Question

a. The principal equilibrium in a solution of \(\mathrm{NaHCO}_{3}\) is $$ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Calculate the value of the equilibrium constant for this reaction. b. At equilibrium, what is the relationship between \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\) and \(\left[\mathrm{CO}_{3}^{2-}\right] ?\) c. Using the equilibrium $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ derive an expression for the pH of the solution in terms of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) using the result from part b. d. What is the pH of a solution of \(\mathrm{NaHCO}_{3} ?\)

Short answer

In summary, the pH of a NaHCO₃ solution can be calculated using the derived expression: \(pH = -\log_{10}\left(\sqrt{\dfrac{K_{a1}\times K_{a2}}{K_{a2}} \cdot [\mathrm{HCO}_{3}^{-}]}\right)\), where Ka1 and Ka2 are the dissociation constants for the two equilibrium reactions, and [HCO₃⁻] is the concentration of bicarbonate ions in the solution. Plug in the known values for Ka1, Ka2, and [HCO₃⁻], and calculate the pH.

Step-by-step solution

Write the principal equilibrium equation with proper dissociation constants.

The given equilibrium equation is: \( \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) \) Instead, we should use the following equilibrium equations with their corresponding dissociation constants: \( \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{HCO}_{3}^{-}(a q) \) This reaction has a dissociation constant Ka1. \( \mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{CO}_{3}^{2-}(a q) \) This reaction has a dissociation constant Ka2. Now, we combine these two equations to match the given equilibrium equation in part a: \( 2 \mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) + \mathrm{CO}_{3}^{2-}(a q) \)

Calculate the equilibrium constant for the principal equilibrium.

To find the equilibrium constant for the principal equilibrium, we multiply the dissociation constants Ka1 and Ka2: \( K = K_{a1} \times K_{a2} \)

Determine the relationship between H2CO3 and CO3^2- at equilibrium.

We are given the equilibrium equation: \( 2 \mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) + \mathrm{CO}_{3}^{2-}(a q) \) At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. Therefore, \(K_{eq} =\dfrac{[\mathrm{H}_{2}\mathrm{CO}_{3}][\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]^{2}}\) Since K = Ka1 × Ka2, we substitute it into the equilibrium equation, \(K_{a1}\times K_{a2} = \dfrac{[\mathrm{H}_{2}\mathrm{CO}_{3}][\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]^{2}}\)

Derive an expression for the pH of the solution using Ka1 and Ka2.

We want to find the pH, so we need to find [H+] concentration. From the first dissociation equilibrium, we can write: \( K_{a1} = \dfrac{[\mathrm{H}^{+}][\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}_{2}\mathrm{CO}_{3}]}\) From the second dissociation equilibrium, we can write: \( K_{a2} = \dfrac{[\mathrm{H}^{+}][\mathrm{CO}_{3}^{2-}]}{[\mathrm{HCO}_{3}^{-}]}\) Now, we will solve the second equation for [CO3^2-], and then substitute it into the equilibrium equation we found in step 3: \([\mathrm{CO}_{3}^{2-}] = \dfrac{K_{a2}[\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}^{+}]}\) The equilibrium equation becomes: \(K_{a1} \times K_{a2} = \dfrac{[\mathrm{H}_{2}\mathrm{CO}_3] K_{a2}[\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}^{+}][\mathrm{HCO}_{3}^{-}]^2}\) Rearranging for pH = -log10([H+]), we get: \(\dfrac{K_{a1} \times K_{a2}}{K_{a2}} = \dfrac{[\mathrm{H}^{+}]^2}{[\mathrm{HCO}_{3}^{-}]}\) \([\mathrm{H}^{+}] = \sqrt{\dfrac{K_{a1} \times K_{a2}}{K_{a2}} \cdot [\mathrm{HCO}_{3}^{-}]}\) \(pH = -\log_{10}\left(\sqrt{\dfrac{K_{a1}\times K_{a2}}{K_{a2}} \cdot [\mathrm{HCO}_{3}^{-}]}\right)\)

Calculate the pH of a NaHCO3 solution.

Given the concentrations of HCO3- in the solution (from sodium bicarbonate), we can use the derived expression for pH: \(pH = -\log_{10}\left(\sqrt{\dfrac{K_{a1}\times K_{a2}}{K_{a2}} \cdot [\mathrm{HCO}_{3}^{-}]}\right)\) Plug in the known values for Ka1, Ka2 and [HCO3-], and calculate the pH.

Key concepts

Acid-Base Equilibrium

Understanding acid-base equilibrium is essential when dealing with chemical reactions involving acids and bases. In the context of sodium bicarbonate (NaHCO₃), we explore how bicarbonate ions are part of this equilibrium. These ions can both donate and accept protons, making the system an amphoteric one. This dynamic balance involves the interactions of hydrogen ions (H⁺), bicarbonate ions (HCO₃⁻), and carbonate ions (CO₃²⁻). By exploring different equilibrium states, we see how the system can shift, adapting to changes in concentration and environmental conditions. This delicate balance is foundational in pH regulation and many biological processes.

Dissociation Constants

Dissociation constants (Ka) are crucial for predicting the strengths and behavior of acids and bases in solution. For carbonic acid, we have two constants, Ka1 and Ka2, representing its two-step dissociation:

• First, \[\mathrm{H}_2\mathrm{CO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{HCO}_3^-\]
• Second, \[\mathrm{HCO}_3^- \rightleftharpoons \mathrm{H}^+ + \mathrm{CO}_3^{2-}\]

Ka1 represents the dissociation of carbonic acid into bicarbonate and hydrogen ions. Ka2 represents the further dissociation of bicarbonate into carbonate and hydrogen ions. The values of Ka allow us to understand how acidic or basic a substance is, as well as how it will behave under different conditions.

pH Calculation

Calculating pH involves understanding the concentration of hydrogen ions \([\mathrm{H}^+]\) in a solution. In this exercise, we utilize the equilibrium constants to derive the pH equation:

• From the relationship \(K = K_{a1} \times K_{a2}\),
• We derive: \[\left[\mathrm{H}^+\right] = \sqrt{\frac{K_{a1} \times K_{a2}}{K_{a2}} \cdot \left[\mathrm{HCO}_3^-\right]}\]
• Thus, \[pH = -\log_{10}\left(\sqrt{\frac{K_{a1} \times K_{a2}}{K_{a2}} \cdot \left[\mathrm{HCO}_3^-\right]}\right)\]

This gives us a clear pathway to determine the pH based on known quantities. The pH tells us the acidity or basicity of the solution, helping to predict its reactivity and stability.

Equilibrium Expression

The equilibrium expression is fundamental to quantifying the position of a chemical equilibrium. For the equilibrium between bicarbonate ions, carbonic acid, and carbonate ions, we write:

• The expression is \[K_{eq} = \frac{[\mathrm{H}_2\mathrm{CO}_3][\mathrm{CO}_3^{2-}]}{[\mathrm{HCO}_3^-]^2}\]
• It is based on the balanced chemical equation \[2 \mathrm{HCO}_3^- \rightleftharpoons \mathrm{H}_2\mathrm{CO}_3 + \mathrm{CO}_3^{2-}\]

Incorporating the dissociation constants, this becomes \(K = K_{a1} \times K_{a2}\), allowing us to relate concentrations to reaction extent.
This expression helps predict how changes in concentration or other conditions affect the system's equilibrium, guiding us in controlling reactions effectively.