Problem 89 - Chemistry: The Central Science in SI Units, Global Edition

Question

The Henry's law constant for \(\mathrm{CO}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is \(3.4 \times 10^{-4} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}(\mathbf{a})\) What is the solubility of \(\mathrm{CO}_{2}\) in water at this temperature if the solution is in contact with air at normal atmospheric pressure? (b) Assume that all of this \(\mathrm{CO}_{2}\) is in the form of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) produced by the reaction between \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}:\) $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$ What is the pH of this solution?

Step-by-Step Solution

Answer

The solubility of \(\mathrm{CO}_{2}\) in water at \(25^{\circ}\mathrm{C}\) and normal atmospheric pressure can be calculated as \(C = (3.4 \times 10^{-4} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}) \times 101325\,\text{Pa}\). Assuming all \(\mathrm{CO}_{2}\) is in the form of \(\mathrm{H}_{2}\mathrm{CO}_{3}\), the pH of the solution can be calculated using the dissociation constant, \(\text{pH} = -\log_{10}([\mathrm{H}^+])\), where \([\mathrm{H}^+] = \sqrt{K_a \times [\mathrm{H}_{2}\mathrm{CO}_{3}]}\), and \(K_a\) is the dissociation constant of \(\mathrm{H}_{2}\mathrm{CO}_{3}\), which is approximately \(4.3 \times 10^{-7}\).

Step 1: Calculate the solubility of \(\mathrm{CO}_{2}\) in water

First, we use the Henry's law formula to calculate the solubility of \(\mathrm{CO}_{2}\) in water at \(25^{\circ}\mathrm{C}\) and normal atmospheric pressure. The formula for Henry's law is: $$ C = k_H \times P $$ Where \(C\) is the solubility of the gas in the solution, \(k_H\) is the Henry's law constant for the gas, and \(P\) is the partial pressure of the gas. For this exercise, we have the Henry's law constant for \(\mathrm{CO}_{2}\) (\(k_H\)) as \(3.4 \times 10^{-4} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\), and the partial pressure of \(\mathrm{CO}_{2}\) in the atmosphere can be approximated as normal atmospheric pressure, which is 101325 Pa. Now, we can calculate the solubility of \(\mathrm{CO}_{2}\) in water: $$ C = (3.4 \times 10^{-4} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}) \times 101325\,\text{Pa} $$

Step 2: Find the concentration of \(\mathrm{H}_{2}\mathrm{CO}_{3}\)

As we assumed that all of the \(\mathrm{CO}_{2}\) is in the form of \(\mathrm{H}_{2}\mathrm{CO}_{3}\), the concentration of \(\mathrm{H}_{2}\mathrm{CO}_{3}\) in the solution is the same as the solubility of \(\mathrm{CO}_{2}\) in water (\(C\)). So: $$ [\mathrm{H}_{2}\mathrm{CO}_{3}] = C $$

Step 3: Calculate the pH of the solution

To calculate the pH of the solution, we first need to know the dissociation constant (\(K_a\)) for \(\mathrm{H}_{2}\mathrm{CO}_{3}\). The \(K_a\) value for \(\mathrm{H}_{2}\mathrm{CO}_{3}\) is approximately \(4.3 \times 10^{-7}\). The \(K_a\) expression for \(\mathrm{H}_{2}\mathrm{CO}_{3}\) is: $$ K_a = \frac{[\mathrm{H}^+][\mathrm{HCO}_3^-]}{[\mathrm{H}_{2}\mathrm{CO}_{3}]} $$ Since the solution is quite dilute, we can assume that \([\mathrm{H}^+] \approx [\mathrm{HCO}_3^-]\). Thus, the equation simplifies to: $$ K_a = \frac{[\mathrm{H}^+]^2}{[\mathrm{H}_{2}\mathrm{CO}_{3}]} $$ We can solve for \([\mathrm{H}^+]\): $$ [\mathrm{H}^+] = \sqrt{K_a \times [\mathrm{H}_{2}\mathrm{CO}_{3}]} $$ Finally, we can calculate the pH of the solution using the formula: $$ \text{pH} = -\log_{10}([\mathrm{H}^+]) $$

