Problem 58 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The average concentration of \(\mathrm{SO}_{2}\) in the atmosphere over a city on a certain day is \(10 \mathrm{ppm}\), when the average temperature is \(298 \mathrm{~K}\). Given that the solubility of \(\mathrm{SO}_{2}\) in water at \(298 \mathrm{~K}\) is \(1.3653\) moles litre \(^{-1}\) and the \(\mathrm{p} K_{a}\) of \(\mathrm{H}_{2} \mathrm{SO}_{3}\) is \(1.92\), estimate the \(\mathrm{pH}\) of rain on that day.

Step-by-Step Solution

Answer

The pH of the rain is approximately 1.8.

Step 1: Understanding the Problem

We need to estimate the pH of rainwater given the concentration of \(\mathrm{SO}_2\) gas in the air, which dissolves in rainwater. We're also given the solubility of \(\mathrm{SO}_2\) in water and the \(\mathrm{p}K_a\) of the resulting \(\mathrm{H_2SO_3}\) (sulfurous acid) in water.

Step 2: Calculate Dissolved \(\mathrm{SO}_2\) Concentration

Since \(\mathrm{SO}_2\) in air is \(10 \mathrm{ppm}\), we have \(10 \times 10^{-6}\) \(\text{moles/liter}\) in water (since \(1 \text{ppm} = 1 \text{mg/L}\) approximately and \(\text{SO}_2\) molar mass is \(64 \text{g/mol}\)). Using the solubility data, \(1.3653 \text{ moles/L}\), this implies that's the total concentration of \(\mathrm{SO}_2\) in water.

Step 3: Understand the Reaction

When \(\mathrm{SO}_2\) dissolves in water, it forms \(\mathrm{H_2SO_3}\), a weak acid. The dissociation of \(\mathrm{H_2SO_3}\) is given by the equation: \(\mathrm{H_2SO_3} \rightleftharpoons \mathrm{H^+} + \mathrm{HSO_3^-}\).

Step 4: Use \(\mathrm{pK_a}\) to Find \([\mathrm{H^+}]\)

Use the \(\mathrm{pK_a}\) value to find the equilibrium concentration of \(\mathrm{H^+}\). Since the first dissociation is controlled by \(pK_a = 1.92\), we use \(pKa = -\log K_a\) to find \(K_a = 10^{-1.92}\).

Step 5: Apply Equilibrium Expression

Using \([\mathrm{H^+}] = [\mathrm{HSO_3^-}] = \sqrt{K_a \times c_0}\), where \(c_0\) is the initial \(\mathrm{SO_2}\) concentration of \(1.3653 \text{ moles/L}\). Calculate \([\mathrm{H^+}] = \sqrt{10^{-1.92} \times 1.3653}\).

Step 6: Calculate pH

Use the expression \(\mathrm{pH} = -\log[\mathrm{H^+}]\) to find the pH of the solution once \([\mathrm{H^+}]\) is known from the previous step.

Key Concepts

SO2 Solubility

Sulfur dioxide ( SO_2 ) plays an important role in the chemistry of our atmosphere, especially when it comes to air pollution and acid rain formation. SO_2 solubility refers to how well SO_2 gas can dissolve in water, such as rainwater. Understanding solubility is crucial when trying to predict chemical behavior in different environmental settings. When SO_2 dissolves in water, it interacts with water molecules to form sulfurous acid ( H_2SO_3 ). The solubility of SO_2 in water at a given temperature indicates how much SO_2 can be held in a liter of water until it is saturated. In this exercise, the solubility given is 1.3653 ext{ moles/L} at 298 ext{ K} , which means that this is the maximum concentration of SO_2 that can exist in the rainwater under these conditions.

• High solubility indicates that a lot of SO_2 can dissolve in water, increasing potential acidity.
• Low solubility means less SO_2 can dissolve, thus less acidity.

The given concentration of 10 ext{ ppm} of SO_2 in the air suggests that enough SO_2 is available to reach the solubility limit when the rain collects this gas. Understanding solubility helps us anticipate and calculate subsequent chemical reactions in natural environmental processes.

Acid Dissociation

Once SO_2 is dissolved in water, it forms sulfurous acid (H_2SO_3). This is where acid dissociation comes into play. Dissociation refers to the splitting of a compound into smaller components, such as ions, in water. In this context, sulfurous acid dissociates into hydrogen ions (H^+) and bisulfite ions (HSO_3^-):\[\mathrm{H_2SO_3} \rightleftharpoons \mathrm{H^+} + \mathrm{HSO_3^-}\]The amount that H_2SO_3 dissociates into these ions depends on the acid's strength, represented by its \mathrm{p}K_a value, which is typically derived from \log ext{-}based measure of how easily an acid gives up its H^+ ion. A low \mathrm{p}K_a value, such as 1.92 here, indicates a stronger tendency for the acid to dissociate.
Knowing the dissociation tendency allows us to calculate the concentration of H^+ ions, which directly affects the acidity (or pH) of the solution. This dissociated H^+ ion concentration can be determined using the equation:\[\mathrm{[H^+]} = \sqrt{K_a \times c_0}\]where K_a is the acid dissociation constant correlating to the given \mathrm{p}K_a, and \(c_0\) is the initial concentration of the acid. Thus, understanding acid dissociation is a key step in determining the pH of the rainwater.

Equilibrium Constant

The equilibrium constant (K_a) is a critical concept in chemistry, representing the balance point of a reversible chemical reaction. In the case of H_2SO_3 dissociation, K_a defines how much the sulfurous acid will dissociate into H^+ and HSO_3^- ions at equilibrium. For the reaction:\[\mathrm{H_2SO_3} \rightleftharpoons \mathrm{H^+} + \mathrm{HSO_3^-}\]The equilibrium constant K_a is given as:\[K_a = 10^{-\mathrm{p}K_a}\]In our example, with a \mathrm{p}K_a of 1.92, we calculate K_a to be 10^{-1.92}, describing the tendency of the acid to dissociate.
In general, a large K_a value indicates strong acid dissociation, while a small K_a value indicates weak dissociation. For weak acids like H_2SO_3, understanding this constant helps predict pH changes in the solution. Using K_a and the concentration of undissociated acid, we calculate [H^+] concentration, leading to the subsequent pH.
To find the pH, calculate the hydrogen ion concentration from K_a and the remaining concentration of the initial solution, then use the formula \[pH = -\log{[\mathrm{H^+]}}\]. Thus, the equilibrium constant is essential to understanding and predicting the course of chemical reactions and concentrations in solutions.