Problem 52 - A Complete Resource Book in Chemistry for JEE Main

Question

\(0.1914 \mathrm{~g}\) of an organic acid is dissolved in about 20 \(\mathrm{mL}\) of water \(25 \mathrm{~mL}\) of \(0.12 \mathrm{~N} \mathrm{NaOH}\) is required for the complete neutralization of the acid solution. The equivalent weight of the acid is (a) \(63.8\) (b) \(64.0\) (c) \(65.8\) (d) \(66.2\)

Step-by-Step Solution

Answer

The equivalent weight of the acid is 63.8, option (a).

Step 1: Understanding Neutralization

When an acid reacts with a base, it forms water and a salt in a neutralization reaction. Here, the organic acid is completely neutralized by the sodium hydroxide (NaOH) solution.

Step 2: Calculate Moles of NaOH

Normality (N) refers to the gram equivalent per liter of solution. Since we have 25 mL of 0.12 N NaOH, we first convert the volume of NaOH to liters:\[ 25 \, \text{mL} = 0.025 \, \text{L} \]Then, we calculate the equivalents of NaOH:\[ \text{Equivalents of NaOH} = 0.025 \, \text{L} \times 0.12 \, \text{N} = 0.003 \, \text{equivalents} \]

Step 3: Relate Equivalents of NaOH to Acid

Since the acid is completely neutralized, the equivalents of NaOH used is equal to the equivalents of the acid.

Step 4: Calculate Equivalent Weight of the Acid

The equivalent weight of the acid is calculated by dividing the mass of the acid by the number of equivalents:\[ \text{Equivalent weight of acid} = \frac{0.1914 \, \text{g}}{0.003 \, \text{equivalents}} = 63.8 \, \text{g/equiv} \]

Step 5: Determine the Correct Option

Compare the calculated equivalent weight of the acid (63.8) with the given options. The correct match is (a) \(63.8\).

Key Concepts

Neutralization Reaction

In chemistry, a neutralization reaction occurs when an acid and a base react together to form water and a salt. This reaction is important for many applications, including titration processes where the concentration of a solution is determined.
For example, when an organic acid is neutralized by sodium hydroxide (\( ext{NaOH}\)), the reaction can be seen as:

• \( ext{Acid} + ext{NaOH} \rightarrow ext{Salt} + ext{Water}\)

During this process, the hydrogen ions (\( ext{H}^+\)) from the acid combine with the hydroxide ions (\( ext{OH}^-\)) from the base to form water (\( ext{H}_2 ext{O}\)). This is the basic principle that dictates how acids can neutralize bases.
In the exercise given, \(25 ext{ mL}\) of \(0.12 ext{ N NaOH}\) is used to neutralize the organic acid. This complete reaction ensures that all the acid's acidic components are balanced by the base, which allows for an accurate determination of the acid's equivalent weight.

Mole Concept

The mole concept is fundamental in chemistry, as it helps relate the amount of substance to its chemical properties. One mole is equivalent to Avogadro's number, which is \(6.022 imes 10^{23}\) of any chemical entity, whether atoms, molecules, or ions.
In this exercise, we use the mole concept in terms of equivalents. The normality (\( ext{N}\)) of a solution is a way of expressing concentration that considers this concept, as it is defined as the number of equivalent units per liter of solution.
When you calculate the number of equivalents of \( ext{NaOH}\) used, you convert milliliters to liters, then multiply by the normality:

• \( ext{Equivalents of NaOH} = 0.025 ext{ L} imes 0.12 ext{ N} = 0.003 ext{ equivalents}\)

This key calculation cleverly utilizes the mole concept to help find out how many moles of acid have reacted in the neutralization process. This relationship ultimately assists in determining the equivalent weight of the organic acid.

Organic Acid Titration

Titration is a common laboratory method of quantitative chemical analysis used to determine the concentration of an identified analyte. In organic chemistry, it is often used for measuring the amount of acid present in a given substance.
When performing an organic acid titration with \( ext{NaOH}\), we gradually add the \( ext{NaOH}\) solution to the acid until we reach the endpoint, where the acid is completely neutralized.
This titration allows you to calculate the unknown concentration of the acid. In our example, \(0.1914 ext{ g}\) of the acid was neutralized by \(0.003 ext{ equivalents}\) of \( ext{NaOH}\). Using those values, you calculate the acid's equivalent weight:

• \( ext{Equivalent weight of acid} = rac{0.1914 ext{ g}}{0.003 ext{ equivalents}} = 63.8 ext{ g/equiv}\)

Titration is the tool that helped find this figure. By monitoring how much titrant is required to neutralize the analyte, we gain detailed insights about the properties of the acid itself. This procedure is beneficial for a variety of organic acids and applications.