Key Concepts

Solubility

Solubility refers to the ability of a gas, liquid, or solid to dissolve in a solvent, forming a homogeneous solution. In our case, it's about carbon dioxide (\(\mathrm{CO}_{2}\)) dissolving in water. The solubility of \(\mathrm{CO}_{2}\) in water is governed by Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Therefore, solubility (\(C\)) can be calculated using the equation:

• \(C = k_H \times P\)

where \(k_H\) is the Henry's Law constant, and \(P\) is the partial pressure of the gas. In our example, with \(k_H = 3.4 \times 10^{-4} \, \mathrm{mol}/\mathrm{m}^{3}-\mathrm{Pa}\) and atmospheric pressure of 101325 Pa, the solubility becomes:\[C = (3.4 \times 10^{-4} \, \mathrm{mol}/\mathrm{m}^{3}-\mathrm{Pa}) \times 101325\,\text{Pa}\]Breaking it down, this calculation gives us the amount of \(\mathrm{CO}_{2}\) that can be dissolved in water under standard conditions.

pH Calculation

pH is a measure of how acidic or basic a solution is, based on the concentration of hydrogen ions \([\mathrm{H}^+]\). In our exercise, we calculated the pH after determining the concentration of hydrogen ions formed when carbonic acid \(\mathrm{H}_{2}\mathrm{CO}_{3}\) dissociates.First, knowing the dissociation constant \(K_a\) for carbonic acid is essential. We found that:

• \(K_a = 4.3 \times 10^{-7}\)

Using the approximate equality \([\mathrm{H}^+] \approx [\mathrm{HCO}_3^-]\), we simplify the expression:\[K_a = \frac{[\mathrm{H}^+]^2}{[\mathrm{H}_{2}\mathrm{CO}_{3}]}\]This allows us to directly find \([\mathrm{H}^+]\):\[[\mathrm{H}^+] = \sqrt{K_a \times [\mathrm{H}_{2}\mathrm{CO}_{3}]}\]Finally, pH is calculated using:\[\text{pH} = -\log_{10}([\mathrm{H}^+])\]This formula provides an understanding of the acidity level of the solution based on the dissociated hydrogen ions.

Dissociation Constant

The dissociation constant, denoted as \(K_a\), is crucial in determining the degree to which carbonic acid \(\mathrm{H}_{2}\mathrm{CO}_{3}\) dissociates in water. It quantifies the strength of an acid in a solution. A smaller \(K_a\) value, such as \(4.3 \times 10^{-7}\) for carbonic acid, indicates a weak acid, meaning only a small fraction of the acid molecules dissociate to release hydrogen ions and bicarbonate ions.The equation representing the dissociation is:

• \(\mathrm{H}_{2}\mathrm{CO}_{3} \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^-\)

The \(K_a\) expression, \(\frac{[\mathrm{H}^+][\mathrm{HCO}_3^-]}{[\mathrm{H}_{2}\mathrm{CO}_{3}]}\), helps us understand the equilibrium concentrations of the ions and the undissociated acid. It's a valuable component in calculating the pH of the solution as it directly influences hydrogen ion concentration. Understanding \(K_a\) is key to predicting how an acid will behave in an aqueous environment.

Carbonic Acid

Carbonic acid \(\mathrm{H}_{2}\mathrm{CO}_{3}\) forms when carbon dioxide \(\mathrm{CO}_{2}\) is dissolved in water. It's a weak acid with important roles in the carbon cycle and in living organisms. Its formation and breakdown are crucial dynamics in the chemistry of carbonated beverages and biological systems for maintaining pH balance.When \(\mathrm{CO}_2\) from the air contacts water, it reacts to form carbonic acid:

• \(\mathrm{CO}_2(aq) + \mathrm{H}_2O(l) \rightarrow \mathrm{H}_2\mathrm{CO}_3(aq)\)

This reaction is reversible and largely governs the presence of carbonic acid in aqueous solutions.As a weak acid, carbonic acid doesn't fully dissociate, instead establishing an equilibrium, which is represented by its dissociation constant \(K_a\). Its presence affects the pH of a solution significantly due to its capacity to release hydrogen ions. Carbonic acid is pivotal in understanding both natural processes and applications like carbon capture and sequestration.