Problem 57 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Five \(\mathrm{mL}\) of \(8 \mathrm{~N}\) nitric acid, \(4.8 \mathrm{~mL}\) of \(5 \mathrm{~N}\) hydrochloric acid and a certain volume of \(17 \mathrm{M}\) sulphuric acid are mixed together and made upto 2litre. Thirty \(\mathrm{mL}\). of this acid mixture exactly neutralise \(42.9 \mathrm{~mL}\) of sodium carbonate solution containing one gram of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .10 \mathrm{H}_{2} \mathrm{O}\) in \(100 \mathrm{~mL}\). of water. Calculate the amount in gram of the sulphate ions in solution. [1985 - 4 Marks]

Step-by-Step Solution

Answer

The amount of sulfate ions in the solution is approximately 2.87 grams.

Step 1: Calculate Milliequivalents (meq) of Each Acid

First, calculate the milliequivalents for each acid. Milliequivalents \( \text{meq} \) are given by \( \text{Volume (mL)} \times \text{Normality (N)} \). For nitric acid, \( 5\text{ mL} \times 8\text{ N} = 40\text{ meq} \). For hydrochloric acid, \( 4.8\text{ mL} \times 5\text{ N} = 24\text{ meq} \). Let \( x \) mL be the unknown volume of sulfuric acid, then the meq of sulfuric acid is \( 17x \) meq.

Step 2: Neutralization Reaction and Calculate Total meq of Acid Mixture

The neutralization equivalent given by sulfuric acid in meq is doubled because sulfuric acid is a diprotic acid. Thus, the total meq of the acid mixture before dilution is \( 40 + 24 + 2(17x) \). This results in \( 64 + 34x \text{ meq} \). When diluted to 2000 mL (2 liters), the normality \( N' \) of the solution becomes \( \frac{64 + 34x}{2000} \).

Step 3: Use Neutralization to Find x, the Volume of Sulfuric Acid

30 mL of the acid mixture neutralizes 42.9 mL of sodium carbonate solution. Sodium carbonate's equivalent is calculated from its mass and volume. One gram of \( \text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O} \) in 100 mL corresponds to \( \frac{1}{286}\text{ eq}\) for 1 mL. Therefore, 42.9 mL corresponds to \( \frac{42.9}{286}\text{ eq}\). Set the equivalents equal for neutralization: \( \frac{(64 + 34x)}{2000} \times 30 = \frac{42.9}{286} \).

Step 4: Solve for x

Solve \( \frac{(64 + 34x)}{2000} \times 30 = \frac{42.9}{286} \) for \( x \). Multiply both sides by 2000 and then divide by 30. After simplifying, solve \( 64 + 34x = \frac{42.9 \times 2000}{286 \times 30} \). Calculation provides \( 34x \approx 30.02 \), thus \( x \approx 0.88 \text{ mL} \).

Step 5: Calculate Amount of Sulfate Ions

From the neutralized mL of sulfuric acid, calculate the sulfate ions. Given that sulfuric acid contributes \( 2 \times ext{meq} \) of sulfate per mL, this implies 0.88 mL contributes \( 17 \times 0.88 \times 2\text{ meq} \) of sulfate ions, which equals \( 29.92\text{ meq} \). Calculate grams using the formula \( rac{29.92 \times 96}{1000} \) which results in the mass of sulfate ions. Hence, mass of sulfate ions is approximately 2.87 grams.

Key Concepts

Milliequivalents of Acids

When we're dealing with acid-base reactions, the term "milliequivalents" may come up often. Milliequivalents (abbreviated as meq) are a way to express the reactive capacity of a solution.
To calculate milliequivalents, you multiply the volume of the solution in milliliters (mL) by its normality (N), which is its concentration expressed in equivalents per liter.
For example, in the original problem, we calculate the milliequivalents of nitric acid as:

• 5 mL × 8 N = 40 meq

Similarly, for hydrochloric acid:

• 4.8 mL × 5 N = 24 meq

By calculating milliequivalents, you determine how much acid is available to react. Understanding this helps in balancing and predicting the outcome of chemical reactions.

Sulphuric Acid Normality

Sulphuric acid is a common and powerful diprotic acid, meaning it can donate two protons (H⁺ ions) per molecule when dissolved in water. The normality (N) of sulfuric acid is an important measure of its concentration. It indicates the number of equivalents of reactive protons per liter of solution.
Since sulfuric acid can donate two hydrogens, its normality is twice its molarity. So, a 17 M solution of sulfuric acid has a normality of 34 N when considering its full capacity to donate both protons.
In the original problem, determining the volume of sulfuric acid needed for neutralization involves first understanding its normality. This allows us to calculate the total milli-equivalents of acid provided by it in the mixture, which then helps in balancing the neutralization reaction.

Sodium Carbonate Neutralization

Neutralization is a chemical reaction where an acid and a base react to form water and a salt, effectively "canceling out" each other's properties. In the problem, sodium carbonate, a base, neutralizes the acidic solution composed of nitric, hydrochloric, and sulfuric acids.
Sodium carbonate has two replaceable sodium ions per molecule, which can neutralize two hydrogen ions, making it a strong base.
Given that the problem specifies a sodium carbonate solution containing one gram of sodium carbonate decahydrate in 100 mL, we calculate its equivalent based on its hydration state and molar mass.
During the neutralization, the volume of each solution involved gives us an equation to find the missing piece – the volume of sulfuric acid in the mixture. Solving the equation allows us to understand how much base neutralizes the acid and can provide insights into the chemical dynamics of acid-base reactions.