Problem 113 - 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 volume of NaOH required is 40 mL, so option (a) is correct.

Step 1: Calculate Moles of Oxalic Acid

First, determine the number of moles of oxalic acid dihydrate. The molar mass of oxalic acid dihydrate (H2C2O4·2H2O) is calculated as: \[ 2(1.01) + 2(14.01) + 4(16.00) + 4(18.02) = 126.07 \, \text{g/mol} \]. Therefore, the moles of oxalic acid dihydrate in 6.3 g are \[ \frac{6.3}{126.07} \approx 0.05 \, \text{mol} \].

Step 2: Calculate Molarity of the Oxalic Acid Solution

Since the volume of the solution is 250 mL (or 0.25 L), the molarity (M) of the oxalic acid solution is given by: \[ M = \frac{0.05 \text{ moles}}{0.25 \text{ L}} = 0.2 \, \text{M} \].

Step 3: Determine Moles in 10 mL of Solution

To find the number of moles of oxalic acid in 10 mL of the solution, use the molarity calculated: \[ 0.2 \, \text{M} \times \frac{10}{1000} \text{ L} = 0.002 \, \text{moles} \].

Step 4: Determine Equivalent Weight of Oxalic Acid

Oxalic acid is a diprotic acid, meaning it can donate two protons. Therefore, the equivalent weight for reaction is half the molar mass: \[ \frac{126.07}{2} = 63.035 \, \text{g/equiv} \]. However, since we already calculated the number of moles, we know each mole reacts with 2 equivalents of NaOH.

Step 5: Calculate Normality of Oxalic Acid Solution

The normality of the oxalic acid solution, which has two replaceable hydrogen ions, is twice its molarity: \[ 0.2 \, \text{M} \times 2 = 0.4 \, \text{N} \].

Step 6: Apply the Neutralization Equation

According to the neutralization reaction \( \text{N}_1 \times \text{V}_1 = \text{N}_2 \times \text{V}_2 \), where \( \text{N}_1 = 0.4 \, \text{N}, \text{V}_1 = 10 \, \text{mL}, \text{N}_2 = 0.1 \, \text{N} \), solve for \( \text{V}_2 \): \[ 0.4 \times 10 = 0.1 \times \text{V}_2 \].

Step 7: Calculate Volume of NaOH Required

Solving the equation \( 0.4 \times 10 = 0.1 \times \text{V}_2 \) yields: \[ \text{V}_2 = \frac{4}{0.1} = 40 \, \text{mL} \].

Key Concepts

Oxalic Acid Dihydrate

Oxalic acid dihydrate, represented as \( \text{H}_2\text{C}_2\text{O}_4 \cdot 2\text{H}_2\text{O} \), is a crystalline solid that often appears in its hydrated form. This means that each molecule of oxalic acid comes with two water molecules attached. You'll find oxalic acid in various applications, such as in bleaching or cleaning compounds. It's particularly interesting in chemistry due to its dual functionality as an acid. This functionality allows it to effectively donate protons in reactions, making it a diprotic acid. In calculations, the inclusion of water in the molar mass is essential, contributing significantly to the compound's weight, which must be considered when finding moles or equivalents.

Molarity Calculation

To comprehend molarity, think of it as a measure of concentration. It's the number of moles of a solute in one liter of solution. In our exercise, calculating the molarity begins with the moles of oxalic acid. We used the molar mass of oxalic acid dihydrate \(126.07 \, \text{g/mol}\) to find moles from the given mass. Once the moles were found (0.05 mol in this case), we divided them by the volume of the solution in liters (0.25 L) to find the molarity: \( M = \frac{0.05 \, \text{mol}}{0.25 \, \text{L}} = 0.2 \, \text{M} \). Remember, molarity offers a clear understanding of how concentrated a solution is.

Normality Calculation

Normality offers another way to express concentration, focusing on the equivalents of a reactive species in a liter of solution. For acids like oxalic acid dihydrate, which can donate more than one proton, normality provides a more accurate representation of their reactive capacity. Since oxalic acid is diprotic, its ability to donate two hydrogen ions makes its normality twice its molarity. So, for our solution with a molarity of 0.2 M, the normality becomes: \( 0.2 \, \text{M} \times 2 = 0.4 \, \text{N} \). Normality is particularly useful in titration calculations where the reaction taking place involves these multiple protons or electrons.

Equivalent Weight

Equivalent weight can be a bit of a confusing concept at first glance, but it's crucial for stoichiometric calculations. It's the weight of a substance that supplies or reacts with one mole of hydrogen ions in acid-base reactions, or electrons in redox reactions. Since oxalic acid dihydrate is diprotic, its equivalent weight is half its molecular weight. This is calculated as: \( \frac{126.07}{2} = 63.035 \, \text{g/equiv} \). This value tells us how much of oxalic acid is needed to have the same proton-donating capacity as 1 gram-equivalent of any monobasic acid.

Stoichiometry

Stoichiometry is the math behind chemistry that lets us calculate how much of each reactant is needed or how much product is formed in a reaction. In the context of our neutralization reaction, stoichiometry uses the normality of solutions to determine reactant volumes. The formula \( \text{N}_1 \times \text{V}_1 = \text{N}_2 \times \text{V}_2 \) helps us solve for unknowns in these calculations. Given that \( \text{N}_1 = 0.4 \, \text{N} \), \( \text{V}_1 = 10 \, \text{mL} \), and \( \text{N}_2 = 0.1 \, \text{N} \), we applied it to find \( \text{V}_2 \) necessary to neutralize the acid. The result \( \text{V}_2 = 40 \, \text{mL} \) showed exactly how much NaOH is needed. Understanding stoichiometry unlocks the ability to predict reaction outcomes accurately.