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

The answer is (a) 40 mL.

Step 1: Find the molarity of oxalic acid solution

The molar mass of oxalic acid dihydrate \( \text{H}_2\text{C}_2\text{O}_4\cdot2\text{H}_2\text{O} \) is calculated as follows: \( 2(1+12+16) + 2(18) = 126 \text{ g/mol} \). The amount of oxalic acid in grams is given as \( 6.3 \text{ g} \). Moles of oxalic acid = \( \frac{6.3}{126} = 0.05 \text{ mol} \). Since this is dissolved in \( 250 \text{ mL} \), the molarity \( M \) is \( \frac{0.05}{0.25} = 0.2 \text{ M} \).

Step 2: Determine moles of oxalic acid in 10 mL of solution

Since we have a \( 0.2 \text{ M} \) solution, the moles of oxalic acid present in \( 10 \text{ mL} \) (which is \( 0.01 \text{ L} \)) can be calculated as: \( 0.2 \times 0.01 = 0.002 \text{ moles} \).

Step 3: Ratio of neutralization with NaOH

Oxalic acid \((\text{H}_2\text{C}_2\text{O}_4)\) is a diprotic acid, meaning each mole can furnish two moles of \( \text{H}^+ \). Therefore, \( 0.002 \text{ moles} \) of oxalic acid will react with \( 0.004 \text{ moles} \) of \( \text{NaOH} \).

Step 4: Calculate volume of NaOH needed

The normality of \( \text{NaOH} \) solution is given as \( 0.1 \text{ N} \), which is the same as \( 0.1 \text{ M} \) for a monoprotic base like \( \text{NaOH} \). Use the relationship: \[ \text{Moles} = \text{Normality} \times \text{Volume in L} \]. We have \( 0.004 \text{ moles} \) of \( \text{NaOH} \) required, so \( \text{Volume} = \frac{0.004}{0.1} = 0.04 \text{ L} \) or \( 40 \text{ mL} \).

Key Concepts

Molarity Calculation

Molarity is a key concept in chemistry, especially when dealing with solutions and reactions. It expresses the concentration of a solute in a solution. Molarity (M) is defined as the number of moles of solute dissolved in one liter of solution.

To calculate molarity, you first need to determine the number of moles of the solute. This is done by taking the mass of the solute in grams and dividing it by its molar mass, which is the mass of one mole of that substance. For oxalic acid dihydrate, we calculated the molar mass as 126 g/mol. With 6.3 g of this compound, we found the moles by dividing 6.3 by 126, giving us 0.05 moles.

Next, since the solution's final volume is 250 mL, which is 0.25 L, the molarity is calculated by dividing the moles by the volume in liters: 0.05 moles / 0.25 L = 0.2 M. This molarity indicates the concentration of oxalic acid in the solution.

Neutralization Reaction

A neutralization reaction is a chemical reaction between an acid and a base, resulting in the formation of water and a salt. This is an important reaction in chemistry, and understanding it is crucial for solving titration problems.

In this exercise, oxalic acid reacts with sodium hydroxide (NaOH). Oxalic acid is a diprotic acid, which means each molecule can donate two protons (H⁺ ions). Because of this, one mole of oxalic acid can react with two moles of NaOH. When dealing with such reactions, it's vital to consider how many acidic protons are available and how many moles of the base each can neutralize.

For example, in our exercise, 0.002 moles of oxalic acid were calculated from a 10 mL sample of a 0.2 M solution. Given its diprotic nature, it can react with twice as many moles of NaOH, totaling 0.004 moles of NaOH needed for complete neutralization.

Normality

Normality is another concept of concentration that is particularly useful when dealing with reactions involving acids and bases. It conveys the gram equivalent weight of a solute per liter of solution. Normality (N) is especially effective in titration calculations where the reaction involves equivalent exchanges of electrons or protons.

Normality relates directly to molarity but takes into account the number of equivalents per mole of solute. For monoprotic acids and bases, the normality is equivalent to molarity. However, for diprotic acids like oxalic acid, wherein each mole contributes two equivalents (because it can donate two protons), the relationship becomes crucial.

In the exercise, a NaOH solution of 0.1 N is used. This is the same as 0.1 M in NaOH's case because it is a monoprotic base, providing one equivalent per mole. Knowing the normality helps in calculating the exact volume of titrant needed to reach neutralization, as we used to determine the 40 mL requirement in the original problem.