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 Ka (acid dissociation constant) of a monoprotic acid, we use the information provided:
- The concentration of the acid solution: 0.0192 M
- The pH of the solution: 2.53
Step 1: Determine the [H⁺] from pH
The pH is defined as: pH=−log[H+]\text{pH} = -\log [\text{H}^+]pH=−log[H+]
Rearranging: [H+]=10−pH=10−2.53≈2.95×10−3 M[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \text{ M}[H+]=10−pH=10−2.53≈2.95×10−3 M
Step 2: Set up the ICE Table
A monoprotic acid (HA) dissociates in water as follows: HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-HA⇌H++A−
Initial concentration:
- [HA] = 0.0192 M
- [H⁺] = 0
- [A⁻] = 0
Change:
- [HA] decreases by xxx
- [H⁺] increases by xxx
- [A⁻] increases by xxx
Since we know that [H+]=x=2.95×10−3[H^+] = x = 2.95 \times 10^{-3}[H+]=x=2.95×10−3 M,
Equilibrium concentrations:
- [HA] = 0.0192 – 0.00295 = 0.01625 M
- [H⁺] = 0.00295 M
- [A⁻] = 0.00295 M
Step 3: Use the Ka Expression
Ka=[H+][A−][HA]K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}Ka=[HA][H+][A−]
Substitute: Ka=(2.95×10−3)20.01625=8.70×10−60.01625≈5.35×10−4K_a = \frac{(2.95 \times 10^{-3})^2}{0.01625} = \frac{8.70 \times 10^{-6}}{0.01625} \approx 5.35 \times 10^{-4}Ka=0.01625(2.95×10−3)2=0.016258.70×10−6≈5.35×10−4
✅ Final Answer:
Ka=5.35×10−4\boxed{K_a = 5.35 \times 10^{-4}}Ka=5.35×10−4
Explanation (300+ words)
The Ka of an acid is a quantitative measure of its strength in water. It tells us how much the acid ionizes. A monoprotic acid like HA donates one proton (H⁺) per molecule when dissolved in water. In this problem, we are given the initial concentration of the acid solution (0.0192 M) and the pH (2.53). From the pH, we first calculate the concentration of hydrogen ions [H⁺] using the formula: [H+]=10−pH[\text{H}^+] = 10^{-\text{pH}}[H+]=10−pH
This results in [H⁺] = 2.95 × 10⁻³ M. Since the acid is monoprotic and weak, it does not completely dissociate. Therefore, at equilibrium, the concentration of the undissociated acid (HA) is less than the initial amount. The amount of HA that dissociated is equal to the [H⁺] (and [A⁻]) formed.
Using the ICE table method, we calculate the remaining [HA] at equilibrium. Then, using the Ka expression: Ka=[H+][A−][HA]K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}Ka=[HA][H+][A−]
we plug in the equilibrium values. This gives us: Ka=(2.95×10−3)20.01625≈5.35×10−4K_a = \frac{(2.95 \times 10^{-3})^2}{0.01625} \approx 5.35 \times 10^{-4}Ka=0.01625(2.95×10−3)2≈5.35×10−4
This value of Ka indicates a moderately weak acid, since stronger acids have Ka values closer to 1 or higher. The low pH confirms some ionization, but the small Ka shows the ionization is limited—typical for weak acids like acetic acid.
