Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.
The Correct Answer and Explanation is:
To find the Ka (acid dissociation constant) for the monoprotic acid, we need to use the information given:
- Concentration of the acid (HA): 0.0192 M
- pH of the solution: 2.53
- The acid is monoprotic, meaning it donates 1 proton (H⁺) per molecule.
Step 1: Determine [H⁺] from pH
We use the pH equation: pH=−log[H+]\text{pH} = -\log[H^+]
Rearranging: [H+]=10−pH=10−2.53≈2.95×10−3 M[H^+] = 10^{-\text{pH}} = 10^{-2.53} ≈ 2.95 \times 10^{-3} \text{ M}
Step 2: Write the dissociation equation
Since the acid is monoprotic: HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
Initially, the concentration of HA is 0.0192 M, and [H⁺] = [A⁻] due to 1:1 molar ratio. When the acid dissociates, x = [H⁺] = 2.95 × 10⁻³ M dissociates, so:
- [HA] at equilibrium = 0.0192 – 0.00295 = 0.01625 M
- [H⁺] = [A⁻] = 0.00295 M
Step 3: Use the Ka expression
Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]} Ka=(2.95×10−3)20.01625K_a = \frac{(2.95 \times 10^{-3})^2}{0.01625} Ka=8.70×10−60.01625≈5.35×10−4K_a = \frac{8.70 \times 10^{-6}}{0.01625} ≈ 5.35 \times 10^{-4}
✅ Final Answer:
Ka≈5.35×10−4K_a ≈ \boxed{5.35 \times 10^{-4}}
Explanation (300+ words)
The dissociation constant (Ka) is a measure of the strength of an acid in solution. It quantifies how much an acid dissociates into hydrogen ions (H⁺) and its conjugate base. A monoprotic acid releases only one hydrogen ion per molecule, making the stoichiometry straightforward.
We begin the problem by using the pH to find the concentration of hydrogen ions. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration. Using the given pH of 2.53, we calculate the [H⁺] as 10−2.5310^{-2.53}, which equals approximately 2.95 × 10⁻³ M.
This value also represents how much of the original acid dissociated, because for every mole of HA that dissociates, one mole of H⁺ and one mole of A⁻ are produced. That means both [H⁺] and [A⁻] equal 2.95 × 10⁻³ M. The remaining amount of the undissociated acid is the original concentration (0.0192 M) minus the amount that dissociated, giving 0.01625 M.
Now, we apply the definition of Ka: the product of the concentrations of the dissociation products (H⁺ and A⁻) divided by the concentration of the undissociated acid. Plugging in the equilibrium values into the formula yields a Ka of approximately 5.35 × 10⁻⁴.
This moderate Ka value suggests that the acid is weak, as it only partially dissociates in water. Strong acids, in contrast, would have Ka values much greater than 1, indicating nearly complete dissociation