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 a monoprotic weak acid, we follow these steps:
Step 1: Use the pH to find [H⁺]
The pH is given as 2.53. Use the formula: [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}
Step 2: Set up an ICE Table
For a monoprotic weak acid HA, the dissociation in water is: HA⇌H++A−\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HA | 0.0192 | –x | 0.0192 – x |
| H⁺ | 0 | +x | x (we found this to be 2.95 × 10⁻³) |
| A⁻ | 0 | +x | x (same as [H⁺]) |
Step 3: Plug into the Ka expression
Ka=[H+][A−][HA]K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
At equilibrium:
- [H+]=x=2.95×10−3[H^+] = x = 2.95 \times 10^{-3}
- [A−]=x=2.95×10−3[A^-] = x = 2.95 \times 10^{-3}
- [HA]=0.0192−x≈0.0192−0.00295=0.01625 M[HA] = 0.0192 – x \approx 0.0192 – 0.00295 = 0.01625 \, \text{M}
Now plug in: 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}
✅ Final Answer:
Ka≈5.35×10−4K_a \approx \boxed{5.35 \times 10^{-4}}
🧠 Explanation (300+ words)
To find the acid dissociation constant (Ka) of a weak acid, we rely on the relationship between pH and hydrogen ion concentration. In this case, the acid is monoprotic, meaning it donates one proton (H⁺) per molecule when dissolved in water. The solution has a concentration of 0.0192 M, and a pH of 2.53.
First, we determine how many hydrogen ions are in solution. Since pH is the negative logarithm of the hydrogen ion concentration, we reverse this using the formula [H+]=10−pH[H^+] = 10^{-\text{pH}}. This gives us the concentration of hydrogen ions as approximately 2.95 × 10⁻³ M.
Using an ICE table, we track how the weak acid (HA) dissociates in water. Initially, we have 0.0192 M of HA and 0 M of H⁺ and A⁻. At equilibrium, HA has lost some concentration (x), and equal amounts of H⁺ and A⁻ have been formed. Since the acid is monoprotic, the increase in H⁺ is the same as the increase in A⁻, and is equal to the amount dissociated from HA.
We then substitute the equilibrium concentrations into the Ka expression: [H+][A−][HA]\frac{[H^+][A^-]}{[HA]}. Plugging in our values, we square the [H⁺] and divide by the remaining [HA]. This calculation gives us the Ka as approximately 5.35 × 10⁻⁴, which reflects the acid’s strength: a relatively small value, indicating a weak acid that does not dissociate completely in water.
This process illustrates the fundamental principles of chemical equilibrium, pH, and weak acid behavior, all essential for understanding acid-base chemistry in biology, medicine, and environmental science.