Problem 56 - A Complete Resource Book in Chemistry for JEE Main

Question

\(50 \mathrm{~mL}\) of \(10 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}, 25 \mathrm{~mL}\) of \(12 \mathrm{~N} \mathrm{HCl}\) and \(40 \mathrm{~mL}\) of \(5 \mathrm{~N} \mathrm{HNO}_{3}\) are mixed and the volume of the mixture is made \(1000 \mathrm{~mL}\) by adding water. The normality of resulting solution will be (a) \(9 \mathrm{~N}\) (b) \(4 \mathrm{~N}\) (c) \(1 \mathrm{~N}\) (d) \(2 \mathrm{~N}\)

Step-by-Step Solution

Answer

The normality of the resulting solution is \(1 \mathrm{~N}\).

Step 1: Determine total milliequivalents of each acid

To find milliequivalents, multiply the normality of each acid by its volume in mL. For \( \mathrm{H}_2\mathrm{SO}_4 \), the milliequivalents are \( 10 \times 50 = 500 \). For \( \mathrm{HCl} \), it is \( 12 \times 25 = 300 \). For \( \mathrm{HNO}_3 \), it is \( 5 \times 40 = 200 \).

Step 2: Calculate total milliequivalents in the mixture

Sum the milliequivalents from each acid to get the total milliequivalents in the mixture: \( 500 + 300 + 200 = 1000 \) milliequivalents.

Step 3: Calculate the normality of the resulting solution

Normality is defined as the number of gram-equivalents of solute per liter of solution. Use the formula: \( N = \frac{\text{Total milliequivalents}}{\text{Total volume in mL}} \times 1000 \). Substitute the known values: \( N = \frac{1000}{1000} \times 1000 = 1 \mathrm{~N} \).

Step 4: Confirm and choose the correct answer

The calculated normality of the resulting solution is \(1 \mathrm{~N}\), which matches option (c). Therefore, the correct answer is (c) \(1 \mathrm{~N}\).

Key Concepts

Milliequivalents

Understanding milliequivalents is essential in chemistry, especially when calculating the normality of solutions. Milliequivalents (usually noted as meq) are a way of measuring the reactive capacity of ions or molecules in a solution. They are commonly used because they allow chemists to compare the eq of different solutions easily.
To find the milliequivalents, you need two pieces of information: the normality (N) of the solution and its volume in milliliters (mL). Multiply these two to get milliequivalents.

• Formula: Milliequivalents = Normality × Volume in mL

Using this formula, you can calculate the contribution of each component in a solution mixture. This step simplifies the process of determining the resulting normality of a final mixture. Remember to sum the milliequivalents of all solutes involved to get the total milliequivalents for the comprehensive solution.
Understanding milliequivalents helps in various fields, including pharmacy, medicine, and chemistry, where precise measurements of reactive capacities are crucial.

Sulfuric Acid (H2SO4)

Sulfuric acid, chemical formula H₂SO₄, is a highly important industrial acid used in various applications, from fertilizers to car batteries. Its strong acidic properties make it a component of interest when calculating normality.
In solutions like the one in our problem, H₂SO₄ contributes significantly to the total milliequivalents due to its high normality.

• For example, with a normality of 10 N and a volume of 50 mL, as used in the solution, you find the milliequivalents to be 500.

This contribution is crucial in determining the overall acidity or basicity of mixtures.
Sulfuric acid's dissociation in water involves two protons, adding complexity to calculations of the total number of equivalents. This factor of dissociation is part of why H₂SO₄ is a vital component in normality chemistry problems.

Hydrochloric Acid (HCl)

Hydrochloric acid (HCl) is a common strong acid used in both industrial and laboratory settings. Its chemical simplicity, with just one hydrogen ion per molecule, makes calculations involving HCl straightforward.
In normality calculations, HCl's contribution is based on its concentration and volume.

• For instance, in our scenario, a 25 mL solution of 12 N HCl amounts to 300 milliequivalents.

This computation is a critical step in understanding the overall acidity of mixed solutions.
HCl is often used as a benchmark for comparing the strength of other acids, due to its complete ionization in water. This property ensures that all the HCl molecules are fully available to contribute to reactions, making it a predictable and reliable component in normality calculations.

Nitric Acid (HNO3)

Nitric acid (HNO₃) is another strong acid commonly encountered in chemistry. Known for its applications in making fertilizers and explosives, HNO₃'s normality calculations are crucial for properly understanding its role in reactions.
In the given solution scenario, a 40 mL of 5 N HNO₃ results in 200 milliequivalents.

• This represents a part of the total reactive capacity of the chemical mixture.

Nitric acid's reactivity is primarily due to its complete dissociation in water, which ensures that each molecule contributes wholeheartedly to the solution's overall acidity.
The formula used for HNO₃ in solution calculations highlights the importance of understanding acid strengths and their respective roles in chemical reactions. Its milliequivalent contribution sees it as a partner with other acids to heighten overall solution normality.