Problem 128 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(\mathrm{CO}_{2}\) dissolves in water, the following equilibrium is established. \(\mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CO}_{3}^{-}\) The equilibrium constant is \(3.8 \times 10^{-7}\) and at \(\mathrm{pH}=6.0\), the ratio \(\frac{\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{CO}_{2}\right]}\) will be (a) \(3.8 \times 10^{-13}\) (b) \(3.8 \times 10^{-1}\) (c) \(6.0\) (d) \(13.4\)

Step-by-Step Solution

Answer

The ratio is \(3.8 \times 10^{-1}\), so the answer is (b).

Step 1: Understanding the Equilibrium Expression

Firstly, understand that the equilibrium constant (\( K_a \)) for the reaction is given by \( K_a = \frac{[ ext{H}_3 ext{O}^+] [ ext{CO}_3^{-}]}{[ ext{CO}_2] imes [ ext{H}_2 ext{O}]^2} \). However, given the context, we simplify the term with water.

Step 2: Using Given pH to Find Hydronium Ion Concentration

Given that the pH is 6.0, we can calculate the hydronium ion concentration as \([ ext{H}_3 ext{O}^+] = 10^{-pH} = 10^{-6}\).

Step 3: Re-evaluate Equilibrium Expression

Since the equilibrium involves the concentrations of \([ ext{H}_3 ext{O}^+]\) and \([ ext{HCO}_3^-]\), we can rewrite the equilibrium expression to focus on the bicarbonate ion: \( [ ext{HCO}_3^-] = K_a \times \frac{[ ext{CO}_2] imes [ ext{H}_2 ext{O}]^2}{[ ext{H}_3 ext{O}^+]} \).

Step 4: Simplifying with Given Constants

Since water concentration remains effectively constant, we can treat it as part of the equilibrium constant for simplification: \( K_a = 3.8 \times 10^{-7} \). We now use \( [ ext{H}_3 ext{O}^+] = 10^{-6} \) to find the ratio: \( \frac{[ ext{HCO}_3^-]}{[ ext{CO}_2]} = \frac{K_a}{[ ext{H}_3 ext{O}^+]} = \frac{3.8 \times 10^{-7}}{10^{-6}} \).

Step 5: Calculate the Ratio

Perform the division: \( \frac{3.8 \times 10^{-7}}{10^{-6}} = 3.8 \times 10^{-1} \).

Step 6: Selecting the Correct Answer

Compare the calculated ratio with the provided options. The correct answer that matches \( 3.8 \times 10^{-1} \) is option (b).

Key Concepts

Equilibrium Constant

When we talk about chemical reactions in equilibrium, we're referring to a state where the forward and reverse reactions occur at the same rate. This means that the concentrations of the reactants and products remain stable over time. The equilibrium constant, symbolized as \( K \), is a numerical value that helps us understand this balance. It is expressed based on the concentration of products over reactants, each raised to the power of their respective coefficients in the balanced chemical equation.

For the reaction \( \mathrm{CO}_2 + 2 \mathrm{H}_2O \rightleftharpoons \mathrm{H}_3O^+ + \mathrm{CO}_3^- \), the equilibrium constant \( K_a \) is defined as follows:

• \( K_a = \frac{[\mathrm{H}_3\mathrm{O}^+] [\mathrm{CO}_3^{-}]}{[\mathrm{CO}_2]} \)

However, in this reaction, water is both a reactant and the solvent. Since its concentration doesn't change significantly, we often don't include it when calculating the equilibrium constant. Thus, the equilibrium constant simplifies, helping to analyze situations better, like in this case where we focus only the main reactants and products.

Understanding \( K \) allows you to predict the direction in which the reaction will proceed to achieve equilibrium. If \( K \) is much greater than 1, the products are favored at equilibrium; if much less than 1, the reactants are favored. Here, with \( K_a = 3.8 \times 10^{-7} \), the result suggests the equilibrium strongly favors the reactants.

pH Calculation

The pH of a solution is a measure of its acidity or basicity. It is calculated using the concentration of hydrogen ions (\( [\mathrm{H}_3\mathrm{O}^+] \)), which are sometimes simplified from \( [\mathrm{H}^+] \). To find the pH, you use the formula:

• \( \mathrm{pH} = -\log_{10}[\mathrm{H}_3\mathrm{O}^+] \)

In the given problem, the solution has a pH of 6.0, which is considered mildly acidic. Using this pH, we can compute the concentration of hydronium ions:

\[ [\mathrm{H}_3\mathrm{O}^+] = 10^{-\mathrm{pH}} = 10^{-6} \]

Here, the hydronium ion concentration is \( 10^{-6} \) M, a key piece of information for further calculations. By understanding how to calculate the pH, one can gauge acidity changes in a solution. It's a fundamental concept in chemistry important for interpreting reactions like the equilibrium of dissolved \( \mathrm{CO}_2 \) and the formation of bicarbonate and carbonate ions.

Bicarbonate Ion Concentration

To find the concentration of bicarbonate ions \( [\mathrm{HCO}_3^-] \), given the equilibrium constant and the pH of the solution, you need to adjust the equilibrium expression. Here, the focus should be re-directed onto the bicarbonate ion and carbon dioxide as opposed to the complete reaction set.Recalling the equilibrium expression:

• \( [\mathrm{HCO}_3^-] = K_a \times \frac{[\mathrm{CO}_2]}{[\mathrm{H}_3\mathrm{O}^+]} \)

Given:

• \( K_a = 3.8 \times 10^{-7} \)
• \( [\mathrm{H}_3\mathrm{O}^+] = 10^{-6} \)

We rearrange this to find the ratio of bicarbonate to carbon dioxide:\[ \frac{[\mathrm{HCO}_3^-]}{[\mathrm{CO}_2]} = \frac{K_a}{[\mathrm{H}_3\mathrm{O}^+]} = \frac{3.8 \times 10^{-7}}{10^{-6}} = 3.8 \times 10^{-1} \]
This calculation shows the relationship between the bicarbonate ion concentration and carbon dioxide, as given by the options. By calculating and comparing, option (b) \( 3.8 \times 10^{-1} \) is identified as the correct choice. Such calculations are essential for understanding buffer systems and predicting the behavior of biological systems like blood, where bicarbonate ions play a vital role.