HCHO2(aq) + H2O(l) <-> H3O+(aq) + CHO2-(aq) HCHO2(aq), a weak acid (Ka = 2 x 10^-4), dissociates in — Acids and Bases Chemistry Question

STEPS:

1. Identify the reaction and the given values:
Formic acid ($\text{HCHO}_2$) dissociates in water according to the equilibrium equation:
$$\text{HCHO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{CHO}_2^-(aq) \quad$$
The given acid dissociation constant is $K_a = 2 \times 10^{-4}$ and the initial concentration of the acid is $0.5\text{ M}$.

2. Set up the equilibrium expression ($K_a$):
$$K_a = \frac{[\text{H}_3\text{O}^+][\text{CHO}_2^-]}{[\text{HCHO}_2]} \quad$$
Let $x$ be the concentration of $\text{H}_3\text{O}^+$ and $\text{CHO}_2^-$ that forms at equilibrium.
* $[\text{H}_3\text{O}^+] = x$
* $[\text{CHO}_2^-] = x$
* $[\text{HCHO}_2] = 0.5 - x$

3. Solve for $[\text{H}_3\text{O}^+]$ using the small^-$x$ approximation:
Because formic acid is a weak acid with a very small dissociation constant ($K_a \ll 1$), it dissociates to a very limited extent in water. Therefore, we can assume that the change in concentration is negligible relative to the starting concentration ($0.5 - x \approx 0.5\text{ M}$).
$$2 \times 10^{-4} = \frac{x^2}{0.5}$$
$$x^2 = 0.5 \times (2 \times 10^{-4}) = 1.0 \times 10^{-4}$$
$$x = \sqrt{1.0 \times 10^{-4}} = 1.0 \times 10^{-2}\text{ M}$$
Thus, the equilibrium concentration of hydronium ions is $[\text{H}_3\text{O}^+] = 1.0 \times 10^{-2}\text{ M}$.

4. Calculate the pH of the solution:
$$\text{pH} = -\log[\text{H}_3\text{O}^+] \quad$$
$$\text{pH} = -\log(1.0 \times 10^{-2}) = 2$$
The best estimate for the pH of the solution is $2$.

5. Determine the species at the highest concentration (excluding $\text{H}_2\text{O}$):
* Compare the equilibrium concentrations of the solute species:
* $[\text{HCHO}_2] = 0.5\text{ M} - 0.01\text{ M} = 0.49\text{ M}$
* $[\text{H}_3\text{O}^+] = 0.01\text{ M}$
* $[\text{CHO}_2^-] = 0.01\text{ M}$
* Because the acid is weak and only a very small fraction (approximately $2\%$) of it ionizes, the vast majority of the formic acid remains in its intact, molecular form.
* Therefore, the concentration of the unionized acid, $\text{HCHO}_2(aq)$, is significantly higher than that of any dissolved product ions ($0.49\text{ M} \gg 0.01\text{ M}$).
* Matching a pH of 2 and $\text{HCHO}_2(aq)$ as the highest concentration solute species identifies Option D as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: A pH of 1 would require a hydronium ion concentration of $[\text{H}_3\text{O}^+] = 0.1\text{ M}$, which is far too high for a weak acid solution of this concentration. Additionally, $\text{H}_3\text{O}^+(aq)$ is a product of a highly suppressed ionization and is present in a very minor amount compared to the intact weak acid molecules.
• Option B is incorrect: While it correctly identifies $\text{HCHO}_2(aq)$ as the species in the highest concentration due to minimal dissociation, a pH of 1 is mathematically incorrect for a weak acid with a $K_a$ of $2 \times 10^{-4}$ at this concentration.
• Option C is incorrect: Although it correctly estimates the pH as 2, it incorrectly asserts that $\text{H}_3\text{O}^+(aq)$ is the most abundant solute species. For any weak acid dissolved in water, the concentration of the unionized weak acid molecule at equilibrium will always be vastly greater than the concentration of the hydronium ion products.