Question

Calculate the $\mathrm{pH}$ at ${25}^{\circ }\mathrm{C}$ of a $0.25-M$ aqueous solution of oxalic acid $\left({\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\right)\cdot (K_{{\mathrm{a}}_1}$ and $K_{{\mathrm{a}}_2}$ for oxalic acid are $6.5\times {10}^{-2}$ and $6.1\times {10}^{-5}$, respectively.)

Short answer

The pH of the 0.25-M oxalic acid solution is approximately 0.90.

Step-by-step solution

Understanding the Problem

Oxalic acid is a diprotic acid, meaning it can donate two protons (H⁺ ions) in solution. It has two dissociation constants ($K_{a_1}$ and $K_{a_2}$) that describe the dissociation of each proton. We need to use these constants to calculate the pH of the solution.

First Dissociation

For the first dissociation, $H_2{C}_2{O}_4$ dissociates as follows: $$H_2{C}_2{O}_{4}ightleftharpoons{H}^{+}+H{C}_2{O}_4^{-}$$The expression for the equilibrium constant $K_{a_1}$ is:$$K_{a_1}=\frac{[H^{+}][H{C}_2{O}_4^{-}]}{[H_2{C}_2{O}_4]}=6.5\times {10}^{-2}$$Initial concentration: [H₂C₂O₄] = 0.25 M, [H⁺] = 0, [HC₂O₄⁻] = 0.Let $x$ be the concentration of $H^{+}$ and $H{C}_2{O}_4^{-}$ at equilibrium: $$[H_2{C}_2{O}_4]=0.25-x,[H^{+}]=x,[H{C}_2{O}_4^{-}]=x.$$

Solving for x in First Dissociation

Using the $K_{a_1}$ expression:$$6.5\times {10}^{-2}=\frac{x^2}{0.25-x}$$Assuming $x\ll 0.25$, simplifies to:$$6.5\times {10}^{-2}\approx \frac{x^2}{0.25}$$$$x^2=6.5\times {10}^{-2}\times 0.25$$$$x^2=1.625\times {10}^{-2}$$$$x=\sqrt{1.625\times {10}^{-2}}\approx 0.127$$

Second Dissociation Negligible

Since $x$ from the first dissociation is relatively large, the concentration of $H{C}_2{O}_4^{-}$ at this point is 0.127 M. The second dissociation:$$H{C}_2{O}_4^{-}ightleftharpoons{H}^{+}+C_2{O}_4^{2-}$$The expression for $K_{a_2}$ is:$$K_{a_2}=\frac{[H^{+}][C_2{O}_4^{2-}]}{[H{C}_2{O}_4^{-}]}=6.1\times {10}^{-5}$$Given $K_{a_2}\ll x$ from $K_{a_1}$, the concentration contribution of $[H^{+}]$ from the second dissociation is negligible compared to the $x=0.127$ M calculated before.

Calculating pH

With $[H^{+}]=0.127$, calculate $$\text{pH}=-{\log }_{10}(0.127)$$$$\text{pH}\approx 0.90$$

Key concepts

Oxalic Acid

Oxalic acid, chemically represented by the formula ${\text{H}}_2{\text{C}}_2{\text{O}}_4$, is an organic compound with notable acidic properties. It occurs naturally in many plants such as spinach and rhubarb. This acid is categorized as a dicarboxylic acid, owing to the two carboxylic acid groups $(\text{-COOH})$ present in its structure. Its prevalent use is in cleaning, bleaching, and as a reducing agent in chemical reactions.
Beyond its practical applications, oxalic acid is crucial in chemistry education because of its ability to lose two protons. As a chemical, it can be deemed relatively strong, easily ionizing in solution, which makes it an excellent case study in exploring concepts such as dissociation and pH in solutions.
Involved in two-step dissociation processes, understanding its behavior enhances comprehension of more complex weak acids and their equilibrium dynamics.

Diprotic Acid

Diprotic acids, like oxalic acid, have the ability to donate two protons or hydrogen ions in a stepwise manner. This characteristic makes them intriguing in acid-base chemistry, as they exhibit distinct dissociation processes for each hydrogen ion. Diprotic acids generally follow two separate equilibria for dissociation:

• The first dissociation involves the transfer of the first proton, resulting in what is typically a stronger acidic reaction compared to the second step.
• The second dissociation involves the loss of the second proton from the intermediate form created after the first dissociation, commonly a weaker process.

The distinction between these steps is due to varying stability and solubility of the resulting ions, and this staged dissociation influences calculations related to their equilibrium constants and the resulting pH of solutions.

Dissociation Constants

Dissociation constants $(K_a)$ provide a numerical indication of the strength of an acid in a solution. For diprotic acids like oxalic acid, there are two dissociation constants $K_{a_1}$ and $K_{a_2}$, each corresponding to one of the proton release steps.
- $K_{a_1}$ represents the equilibrium constant for the first dissociation of an acid, commonly larger due to the typically stronger dissociation reaction. - $K_{a_2}$ is the constant for the second dissociation, reflecting weaker acid behavior as the acid ionizes further.
In equilibrium expressions, these constants help calculate concentrations of various ions in solution.
Such constants are essential for predicting pH and understanding the degree of ionization under different conditions.

Acid Dissociation

Acid dissociation refers to the reversible process by which an acid releases its protons into an aqueous solution to form ions. For diprotic acids, this involves two separate dissociation stages. These stages contribute different amounts of hydrogen ions to the solution, impacting its overall acidity and therefore the pH.
The dissociation of oxalic acid, for instance, can be described through:

• First dissociation: ${\text{H}}_2{\text{C}}_2{\text{O}}_4\rightleftharpoons {\text{H}}^{+}+{\text{HC}}_2{\text{O}}_4^{-}$
• Second dissociation: ${\text{HC}}_2{\text{O}}_4^{-}\rightleftharpoons {\text{H}}^{+}+{\text{C}}_2{\text{O}}_4^{2-}$

Understanding these dissociation pathways is crucial for calculating the solution's properties, such as pH, concentration of various species, and the acid's overall behavior in different environments.

Equilibrium Constant

The equilibrium constant in the context of acid-base chemistry is a measure of the extent to which a reversible reaction proceeds. Specifically, for the dissociation of acids in solution, it is known as the acid dissociation constant $K_a$. This constant reflects the balance between the dissociated ions and the undissociated form of the acid.
- When an acid dissociates, the equilibrium constant helps determine how much of the acid remains in each form at equilibrium.- Higher $K_a$ values typically indicate stronger acids with greater tendencies to donate protons, whereas lower $K_a$ reflects weaker acids with less dissociation.
For oxalic acid, separate equilibrium constants $K_{a_1}$ and $K_{a_2}$ provide insight into each dissociation stage, indicating the ionization strength of each proton donative step.