Question

Calculate the \(\mathrm{pH}\) and the concentrations of all species present \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}, \mathrm{HCO}_{3}^{-}, \mathrm{CO}_{3}^{2-}, \mathrm{H}_{3} \mathrm{O}^{+}\right.\), and \(\left.\mathrm{OH}^{-}\right)\) in \(0.010 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}\left(K_{\Delta 1}=4.3 \times 10^{-7} ; K_{\Delta 2}=5.6 \times 10^{-11}\right)\)

Short answer

pH is 4.68; \([\mathrm{HCO}_3^-] = 2.07 \times 10^{-5} \), \([\mathrm{CO}_3^{2-}] = 5.6 \times 10^{-11}\) M.

Step-by-step solution

Understanding the problem

Begin by understanding that we need to find the concentrations of the various species in a 0.010 M solution of \( \mathrm{H}_2\mathrm{CO}_3 \). Given dissociation constants \( K_{a1} = 4.3 \times 10^{-7} \) and \( K_{a2} = 5.6 \times 10^{-11} \), these indicate two dissociation steps. We also need to determine the \( \mathrm{pH} \) of the solution.

First dissociation step

For the first dissociation \( \mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^- \), apply the equilibrium expression: \[ K_{a1} = \frac{[\mathrm{H}^+][\mathrm{HCO}_3^-]}{[\mathrm{H}_2\mathrm{CO}_3]} \]Assume initial concentration of \( \mathrm{H}_2\mathrm{CO}_3 \) is 0.010 M and small 'x' dissociates, leading to: \( (0.010 - x) \approx 0.010 \), \[ K_{a1} = \frac{x^2}{0.010} \] Solving \( x^2 = 4.3 \times 10^{-9} \), we get \( x \approx 2.07 \times 10^{-5} \). This is \([\mathrm{H}^+] = [\mathrm{HCO}_3^-] \).

Second dissociation step

For the second dissociation \( \mathrm{HCO}_3^- \rightleftharpoons \mathrm{H}^+ + \mathrm{CO}_3^{2-} \), apply:\[ K_{a2} = \frac{[\mathrm{H}^+][\mathrm{CO}_3^{2-}]}{[\mathrm{HCO}_3^-]} \]Given \( [\mathrm{H}^+] = 2.07 \times 10^{-5} \), the contribution from the second dissociation will be negligible compared to the first dissociation. Assume \([\mathrm{CO}_3^{2-}] = y\) is small:\[ K_{a2} = \frac{(2.07 \times 10^{-5})y}{2.07 \times 10^{-5}} \approx y\]Thus, \( y \approx K_{a2} = 5.6 \times 10^{-11} \).

Determine the pH of the solution

Using the \([\mathrm{H}^+]\) from the first dissociation, calculate the \( \mathrm{pH} \):\[ \mathrm{pH} = -\log(2.07 \times 10^{-5}) \approx 4.68 \].

Calculate hydroxide ion concentration

Use the relation \([\mathrm{H}^+][\mathrm{OH}^-] = 10^{-14} \):\[ [\mathrm{OH}^-] = \frac{10^{-14}}{2.07 \times 10^{-5}} \approx 4.83 \times 10^{-10} \]

Summarize concentrations of all species

Summarize the concentrations: \([\mathrm{H}_2\mathrm{CO}_3] = 0.010 \), \([\mathrm{HCO}_3^-] \approx 2.07 \times 10^{-5} \), \([\mathrm{CO}_3^{2-}] \approx 5.6 \times 10^{-11} \), \([\mathrm{H}^+] = 2.07 \times 10^{-5}, \mathrm{pH} = 4.68 \), \([\mathrm{OH}^-] \approx 4.83 \times 10^{-10}\).

Key concepts

Carbonic Acid Dissociation

Carbonic acid (\(\mathrm{H}_2\mathrm{CO}_3\)) is a weak acid that undergoes stepwise dissociation in water. It dissociates in two steps to release protons (\(\mathrm{H}^+\)), making it an important component in acid-base equilibria. Understanding the dissociation process is essential to calculate the concentrations of various species in solution and their corresponding \(\mathrm{pH}\) values.

• **First Dissociation:** In the initial step, carbonic acid dissociates into bicarbonate (\(\mathrm{HCO}_3^-\)) and a proton. The equilibrium for this reaction can be expressed with an equilibrium constant \(K_{a1}=4.3 \times 10^{-7}\):\[\mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^-\]
• **Second Dissociation:** In the second step, the bicarbonate ion further dissociates to form carbonate (\(\mathrm{CO}_3^{2-}\)) and another proton. The equilibrium constant \(K_{a2}=5.6 \times 10^{-11}\) is used here:\[\mathrm{HCO}_3^- \rightleftharpoons \mathrm{H}^+ + \mathrm{CO}_3^{2-}\]

The strength of each dissociation step is determined by its equilibrium constant, and since \(K_{a2}\) is much smaller than \(K_{a1}\), the first dissociation is more predominant in affecting the \(\mathrm{pH}\) of the solution.

pH Calculation

Calculating the pH of a solution involves determining the concentration of hydrogen ions (\(\mathrm{H}^+\)). In the context of carbonic acid dissociation, the primary source of \(\mathrm{H}^+\) ions is the first dissociation of \(\mathrm{H}_2\mathrm{CO}_3\).

• **Calculating Initial \(\mathrm{H}^+\) Concentration:** Assume a small 'x' amount of \(\mathrm{H}_2\mathrm{CO}_3\) dissociates at equilibrium, leading to the equation\[K_{a1} = \frac{x^2}{0.010 - x}\]Since \(x\) is small, we approximate \(0.010 - x \approx 0.010\) and solve to find \(x \approx 2.07 \times 10^{-5}\).
• **Computing \(\mathrm{pH}\):** With \(x\) representing the \(\mathrm{H}^+\) concentration from the first dissociation, we can determine the \(\mathrm{pH}\) by taking the negative logarithm:\[\mathrm{pH} = -\log(2.07 \times 10^{-5}) \approx 4.68\]

This procedure illustrates that the \(\mathrm{pH}\) primarily results from the first dissociation step, as subsequent dissociations contribute minimally to the \(\mathrm{H}^+\) concentration.

Equilibrium Constants

Equilibrium constants play a crucial role in understanding and predicting chemical equilibria, especially in acid-base reactions. When dealing with carbonic acid, we look at its two dissociation constants \(K_{a1}\) and \(K_{a2}\) to predict species concentrations and \(\mathrm{pH}\).

• **Role of \(K_{a}\):** The equilibrium constant reflects the ratio of product to reactant concentrations at equilibrium. A larger \(K_{a1}\) suggests more dissociation of the original acid, creating a more acidic solution.
• **Comparison between \(K_{a1}\) and \(K_{a2}\):** Because \(K_{a2}\) is significantly smaller than \(K_{a1}\), the second dissociation contributes less to the overall concentration of ions in solution. This means that carbonate ions (\(\mathrm{CO}_3^{2-}\)) will have minimal impact on the pH.

Understanding these constants enables us to determine not only the pH but also the concentration balance among different species in a solution of carbonic acid. Consistently considering equilibrium constants is essential for accurate chemical predictions in a variety of applications.