Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

A 5 molar solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is diluted from 1 litre to a volume of 10 litres, the normality of the solution will be (a) \(0.5 \mathrm{~N}\) (b) \(1 \mathrm{~N}\) (c) \(2.5 \mathrm{~N}\) (d) \(5 \mathrm{~N}\)

Step-by-Step Solution

Answer

The normality of the solution after dilution is (b) 1 N.

Step 1: Understanding Molarity

Molarity (M) indicates the number of moles of solute (here, \(\mathrm{H}_{2}\mathrm{SO}_{4}\)) per litre of solution. In this problem, the molarity of the \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution is 5 M, meaning there are 5 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) per litre.

Step 2: Understanding Normality

Normality (N) relates to the equivalent concentration of the solution. For \(\mathrm{H}_{2}\mathrm{SO}_{4}\), which is a diprotic acid (can donate 2 protons per molecule), the normality is twice the molarity. Thus, the original normality is \(5 \text{ M} \times 2 = 10 \text{ N}\).

Step 3: Calculate Dilution Effect

Dilution affects concentration. The formula \(C_1V_1 = C_2V_2\) applies, where \(C_1\) and \(V_1\) are the initial concentration and volume, and \(C_2\) and \(V_2\) are the final concentration and volume. Here, \(C_1 = 10 \text{ N}\), \(V_1 = 1 \text{ L}\), and \(V_2 = 10 \text{ L}\). Solve for \(C_2\):\[C_2 = \frac{C_1 \times V_1}{V_2} = \frac{10 \text{ N} \times 1 \text{ L}}{10 \text{ L}} = 1 \text{ N}\]

Step 4: Conclusion

The calculation shows that after the dilution, the new normality of the \(\mathrm{H}_{2}\mathrm{SO}_{4}\) solution is \(1 \text{ N}\).

Key Concepts

Molarity

When we discuss molarity in chemistry, we're looking at a measure of concentration. It's all about how many moles of a substance (the solute) are present in a liter of solution. Imagine you're making lemonade: molarity tells you how concentrated your lemonade is by specifying how much lemon juice is in each liter of the drink.
In the context of the exercise, the molarity of the sulfuric acid (\(\mathrm{H}_{2}\mathrm{SO}_{4}\)) solution is 5 M. This means there are 5 moles of sulfuric acid in every liter. Understanding molarity is crucial because it helps determine how strong or weak a solution is. This knowledge is useful when we need to mix solutions or when reactions depend on concentration levels.
Key points about molarity:

• It is expressed in moles per liter (mol/L).
• High molarity indicates a high concentration of solute.
• It provides insight into how much of a substance is available for a chemical reaction.

Equivalence in Acids

Equivalence in acids is about how much acid reacts in terms of donating protons or hydrogen ions. In chemistry, the concept of normality comes into play here, especially for acids and bases.
For acids like \(\mathrm{H}_{2}\mathrm{SO}_{4}\), which can donate more than one proton, normality is a way of describing concentration based on the reactive capacity or equivalence factor of the acid. This is where the diprotic nature of sulfuric acid comes in: it can release two protons for every molecule. That's twice the action compared to a monoprotic acid that releases one proton per molecule.
To find normality for sulfuric acid, multiply its molarity by the number of protons the acid can donate:

• Normality \(\text{(N)} = \text{Molarity (M)} \times \text{n (number of protons)}\).
• For \(\mathrm{H}_{2}\mathrm{SO}_{4}\), \(\text{N} = 5 \text{ M} \times 2 = 10 \text{ N}\).

Understanding equivalence in acids is crucial for reactions where an exact amount of substance is needed, like titrations or preparing standardized solutions.

Dilution Formula

The dilution formula is a handy tool when we want to make a solution less concentrated. It's often used in labs to ensure that solutions have the right strength for experiments without altering their inherent chemical properties.
Here's the dilution equation: \(C_1V_1 = C_2V_2\). What does this mean?

• \(C_1\) is the initial concentration.
• \(V_1\) is the initial volume.
• \(C_2\) is the final concentration.
• \(V_2\) is the final volume.

Basically, you multiply the initial concentration and volume to give a new, diluted concentration after you've added more solvent (usually water).
For the sulfuric acid solution, you began with a normality of 10 N at 1 liter. Diluting this to 10 liters means using the formula to find the final normality:

• \(C_2 = \frac{C_1 \times V_1}{V_2} = \frac{10 \text{ N} \times 1 \text{ L}}{10 \text{ L}} = 1 \text{ N}\).

In essence, dilution doesn't change the total amount of solute but spreads it out over a larger volume, decreasing the concentration.