Problem 177 - Chemistry

Question

The atmospheric sulfur dioxide \(\left(\mathrm{SO}_{2}\right)\) concentration over a certain region is 0.12 ppm by volume. Calculate the \(\mathrm{pH}\) of the rainwater due to this pollutant. Assume that the dissolution of \(\mathrm{SO}_{2}\) does not affect its pressure. $$ \left(K_{\mathrm{a}} \text { for } \mathrm{H}_{2} \mathrm{SO}_{3}=1.3 \times 10^{-2} .\right) $$

Step-by-Step Solution

Answer

The pH of the rainwater is approximately 3.81.

Step 1: Understanding the Problem

We need to calculate the pH of rainwater given a sulfur dioxide concentration of 0.12 ppm (parts per million) by volume. The dissolution of \(\mathrm{SO}_2\) contributes to the formation of \(\mathrm{H}_2\mathrm{SO}_3\) (sulfurous acid) in water, which is a weak acid with \(K_a = 1.3 \times 10^{-2}\).

Step 2: Converting ppm to Molarity

Since 0.12 ppm by volume is approximately equal to 0.12 mg/L for gases, we convert this concentration to molarity. \(\mathrm{SO}_2\) has a molar mass of 64.07 g/mol, so:\[\text{Molarity} = \frac{0.12 \text{ mg/L}}{64.07 \text{ g/mol}} \times \frac{1 \text{ g}}{1000 \text{ mg}} \approx 1.87 \times 10^{-6} \text{ M}\]

Step 3: Writing the Dissolution and Ionization Equations

\(\mathrm{SO}_2\) dissolves in water and forms sulfurous acid:\[\mathrm{SO}_2 + \mathrm{H_2O} \rightarrow \mathrm{H_2SO_3}\]\(\mathrm{H_2SO_3}\) dissociates in water:\[\mathrm{H_2SO_3} \rightleftharpoons \mathrm{H}^+ + \mathrm{HSO_3}^-\]

Step 4: Using the Acid Dissociation Constant

For the weak acid \(\mathrm{H_2SO_3}\), the dissociation constant is given by:\[K_a = \frac{[\mathrm{H}^+][\mathrm{HSO_3}^-]}{[\mathrm{H_2SO_3}]} = 1.3 \times 10^{-2}\]Assume \([\mathrm{H}^+] = [\mathrm{HSO_3}^-] = x\) and initial \([\mathrm{H_2SO_3}] = 1.87 \times 10^{-6}\text{ M}\).

Step 5: Solving for Hydrogen Ion Concentration

Substitute into the equation to find \(x\):\[K_a = \frac{x^2}{1.87 \times 10^{-6} - x} = 1.3 \times 10^{-2}\]Assuming \(x\) is small compared to \(1.87 \times 10^{-6}\), we solve:\[x^2 \approx 1.3 \times 10^{-2} \times 1.87 \times 10^{-6}\]\[x^2 \approx 2.43 \times 10^{-8}\]\[x \approx \sqrt{2.43 \times 10^{-8}} \approx 1.56 \times 10^{-4} \text{ M}\]

Step 6: Calculating the pH

The pH is given by:\[\text{pH} = -\log_{10}[\mathrm{H}^+]\]Substituting the value of \([\mathrm{H}^+]\):\[\text{pH} = -\log_{10}(1.56 \times 10^{-4}) \approx 3.81\]

Step 7: Conclusion

The pH of the rainwater due to \(\mathrm{SO}_2\) pollution is approximately 3.81, indicating acidic rain.

Key Concepts

Sulfur Dioxide

Sulfur dioxide, abbreviated as \(\mathrm{SO}_2\), is a colorless gas with a sharp, pungent odor. It is released during the burning of fossil fuels and various industrial processes. Once in the atmosphere, it can dissolve in water to form weak acids. Understanding its role in atmospheric chemistry is crucial, especially when discussing environmental issues such as air pollution and climate change.
When \(\mathrm{SO}_2\) enters the atmosphere, it can have direct and indirect effects.

• Direct effect: It can cause respiratory problems and other health issues for living organisms.
• Indirect effect: It contributes to the formation of acid rain when it reacts with water, oxygen, and other chemicals in the sky.

By understanding the behavior of sulfur dioxide, we can better assess and manage its impact on both the environment and human health.

Acid Rain

Acid rain forms when sulfur dioxide (\(\mathrm{SO}_2\)) and nitrogen oxides react with water and other substances in the atmosphere. This reaction produces sulfuric acid and nitric acid, which then mix with rainwater, resulting in precipitation that is more acidic than normal rain.
Normal rain has a slightly acidic pH of about 5.6 due to the presence of carbonic acid from dissolved carbon dioxide. However, when \(\mathrm{SO}_2\) dissolves in raindrops, it forms sulfurous acid (\(\mathrm{H}_2\mathrm{SO}_3\)), lowering the pH further.
Acid rain has significant environmental impacts. It can:

• Damage vegetation by leaching nutrients from the soil.
• Harm aquatic life by lowering the pH of water bodies.
• Corrode buildings and structures.

This makes understanding acid rain formation and its effects vital for environmental conservation efforts.

Weak Acid Dissociation

Weak acids, unlike strong acids, do not completely dissociate in water. Sulfurous acid (\(\mathrm{H}_2\mathrm{SO}_3\)), formed from \(\mathrm{SO}_2\) in rainwater, is a classic example of a weak acid. It partially ionizes in water to produce hydrogen ions (\(\mathrm{H}^+\)) and bisulfite ions (\(\mathrm{HSO}_3^-\)).
The extent of this dissociation is characterized by an acid dissociation constant, \(K_a\). For weak acids, \(K_a\) is usually less than 1 and indicates the acid's strength. The value of \(K_a\) helps us to calculate the hydrogen ion concentration, which in turn dictates the pH of the solution.
The equation \(\mathrm{H}_2\mathrm{SO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HSO}_3^-\) represents the equilibrium between the undissociated acid and its ions. Understanding weak acid dissociation is crucial for calculations related to pH and for predicting the behavior of acids in different environments.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction in a reversible chemical process. In the case of weak acids like \(\mathrm{H}_2\mathrm{SO}_3\), this concept is key to understanding how dissociation happens in equilibrium.
At equilibrium, the concentrations of the reactants (the undissociated acid) and the products (the ions) remain constant. The expression for the acid dissociation constant \(K_a\) is derived from this equilibrium, calculated by dividing the product of the concentrations of the ions by the concentration of the undissociated acid.
Understanding chemical equilibrium involves

• Recognizing that the equilibrium can shift due to changes in concentration, pressure, or temperature, according to Le Chatelier's principle.
• Appreciating that it helps in understanding other concepts like buffer solutions, solubility, and reaction quotients.

In environmental chemistry, equilibrium concepts help us predict the extent to which \(\mathrm{SO}_2\) and other pollutants affect natural systems.