\)): each has a contribution of 18.02 g/mol

Understanding the molar mass is crucial since it helps calculate how much of the substance is involved in a chemical reaction.

In this exercise, oxalic acid dihydrate serves as the substance whose complete neutralization by NaOH is under investigation.

NaOH Titration

In this exercise, we're using a titration method to neutralize oxalic acid dihydrate with sodium hydroxide (NaOH), a strong base. During a titration, one solution is gradually added to another until the chemical reaction is complete.

In the context of this reaction, the balanced equation is:\[\text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O}\]This equation shows that one molecule of oxalic acid reacts with two molecules of NaOH. Therefore, the titration must be carried out until the equivalence point is reached, where both reactants are completely reacted.

Because this exercise involves a 2:1 molar ratio, for every mole of oxalic acid, we require two moles of NaOH. Here, the role of the titration is to precisely determine the volume of NaOH needed to completely neutralize a known amount of oxalic acid.

Understanding how to properly conduct a titration is essential in various scientific fields, particularly in chemistry and biology, as it allows precise determination of concentration and purity of substances.

Molarity Calculations

Molarity is a way to express the concentration of a solute in a solution. It's expressed as the number of moles of solute per liter of solution (mol/L). Understanding molarity is key for carrying out titrations and many chemical reactions.

In this problem, the calculated molarity of the oxalic acid solution is 0.2 M, indicating there are 0.2 moles of oxalic acid per liter of solution.

For calculations:

• First, determine the moles of solute: We have 6.3 g of oxalic acid dihydrate, equating to 0.05 moles.
• Then, convert the volume of the solution from milliliters to liters: 250 mL = 0.250 L.
• Finally, divide the number of moles by the volume in liters: \[ \text{Molarity} = \frac{0.05}{0.250} = 0.2 \text{ M}\]

Such molarity calculations are fundamental in laboratory settings as they ensure reactions are conducted with precise concentrations, resulting in accurate experimental outcomes.