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

(a) 40 mL

Step 1: Determine the Molarity of Oxalic Acid Solution

Oxalic acid dihydrate has a molar mass of \(126 \mathrm{~g/mol}\). The given mass of oxalic acid is \(6.3 \mathrm{~g}\). Hence, the moles of oxalic acid in solution are calculated as follows:\[ \text{Moles of oxalic acid} = \frac{6.3 \mathrm{~g}}{126 \mathrm{~g/mol}} = 0.05 \text{ moles} \]The volume of the solution is \(250 \mathrm{~mL} = 0.25 \mathrm{~L}\), thus the molarity \(M\) of the oxalic acid solution is:\[ M = \frac{0.05 \text{ moles}}{0.25 \mathrm{~L}} = 0.2 \mathrm{~M} \]

Step 2: Calculate Normality of Oxalic Acid Solution

Since oxalic acid (\(H_2C_2O_4\cdot2H_2O\)) is a diprotic acid (it can donate two protons), its normality is double its molarity. Therefore:\[ N = 2 \times 0.2 \mathrm{~M} = 0.4 \mathrm{~N} \]

Step 3: Determine the Volume of NaOH Required

From the normality equation for titration, \(N_1V_1 = N_2V_2\), where- \(N_1 = 0.4 \mathrm{~N}\) (normality of oxalic acid solution),- \(V_1 = 10 \mathrm{~mL}\) (volume of oxalic acid solution),- \(N_2 = 0.1 \mathrm{~N}\) (normality of NaOH),- \(V_2\) is the volume of NaOH required.Plug in the known values:\[ 0.4 \mathrm{~N} \times 10 \mathrm{~mL} = 0.1 \mathrm{~N} \times V_2 \]Solving for \(V_2\), we get:\[ V_2 = \frac{0.4 \times 10}{0.1} = 40 \mathrm{~mL} \]

Step 4: Validate the Final Answer

After the calculations, the volume of \(0.1 \mathrm{~N} \mathrm{NaOH}\) required to completely neutralize \(10 \mathrm{~mL}\) of the oxalic acid solution is determined to be \(40 \mathrm{~mL}\). Thus, the correct option is (a).

Key Concepts

Oxalic Acid

Oxalic acid, known scientifically as \(H_2C_2O_4\cdot2H_2O\), is a compound that plays a significant role in titration studies due to its acidic properties. It is characterized as a diprotic acid, which means that it has the ability to donate two protons (\(H^+\)) in an aqueous solution.
This intrinsic property allows oxalic acid to engage in multiple titration reactions, making it useful in a variety of chemical processes. The solid form of oxalic acid is often used in the form of a dihydrate, meaning it incorporates two water molecules. The molar mass of oxalic acid dihydrate is calculated by adding the molar mass of oxalic acid and the mass of these water molecules, leading to a total molar mass of \(126 \text{ g/mol}\).
Understanding its characteristics is crucial when working to determine concentrations in solutions, as seen in titration experiments where precise measurements are key to obtaining accurate results.

Normality

Normality is a measure of concentration that is primarily used in acid-base chemistry. It accounts for the number of reactive species in a given volume of solution. In the case of oxalic acid, which is a diprotic acid, the normality is twice its molarity because each molecule of oxalic acid can donate two \(H^+\) ions.
This is illustrated in the calculation of normality where, if the molarity is found to be \(0.2 \text{ M}\), the normality becomes \(0.4 \text{ N}\) (\(N = 2 \times 0.2 \text{ M}\)). Normality is especially important in titrations where it helps in finding the stoichiometrically correct amount of a titrant required to complete the reaction.

• Normality offers a straightforward way to match the volumes of the acid and base in titration reactions.
• Using normality simplifies calculations in titrations, especially for polyprotic acids like oxalic acid.

It is a practical measure that underscores the importance of understanding chemical equivalents rather than just molecular concentrations.

Molarity

Molarity is a fundamental concept in chemistry that defines the number of moles of a solute present per liter of solution. It is denoted by \(M\) and is crucial in calculations involving dilutions and reactions, including titration.
In our exercise, to find the molarity of oxalic acid, the number of moles was first determined by dividing the mass of oxalic acid (\(6.3 \text{ g}\)) by its molar mass (\(126 \text{ g/mol}\)). This gave us \(0.05\) moles of oxalic acid. This quantity was then divided by the volume of the solution in liters, yielding a molarity of \(0.2 \text{ M}\).
Molarity is critical because:

• It provides a direct measure of the concentration, allowing comparisons between different solutions.
• It aids in calculating dilutions and stoichiometric relationships in reactions.
• It is used alongside normality in titrations to determine the correct volume of titrant required.

Understanding molarity is key to mastering analytical techniques in chemistry, providing precise instructions to balance chemical equations efficiently.

NaOH Neutralization

Neutralization reactions involve an acid and a base reacting to form water and a salt. Sodium hydroxide (NaOH) is a common base used in these experiments. During the titration of an oxalic acid solution with NaOH, the hydrogen ions from the acid react with the hydroxide ions from the base to yield water.
In this specific exercise, the relationship between the acid and base is quantified using the equation \(N_1V_1 = N_2V_2\). Here, \(N_1\) and \(V_1\) represent the normality and volume of the oxalic acid, respectively, while \(N_2\) and \(V_2\) relate to the NaOH solution.
Applying the principle of normality:

• The calculated normality of oxalic acid solution was \(0.4 \text{ N}\), while the normality of NaOH was \(0.1 \text{ N}\).
• By rearranging parts of the formula, it was determined that 40 mL of NaOH was necessary to achieve complete neutralization of \(10 \text{ mL}\) of oxalic acid solution.

Titration utilizing NaOH is an essential analytical method, helping chemists exactly quantify how much base is needed to neutralize an acid, an invaluable tool in scientific and industrial applications.