Problem 84 - Chemistry: An Atoms-Focused Approach

Question

The \(K_{a_{1}}\) value for oxalic acid (HOOCCOOH) is \(5.9 \times 10^{-2},\) and the \(K_{2,}\) value is \(6.4 \times 10^{-5} .\) What are the values of \(K_{\mathrm{b}_{1}}\) and \(K_{\mathrm{b}_{2}}\) of the oxalate anion \(\left(-\mathrm{OOCCOO}^{-}\right) ?\)

Step-by-Step Solution

Answer

Question: Calculate the values of \(K_{b_1}\) and \(K_{b_2}\) for the oxalate anion, given the values of \(K_{a_1}\) = 5.9 x 10^{-2} and \(K_{a_2}\) = 6.4 x 10^{-5}. Answer: The values of \(K_{b_1}\) and \(K_{b_2}\) for the oxalate anion are \(1.69 \times 10^{-13}\) and \(1.56 \times 10^{-10}\), respectively.

Step 1: Recall the ion product of water

The ion product of water (\(K_w\)) is a constant value at a given temperature. For most calculations, we assume a temperature of 25°C, where \(K_w\) is equal to \(1.0 \times 10^{-14}\).

Step 2: Compute \(K_{b_1}\)

We will use the provided \(K_{a_1}\) value and the ion product of water (\(K_w\)) to calculate the \(K_{b_1}\) value for the oxalate anion. Using the formula \(K_{b_1} = \frac{K_w}{K_{a_1}}\), we get: \(K_{b_1} = \frac{1.0 \times 10^{-14}}{5.9 \times 10^{-2}}\).

Step 3: Solve for \(K_{b_1}\)

Calculate the value of \(K_{b_1}\): \(K_{b_1} = \frac{1.0 \times 10^{-14}}{5.9 \times 10^{-2}} = 1.69 \times 10^{-13}\)

Step 4: Compute \(K_{b_2}\)

Now, we will use the provided \(K_{a_2}\) value and the ion product of water (\(K_w\)) to calculate the \(K_{b_2}\) value for the oxalate anion. Using the formula \(K_{b_2} = \frac{K_w}{K_{a_2}}\), we get: \(K_{b_2} = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-5}}\).

Step 5: Solve for \(K_{b_2}\)

Calculate the value of \(K_{b_2}\): \(K_{b_2} = \frac{1.0 \times 10^{-14}}{6.4 \times 10^{-5}} = 1.56 \times 10^{-10}\) Thus, the values of \(K_{b_1}\) and \(K_{b_2}\) for the oxalate anion are \(1.69 \times 10^{-13}\) and \(1.56 \times 10^{-10}\), respectively.

Key Concepts

Oxalic Acid

Oxalic acid, represented as \( \text{HOOCCOOH} \), is a dicarboxylic acid, which means it contains two carboxylic acid groups \((\text{-COOH})\). This organic compound is sometimes found in plants and is involved in many biological processes.
The compound dissociates in water to donate protons \((\text{H}^+)\), making it a typical example used to study acid-base equilibria.
When oxalic acid dissolves, it dissociates in two stages:

• The first dissociation step: \( \text{HOOCCOOH} \rightarrow \text{H}^+ + \text{HOOC-COO}^- \)
• The second dissociation step: \( \text{HOOC-COO}^- \rightarrow \text{H}^+ + \text{-OOC-COO}^- \)

The equilibrium constants for these reactions are called \( K_{a_1} \) and \( K_{a_2} \), referring to the strength of the acid in its two dissociative states. For oxalic acid, these constants are \(5.9 \times 10^{-2}\) for \(K_{a_1}\) and \(6.4 \times 10^{-5}\) for \(K_{a_2}\).
These values are crucial because they help us understand the acid's behavior in aqueous solutions and compute related constants for the base formed after dissociation.

Ion Product of Water

The ion product of water, noted as \( K_w \), is a fundamental constant in chemistry. It reflects the equilibrium condition of pure water dissociating into hydrogen and hydroxide ions.
At 25°C, the value of \( K_w \) is \( 1.0 \times 10^{-14} \). This small value emphasises that water naturally has low ion concentrations under neutral conditions.
The concept of \( K_w \) is essential when it comes to calculating the equilibria of acids and bases.
For any acid-base reaction occurring in an aqueous solution, \( K_w \) ensures the balance between proton \((\text{H}^+)\) and hydroxide ion \((\text{OH}^-)\) concentrations. This is particularly useful when calculating the base dissociation constants \( K_{b_1} \) and \( K_{b_2} \) from \( K_{a_1} \) and \( K_{a_2} \) using:

• \( K_b = \frac{K_w}{K_a} \)

Thus, understanding \( K_w \) allows the conversion between acid and base constants, aiding in comprehensive studies of acid-base interactions.

Equilibrium Constants

Equilibrium constants are crucial in understanding chemical reactions that reach a state of equilibrium. In acid-base chemistry, these constants provide valuable information about acid or base strength.
For acids, the constant is represented by \( K_a \), and for bases, it is signified as \( K_b \).
These constants are calculated from the concentrations of reactants and products at equilibrium.
Key Points for Equilibrium Constants:

• High \( K_a \) or \( K_b \) values indicate stronger acids or bases and more complete dissociation.
• Low \( K_a \) or \( K_b \) values suggest weaker acids or bases with minimal dissociation.

For oxalic acid dissociation into its anions, knowing \( K_{a_1} \) and \( K_{a_2} \) allows calculation of its base constants \( K_{b_1} \) and \( K_{b_2} \) using the previously mentioned relation:\[ K_b = \frac{K_w}{K_a} \]This calculation is vital for understanding the dual nature of substances like oxalic acid, which exist both as acids and conjugate bases.