Question

The concentration of \(\mathrm{SO}_{2}\) in the troposphere over a certain region is 0.16 ppm by volume. The gas dissolves in rainwater as follows: $$ \mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}^{+}(a q)+\mathrm{HSO}_{3}^{-}(a q) $$ Given that the equilibrium constant for the preceding reaction is \(1.3 \times 10^{-2},\) calculate the \(\mathrm{pH}\) of the rainwater. Assume that the reaction does not affect the partial pressure of \(\mathrm{SO}_{2}\)

Short answer

The pH of the rainwater is approximately 3.53.

Step-by-step solution

Understand the reaction and equilibrium constant

The chemical reaction given is \(\mathrm{SO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}^{+}(aq) + \mathrm{HSO}_{3}^{-}(aq)\). The equilibrium constant \(K_c\) for this reaction is provided as \(1.3 \times 10^{-2}\).

Express the equilibrium relationship

The equilibrium constant expression for the reaction is \(K_c = \frac{[\mathrm{H}^{+}][\mathrm{HSO}_{3}^{-}]}{[\mathrm{SO}_{2}(aq)]}\). Here, \([\mathrm{H}^{+}]\) and \([\mathrm{HSO}_{3}^{-}]\) are concentrations of the ions, and \([\mathrm{SO}_{2}(aq)]\) is the concentration of \(\mathrm{SO}_{2}\) dissolved in water.

Convert ppm to molarity for \(\mathrm{SO}_{2}\)

The concentration of \(\mathrm{SO}_{2}\) in ppm by volume needs to be converted to molarity. Assuming ideal gas behavior, 0.16 ppm by volume is equivalent to 0.16 moles of \(\mathrm{SO}_{2}\) per 1,000,000 liters of air. Considering that in water these volumes correspond directly as a gas volume fraction: \([\mathrm{SO}_{2}] = \frac{0.16}{24.45} = 6.54 \times 10^{-6} \text{ M}\) under standard conditions where 1 mole of gas occupies 24.45 L.

Solve for \([\mathrm{H}^{+}]\) and \([\mathrm{HSO}_{3}^{-}]\)

Assuming \([\mathrm{H}^{+}] = x = [\mathrm{HSO}_{3}^{-}]\) due to the stoichiometry of the reaction, plug into the equilibrium equation: \(1.3 \times 10^{-2} = \frac{x^2}{6.54 \times 10^{-6}}\). Solving yields \(x^2 = 1.3 \times 10^{-2} \times 6.54 \times 10^{-6}\), so \(x \approx 2.93 \times 10^{-4} \text{ M}\).

Calculate the pH of the solution

The pH of a solution is calculated using the formula \(\text{pH} = -\log_{10}([\mathrm{H}^{+}])\). Substitute \([\mathrm{H}^{+}] = 2.93 \times 10^{-4}\), then \(\text{pH} = -\log_{10}(2.93 \times 10^{-4}) \approx 3.53\).

Key concepts

Equilibrium Constant

In chemistry, an equilibrium constant, denoted as \( K_c \), is a number expressing the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. This ratio remains constant at a given temperature.
For the reaction \[ \mathrm{SO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{3}^{-}(aq) \]the equilibrium constant is defined as\[ K_c = \frac{[\mathrm{H}^{+}][\mathrm{HSO}_{3}^{-}]}{[\mathrm{SO}_{2}]} \]This equation illustrates how the concentrations of hydrogen ions and bisulfite ions compare to the sulfur dioxide concentration that is still present in the solution. If you know this constant, it helps predict the direction of the reaction and the position of equilibrium. It's crucial for controlling the conditions under which a reaction takes place effectively.

Sulfur Dioxide

Sulfur dioxide (SO₂) is a colorless gas with a sharp odor. It's produced naturally by volcanic eruptions and as an industrial by-product through the burning of fossil fuels.
- In the atmosphere, it can contribute to pollution, especially when mixed with water vapor.- It's a precursor to acid rain, as it reacts readily with water to form acidic solutions.When dissolved in rainwater, sulfur dioxide undergoes a reversible reaction, producing hydrogen ions \( \mathrm{H}^{+} \) and bisulfite ions \( \mathrm{HSO}_{3}^{-} \).This reaction is crucial because the formation of these ions leads to a decrease in pH, making the rainwater more acidic. Understanding the behavior of sulfur dioxide in the environment helps in environmental protection and reducing acid rain.

Acid Rain

Acid rain is a major environmental issue resulting from the atmospheric pollution of sulfur dioxide and nitrogen oxides. These gases undergo chemical transformations, forming acidic compounds that dissolve in rainwater.
As sulfur dioxide interacts with water vapor, it generates acids that lower the pH of precipitation to harmful levels, affecting ecosystems, structures, and aquatic habitats.

• Acid rain accelerates the decay of buildings and monuments, especially those constructed from limestone and marble.
• It affects forest productivity and aquatic life by altering the natural pH balance of soils and water bodies.

Preventive measures include reducing emissions of sulfur dioxide through cleaner fuels and emission control technologies. Understanding the chemical reactions involved in acid rain formation helps in designing strategies to minimize its impact.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In the given reaction, stoichiometry helps determine the proportions of reactants and products at equilibrium.
In the equation\[ \mathrm{SO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{3}^{-}(aq) \],- The stoichiometry of the reaction indicates that one molecule of sulfur dioxide and one molecule of water yields one hydrogen ion and one bisulfite ion.- Based on this 1:1 ratio, if you know the concentration of sulfur dioxide, you can calculate the concentrations of hydrogen and bisulfite ions at equilibrium, assuming all \( \mathrm{SO}_{2} \) converts.This concept is essential for tasks like determining the pH of solutions, as seen in the given exercise. Mastery of stoichiometry enables you to solve complex chemical problems by understanding these ratios.