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 need to use the given concentration of the acid and the pH of the solution. A monoprotic acid dissociates as follows:
[
HA \rightleftharpoons H^+ + A^-
]
Where:
- HA is the monoprotic acid,
- H+ is the hydrogen ion,
- A- is the conjugate base.
Step 1: Determine the concentration of hydrogen ions ([H^+])
We are given the pH of the solution, which is 2.53. The pH is related to the hydrogen ion concentration ([H^+]) by the formula:
[
\text{pH} = -\log[H^+]
]
We can solve for ([H^+]) by rearranging the equation:
[
[H^+] = 10^{-\text{pH}} = 10^{-2.53} = 2.96 \times 10^{-3} \, \text{M}
]
Step 2: Set up the ICE table (Initial, Change, Equilibrium)
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HA | 0.0192 | -x | 0.0192 – x |
| H+ | 0 | +x | x |
| A- | 0 | +x | x |
Where (x) is the amount of acid that dissociates, which is equal to the concentration of (H^+) at equilibrium.
Thus, the concentration of (H^+) at equilibrium is:
[
[H^+] = x = 2.96 \times 10^{-3} \, \text{M}
]
Step 3: Use the equilibrium expression for Ka
The acid dissociation constant (K_a) is given by the expression:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
At equilibrium, ([H^+] = [A^-] = x), and ([HA] = 0.0192 – x). Substituting these values:
[
K_a = \frac{(2.96 \times 10^{-3})(2.96 \times 10^{-3})}{0.0192 – 2.96 \times 10^{-3}}
]
[
K_a = \frac{(8.76 \times 10^{-6})}{0.0162}
]
[
K_a = 5.41 \times 10^{-4}
]
Final Answer:
The acid dissociation constant (K_a) for the monoprotic acid is approximately 5.41 × 10^-4.
Explanation:
To find the Ka of the acid, we used the relationship between pH and the concentration of hydrogen ions ([H^+]), then applied the concept of equilibrium in the dissociation of a monoprotic acid. By using the ICE table, we determined the change in concentration due to dissociation and used the equilibrium concentrations in the Ka expression to find the acid dissociation constant. This process allowed us to determine how strongly the acid dissociates in water.