Question

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is a diprotic acid that occurs in plants such as rhubarb and spinach. Calculate the \(\mathrm{pH}\) and the concentration of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions in \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\left(K_{u 1}=5.9 \times 10^{-2} ; K_{22}=6.4\right.\) \(\left.\times 10^{-5}\right)\)

Short answer

pH = 0.96; [C₂O₄²⁻] = 6.4 × 10⁻⁵ M.

Step-by-step solution

Understanding Diprotic Acids

A diprotic acid like oxalic acid can donate two protons (H\(^+\)) in two sequential steps. In the first ionization, oxalic acid (H\(_2\)C\(_2\)O\(_4\)) loses one proton to form HC\(_2\)O\(_4^-\), and in the second step, HC\(_2\)O\(_4^-\) loses a proton to form C\(_2\)O\(_4^{2-}\).

Write the Ionization Equations and Constants

For the first ionization: \( \text{H}_2\text{C}_2\text{O}_4 \rightleftharpoons \text{H}^+ + \text{HC}_2\text{O}_4^- \), with \( K_{a1} = 5.9 \times 10^{-2} \).For the second ionization: \( \text{HC}_2\text{O}_4^- \rightleftharpoons \text{H}^+ + \text{C}_2\text{O}_4^{2-} \), with \( K_{a2} = 6.4 \times 10^{-5} \).

Set up the First Ionization Equilibrium

Initial concentration of H\(_2\)C\(_2\)O\(_4\) is 0.20 M and we assume that it ionizes to form x amount of H\(^+\) and HC\(_2\)O\(_4^-\). The equilibrium expression is: \[K_{a1} = \frac{[\text{H}^+][\text{HC}_2\text{O}_4^-]}{[\text{H}_2\text{C}_2\text{O}_4]} = \frac{x^2}{0.20 - x} \approx \frac{x^2}{0.20}\]

Solve for x in First Ionization

Assume \( x << 0.20 \), simplifying the equation to \( x^2 = K_{a1} \times 0.20 \).Calculate \( x \):\[x^2 = (5.9 \times 10^{-2}) \times 0.20 = 1.18 \times 10^{-2}\]\[ x = \sqrt{1.18 \times 10^{-2}} = 0.11 \]So, \([\text{H}^+] = 0.11 \text{ M}\).

Calculate pH from First Ionization

Using \([\text{H}^+]\), calculate the pH:\[pH = -\log(0.11) \approx 0.96\]

Set up the Second Ionization Equilibrium

Use the concentration of HC\(_2\)O\(_4^-\) from the first ionization as the initial concentration for the second ionization, \[ [\text{HC}_2\text{O}_4^-] = 0.11 \text{ M} \]and let \( y \) be the concentration of C\(_2\)O\(_4^{2-}\).\[K_{a2} = \frac{[\text{H}^+][\text{C}_2\text{O}_4^{2-}]}{[\text{HC}_2\text{O}_4^-]} \approx \frac{0.11 \cdot y}{0.11}\]

Solve for y in Second Ionization

Using \( K_{a2} = 6.4 \times 10^{-5} \), equate and solve:\[y = 6.4 \times 10^{-5}\]\([\text{C}_2\text{O}_4^{2-}] = 6.4 \times 10^{-5} \text{ M} \).

Key concepts

Diprotic Acid

Diprotic acids are unique because they can donate two protons (H\(^{+}\)) per molecule. This makes them different from monoprotic acids, which only donate one proton. Oxalic acid, present in plants like spinach and rhubarb, is a common example of a diprotic acid.
When oxalic acid (H\(_2\)C\(_2\)O\(_4\)) is dissolved in water, it undergoes two separate ionization steps:

• First, it loses one proton to form the hydrogen oxalate ion (HC\(_2\)O\(_4^-\)).
• Next, the hydrogen oxalate can lose another proton, resulting in the formation of the oxalate ion (C\(_2\)O\(_4^{2-}\)).

Understanding how each proton is released in these steps is crucial for calculating the pH and ion concentrations of such solutions.

Ionization Equilibrium

In any ionization process, there is a balance between the forward and reverse reactions, known as ionization equilibrium. For oxalic acid, this process occurs in two stages.
The first equilibrium involves the conversion of oxalic acid to hydrogen oxalate ion. The equation is:

• \( \text{H}_2\text{C}_2\text{O}_4 \rightleftharpoons \text{H}^+ + \text{HC}_2\text{O}_4^- \)

Here, the equilibrium constant is represented by \( K_{a1} = 5.9 \times 10^{-2} \).
In the second stage, the hydrogen oxalate ion further dissociates into oxalate ion:

• \( \text{HC}_2\text{O}_4^- \rightleftharpoons \text{H}^+ + \text{C}_2\text{O}_4^{2-} \)

The equilibrium constant for this reaction is \( K_{a2} = 6.4 \times 10^{-5} \). Maintaining these equilibria is vital for determining other properties of the solution.

pH Calculation

Calculating the pH of a diprotic acid solution requires understanding its ionization steps. For oxalic acid, this is simplified by focusing first on the initial ionization.
From the ionization, the concentration of hydrogen ions (H\(^{+}\)) can be calculated using the first ionization equilibrium, where:
\[K_{a1} = \frac{[\text{H}^+][\text{HC}_2\text{O}_4^-]}{[\text{H}_2\text{C}_2\text{O}_4]} \approx \frac{x^2}{0.20}\]
By isolating \(x\), which represents the concentration of H\(^+\), we found it to be 0.11 M.
The pH, a measure of the acidity, can then be determined using the formula:
\[pH = -\log(0.11)\]
This results in a pH of approximately 0.96, indicating a strongly acidic solution.

Equilibrium Constants

Equilibrium constants \( K_{a1} \) and \( K_{a2} \) are essential for understanding how much a diprotic acid will ionize in solution. These constants provide a quantitative measure:

• \( K_{a1} \), the constant for the first ionization, is 5.9 \times 10^{-2}, indicating a relatively high level of ionization.
• \( K_{a2} \), for the second ionization, is much smaller at 6.4 \times 10^{-5}, reflecting lesser ionization.

These diminishing values show that the first dissociation step is more favored than the second.
The second ionization is typically much weaker compared to the first,
so these constants are crucial in chemical calculations concerning diprotic acids.

Oxalate Ion Concentration

The concentration of the oxalate ion (C\(_2\)O\(_4^{2-}\)) is crucial for understanding the extent of the second ionization step in an oxalic acid solution. To find this concentration, consider the following equilibrium expression:
\[ K_{a2} = \frac{[\text{H}^+][\text{C}_2\text{O}_4^{2-}]}{[\text{HC}_2\text{O}_4^-]} \]
Given that the concentration of the hydrogen oxalate ion from the first equilibrium stage was determined as 0.11 M, you can use this to solve for \( y \), where \( y \) is the concentration of C\(_2\)O\(_4^{2-}\). With \( y = 6.4 \times 10^{-5} \), this is relatively low due to the weak second ionization.
Such information is invaluable in predicting reactions involving oxalate and understanding its behavior in biological and environmental systems.