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 calculate the acid dissociation constant (Ka) for the monoprotic acid, we follow these steps:
✅ Final Answer:
Ka = 1.83 × 10⁻⁵
🧪 Step-by-Step Explanation:
1. Given Data:
- Initial concentration of acid, [HA] = 0.0192 M
- pH of solution = 2.53
- Monoprotic acid (donates 1 proton per molecule)
2. Find ([H⁺]) from pH:
We use the formula:
[
[H^+] = 10^{-\text{pH}} = 10^{-2.53}
]
[
[H^+] = 2.95 \times 10^{-3} \, \text{M}
]
This is the amount of acid that dissociated into H⁺ and A⁻.
3. ICE Table Setup:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HA | 0.0192 | –x | 0.0192 – x |
| H⁺ | 0 | +x | x = 2.95 × 10⁻³ |
| A⁻ | 0 | +x | x = 2.95 × 10⁻³ |
Since ([H⁺] = x), we can substitute into the Ka expression:
[
K_a = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{0.0192 – x}
]
4. Substitute values:
[
K_a = \frac{(2.95 \times 10^{-3})^2}{0.0192 – 2.95 \times 10^{-3}}
]
[
= \frac{8.70 \times 10^{-6}}{0.01625}
]
[
K_a = 1.83 \times 10^{-4}
]
❗ Correction:
Oops! Wait — there’s a typo. Let’s recalculate that denominator properly:
[
K_a = \frac{8.70 \times 10^{-6}}{0.01625} = 5.35 \times 10^{-4}
]
BUT this contradicts our earlier value. Let’s check once again using precise values:
[
(2.95 \times 10^{-3})^2 = 8.70 \times 10^{-6}
]
[
0.0192 – 0.00295 = 0.01625
]
[
Ka = \frac{8.70 \times 10^{-6}}{0.01625} ≈ 5.35 \times 10^{-4}
]
✔️ Final Corrected Ka = 5.35 × 10⁻⁴
🧠 Explanation:
Ka measures how much the acid dissociates in water. A stronger acid has a higher Ka, meaning more ionization. By knowing the initial concentration and the pH, we calculate how much H⁺ is in solution, and use that to find Ka using the equilibrium expression.
This problem is classic for weak acids—since it’s monoprotic, the math stays cleaner. The most important step is accurately converting pH to [H⁺], and then applying it to the Ka expression.
