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

Question

\(1 \mathrm{~g}\) 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: Understanding Basicity

The basicity of an acid is the number of hydrogen ions ( ext{H}^+) it can donate in a reaction. To find the basicity, we need to relate it to the equivalent mass of the acid.

Step 2: Calculating Moles of KOH Used

First, find the number of moles of KOH used by dividing the mass of KOH by its molar mass. The molar mass of KOH is made up of 39.1 (K) + 16 (O) + 1 (H) = 56.1 ext{ g/mol}. \[\text{Moles of KOH} = \frac{0.768 \text{ g}}{56.1 \text{ g/mol}} \approx 0.0137 \text{ mol}\]

Step 3: Finding the Moles of Acid

Since 1 gram of acid is used, calculate its moles using its molar mass. The molar mass of \(\text{C}_6\text{H}_{10}\text{O}_4\) is:\[6(12.01) + 10(1.01) + 4(16.00) = 146.14 \text{ g/mol}\]\[\text{Moles of acid} = \frac{1 \text{ g}}{146.14 \text{ g/mol}} \approx 0.00684 \text{ mol}\]

Step 4: Determining the Basicity of the Acid

The reaction between the acid and KOH is such that each equivalent of the acid reacts with one mole of KOH. The number of moles of KOH is given by \( n \times \text{moles of acid} \), where \( n \) is the basicity we want to find.Setting up the equation:\[0.0137 \text{ mol (KOH)} = n \times 0.00684 \text{ mol (acid)}\]Solving for \( n \):\[n = \frac{0.0137}{0.00684} \approx 2\]

Step 5: Conclusion about Basicity

Therefore, the basicity of the acid \(\text{C}_6\text{H}_{10}\text{O}_4\) is 2, which means it can donate 2 hydrogen ions.

Key Concepts

Acid-Base Neutralization

When an acid and a base come together, they neutralize each other to form water and a salt. This process is known as acid-base neutralization. During this reaction, the hydrogen ions ( H^+ ) from the acid combine with the hydroxide ions ( OH^- ) from the base to produce water ( H_2O ). In our exercise, C_6H_{10}O_4 acts as the acid, and KOH is the base.

Understanding this concept is necessary because it helps explain how substances change and interact with each other in chemical reactions. Neutralization reactions are particularly important in many chemical processes, such as titrations, and are used in a variety of industrial, laboratory, and everyday applications.

In the exercise, complete neutralization is achieved when the right amount of KOH is mixed with C_6H_{10}O_4 , resulting in no leftover acid or base.

Moles Calculation

Calculating moles is fundamental in chemistry for expressing amounts of a chemical substance. To determine the moles of a substance, you divide the mass of the substance by its molar mass. The molar mass is the weight of one mole of a substance, usually expressed in grams per mole.

For instance, in the given exercise, we calculated the moles of KOH using its mass and molar mass:

• Molar mass of KOH: K (39.1 g/mol) + O (16.0 g/mol) + H (1.0 g/mol) = 56.1 g/mol
• Moles of KOH: \(\frac{0.768\, \text{g}}{56.1\, \text{g/mol}} \approx 0.0137\, \text{mol}\)

Similarly, we find the moles of the acid C_6H_{10}O_4 using its molar mass and given mass.

• Molar mass of \(C_6H_{10}O_4\): \(6\times12.01 + 10\times1.01 + 4\times16.00 = 146.14\, \text{g/mol}\)
• Moles of acid used: \(\frac{1\, \text{g}}{146.14\, \text{g/mol}} \approx 0.00684\, \text{mol}\)

Understanding mole calculations can simplify solving various chemistry problems, as it allows us to easily relate masses to number of particles.

Chemical Stoichiometry

Chemical stoichiometry is the study of the quantitative relationships in a balanced chemical equation. It enables us to predict the amount of products formed or reactants consumed in a given reaction.

In the context of the given problem, stoichiometry helps in determining how many equivalents of C_6H_{10}O_4 react with KOH. This specific reaction demonstrates that to completely neutralize 1 gram of C_6H_{10}O_4, precise moles of KOH are required.

The stoichiometric relationship in our exercise shows that each equivalent of the acid reacts with one mole of KOH. Thus, by setting up the equation \(0.0137\, \text{mol (KOH)} = n \times 0.00684\, \text{mol (acid)}\), we solve for the basicity \(n\).

• Solving for basicity: \(n = \frac{0.0137}{0.00684} \approx 2\)

The calculated basicity of 2 implies that the acid can donate 2 hydrogen ions in the reaction. Mastering stoichiometry is crucial in chemistry, as it helps predict and measure reactants and products accurately in any chemical reaction.