Problem 71 - Chemistry and Chemical Reactivity

Question

Prove that \(K_{\mathrm{a} 1} \times K_{\mathrm{b} 2}=K_{\mathrm{w}}\) for oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) by adding the chemical equilibrium expressions that correspond to first ionization step of the acid in water with the second step of the reaction of the fully deprotonated base, \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-},\) with water.

Step-by-Step Solution

Answer

The product of the equilibrium constants is equal to the ion-product of water: \(K_{a1} \times K_{b2} = K_w\).

Step 1: Identify the Required Equilibrium Expressions

The first ionization of oxalic acid in water is given by the equilibrium \( \text{H}_2\text{C}_2\text{O}_4 \rightleftharpoons \text{H}^+ + \text{HC}_2\text{O}_4^- \). For this reaction, the equilibrium constant is \( K_{a1} = \frac{[\text{H}^+][\text{HC}_2\text{O}_4^-]}{[\text{H}_2\text{C}_2\text{O}_4]} \).

Step 2: Second Reaction: Deprotonated base with water

The reaction for the deprotonated base reacting with water is \( \text{C}_2\text{O}_4^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{HC}_2\text{O}_4^- + \text{OH}^- \), with the equilibrium constant \( K_{b2} = \frac{[\text{HC}_2\text{O}_4^-][\text{OH}^-]}{[\text{C}_2\text{O}_4^{2-}]} \).

Step 3: Consider Water Ion-Product Constant

The ion-product of water is represented by the equilibrium \( \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \), and it has an equilibrium constant represented as \( K_w = [\text{H}^+][\text{OH}^-] \).

Step 4: Multiply Equilibrium Expressions

Multiply \( K_{a1} \) by \( K_{b2} \). This is: \[ K_{a1} \times K_{b2} = \left( \frac{[\text{H}^+][\text{HC}_2\text{O}_4^-]}{[\text{H}_2\text{C}_2\text{O}_4]} \right) \times \left( \frac{[\text{HC}_2\text{O}_4^-][\text{OH}^-]}{[\text{C}_2\text{O}_4^{2-}]} \right) \]This simplifies to:\[ K_{a1} \times K_{b2} = \frac{[\text{H}^+][\text{OH}^-][\text{HC}_2\text{O}_4^-]^2}{[\text{H}_2\text{C}_2\text{O}_4][\text{C}_2\text{O}_4^{2-}]} \]

Step 5: Final Simplification and Conclusion

Notice that the term \([\text{HC}_2\text{O}_4^-]^2\) in the numerator will cancel out as this term expresses both the concentration of \([\text{HC}_2\text{O}_4^-]\) produced in both reactions. Thus, the expression simplifies to:\[ K_{a1} \times K_{b2} = [\text{H}^+][\text{OH}^-] = K_w \]Hence, it is proven that \( K_{a1} \times K_{b2} = K_w \).

Key Concepts

Acid-Base Reactions

Acid-base reactions are fundamental processes in chemistry and occur when an acid donates a proton (H⁺) to a base. In simple terms, an acid is a substance that can donate a proton, while a base is one that can accept it. When oxalic acid (H₂C₂O₄) undergoes ionization, it acts as an acid by donating protons.

• Proton Transfer: The main feature of an acid-base reaction is the transfer of protons from the acid to the base.
• Equilibrium Nature: These reactions can reach a state of equilibrium where the rate of the forward reaction (acid to base) equals the rate of the reverse reaction (base to acid).

Understanding acid-base reactions prepares the ground for studying chemical equilibrium and how equilibrium constants function in such settings.

Equilibrium Constants

Equilibrium constants are pivotal in quantifying the position of equilibrium in a chemical reaction. They provide insight into the amounts of products and reactants present at equilibrium.

• Formula Representation: For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is represented by:\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
• Dependence on Reaction Type: The Ka and Kb are specific equilibrium constants applied to acid and base reactions, respectively.
• Ion Product Constant: For water, the equilibrium constant is the famously known \( K_w \), which is equal to \( [H^+][OH^-] \).

These constants are crucial for predicting the extent of reaction and calculating concentrations of various species in an equilibrium state.

Ionization

Ionization refers to the process through which neutral molecules transform into ions. This is a key concept in understanding chemical equilibria and acid-base behavior.

• Initial Ionization: Oxalic acid (H₂C₂O₄), for instance, undergoes ionization to form H⁺ and HC₂O₄⁻ ions.
• Multiple Steps: Some acids, like oxalic acid, ionize in more than one step, meaning they can donate multiple protons sequentially.
• Impact on Equilibrium: Each ionization step has a specific equilibrium constant \( K_a \), with the first ionization usually being the central focus when talking about overall acid strength.

The completion and balance of these ionizations heavily influence the position of acid-base equilibrium in the system.

Oxalic Acid

Oxalic acid is a naturally occurring, relatively strong organic acid found in plants like rhubarb and spinach.

• Formula and Properties: Its chemical formula is H₂C₂O₄, and it is characterized by its ability to donate protons through ionization, playing the role of an acid in various chemical reactions.
• Multiproton Donor: Oxalic acid can lose two protons, making it a diprotic acid, with separate equilibrium constants \( K_{a1} \) and \( K_{a2} \) for each step of ionization.
• Industrial Use: It has several applications, including as a reducing agent in photography, cleaning agents, and metal polishing.

Grasping the behavior and properties of oxalic acid is essential for understanding more complex reactions involving this acid, especially its role in equilibrium calculations.