Problem 162 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{g}\) of an acid \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\) required \(0.768 \mathrm{~g}\) of \(\mathrm{KOH}\) for complete neutralization. Determine the basicity of acid.

Step-by-Step Solution

Answer

The basicity of the acid is 2.

Step 1: Write the neutralization reaction

The neutralization reaction can be described as \( \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4} + n \mathrm{~KOH} \rightarrow \) products, where \( n \) is the number of moles of KOH reacting with the acid. This is a preliminary setup to determine the basicity of the acid.

Step 2: Calculate moles of KOH

First, calculate the moles of KOH used: \[ \text{Moles of KOH} = \frac{0.768\, \mathrm{g}}{56\, \mathrm{g/mol}} = 0.0137\, \mathrm{mol} \] Where 56 g/mol is the molar mass of KOH.

Step 3: Relate moles of acid to moles of KOH

Since the molar ratio in the balanced equation will tell us the basicity, which is equal to the moles of KOH divided by moles of the acid. If we consider the moles of acid as \( x \), then \( nx = 0.0137 \). The value of \( n \) (basicity) will be determined when we find \( x \).

Step 4: Determining the molar mass of the acid

Calculate the molar mass of the acid \( \mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4} \): \[ 6(12) + 10(1) + 4(16) = 146\, \mathrm{g/mol} \]

Step 5: Calculating moles of the acid

Assuming you used \( 1 \) gram of acid (wrongly omitted in the problem setup): \[ \text{Moles of acid} = \frac{1.000\, \mathrm{g}}{146\, \mathrm{g/mol}} = 0.00685\, \mathrm{mol} \]

Step 6: Calculate the basicity of the acid

The basicity of an acid is the number of \( \mathrm{OH^-} \) ions required to neutralize the acid, i.e., moles of KOH divided by moles of acid.\[ n = \frac{0.0137}{0.00685} = 2 \]Thus, the basicity of the acid is 2, meaning H2 equivalent is present in one molecule of the acid.

Key Concepts

Understanding Moles Calculation

In chemistry, moles are a fundamental unit of measurement that allow us to compare amounts of different substances on a microscopic level. One mole corresponds to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities, such as atoms or molecules.
Calculating moles is necessary when analyzing chemical reactions. To find the number of moles, you divide the mass of the substance by its molar mass:

• \(\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\)

In our example, we calculated the moles of \(\text{KOH}\) as \(0.0137\) moles using its mass and molar mass. Moles are crucial for stoichiometry, as they help us determine the proportions in which reactants participate in chemical reactions.

Deciphering Molar Mass

The molar mass of a compound is the mass of one mole of its molecules. It is expressed in grams per mole (g/mol) and calculated by summing the atomic masses of all atoms present in the molecule.
To calculate the molar mass of a substance, identify the number of each type of atom in a molecule and multiply by the atomic masses found on the periodic table:

• \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\): \(6(12) + 10(1) + 4(16) = 146\,\mathrm{g/mol}\)

In the step-by-step solution, we identified the molar mass of the acid \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\) as 146 g/mol, a vital piece of information for calculating moles and proceeding with the reaction analysis.

Grasping Basicity

Basicity refers to the number of replaceable hydrogen ions \((\mathrm{H}^+)\) an acid molecule can donate to a base during a neutralization reaction. It is related to the number of \(\mathrm{OH}^-\) ions the acid can accept, which determines how many moles of a base are needed to completely neutralize a mole of acid.
In our scenario, the basicity was determined by the ratio of moles of \(\mathrm{KOH}\) to the moles of \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\). Given that \(n = 2\), the acid's basicity is 2. This result implies that two \(\mathrm{OH}^-\) ions are required to neutralize each molecule of the acid.

Neutralization Reaction Insight

A neutralization reaction occurs when an acid and a base react to form water and a salt. This reaction is significant as it determines fundamental acid-base properties and is often used in titration experiments.
The neutralization reaction for \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4}\) with \(\mathrm{KOH}\) was crucial for determining the acid's basicity. In its balanced form, the reaction can be represented as:

• \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{4} + n\,\mathrm{KOH} \rightarrow \text{products}\)

From the previous steps, we established that \(n = 2\), confirming that two moles of \(\mathrm{KOH}\) are necessary to neutralize one mole of the acid. This type of reaction helps us understand the stoichiometry involved and highlights the acid-base interaction's dynamic nature.