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) of the monoprotic acid, we follow these steps:
Step 1: Define the Dissociation Reaction
Let HA represent the monoprotic acid. It dissociates as:
[
HA \rightleftharpoons H^+ + A^-
]
Given:
- Initial concentration of HA = 0.0192 M
- pH = 2.53
Step 2: Determine ([H^+])
The concentration of hydrogen ions ([H^+]) can be found using the pH formula:
[
[H^+] = 10^{-\text{pH}}
]
[
[H^+] = 10^{-2.53} = 2.95 \times 10^{-3} \text{ M}
]
Step 3: Set Up the ICE Table
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HA | 0.0192 | (- x) | 0.0192 – x |
| H⁺ | 0 | (+x) | (2.95 \times 10^{-3}) |
| A⁻ | 0 | (+x) | (2.95 \times 10^{-3}) |
Since ([H^+] = x = 2.95 \times 10^{-3}), the equilibrium concentration of HA is:
[
[HA] = 0.0192 – 2.95 \times 10^{-3} = 0.0163 \text{ M}
]
Step 4: Calculate (K_a)
The acid dissociation constant is given by:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
[
K_a = \frac{(2.95 \times 10^{-3}) (2.95 \times 10^{-3})}{0.0163}
]
[
K_a = \frac{8.70 \times 10^{-6}}{0.0163}
]
[
K_a = 5.34 \times 10^{-4}
]
Final Answer:
[
K_a = 5.34 \times 10^{-4}
]
Explanation (300 Words)
The acid dissociation constant ((K_a)) quantifies the strength of an acid in solution. A monoprotic acid ((HA)) dissociates into hydrogen ions ((H^+)) and conjugate base ((A^-)). Given a 0.0192 M acid solution with a pH of 2.53, we first find ([H^+]) using (10^{-\text{pH}}), yielding (2.95 \times 10^{-3}) M. Since ([H^+]) equals the dissociated amount of acid, we subtract this from the initial concentration to find the remaining ([HA]). The equilibrium concentrations allow us to apply the (K_a) formula: (\frac{[H^+][A^-]}{[HA]}). Substituting values, we calculate (K_a = 5.34 \times 10^{-4}), which is a moderate value, suggesting the acid is weak but partially dissociated in solution.