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 acid, we can use the following approach based on the relationship between the pH and the concentration of the acid in solution.
Given Data:
- Concentration of the acid (([HA])) = 0.0192 M
- pH of the solution = 2.53
Step 1: Calculate the [H⁺] concentration from the pH.
The pH is related to the concentration of hydrogen ions ([H⁺]) by the equation:
[
\text{pH} = -\log[H⁺]
]
Rearranging to solve for ([H⁺]):
[
[H⁺] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.96 \times 10^{-3} \, \text{M}
]
Step 2: Write the dissociation equation for the monoprotic acid.
For a monoprotic acid (HA), the dissociation can be written as:
[
HA \rightleftharpoons H^+ + A^-
]
The acid dissociation constant (Ka) is given by:
[
Ka = \frac{[H^+][A^-]}{[HA]}
]
Since the acid dissociates into equal amounts of ([H^+]) and ([A^-]), at equilibrium, we can assume:
- ([H^+] = [A^-]), and both are approximately equal to (2.96 \times 10^{-3} \, \text{M}) based on the pH.
- The remaining concentration of HA at equilibrium is approximately ([HA] = 0.0192 – 2.96 \times 10^{-3} \approx 0.0162 \, \text{M}).
Step 3: Plug values into the Ka expression.
Now substitute the known values into the Ka equation:
[
Ka = \frac{(2.96 \times 10^{-3})(2.96 \times 10^{-3})}{0.0162} = \frac{8.76 \times 10^{-6}}{0.0162} \approx 5.4 \times 10^{-4}
]
Final Answer:
The acid dissociation constant ( Ka ) for the monoprotic acid is approximately ( 5.4 \times 10^{-4} ).
Explanation:
In this problem, we use the pH to calculate the concentration of hydrogen ions ([H^+]), which provides insight into how much of the acid has dissociated. The dissociation constant (Ka) is a measure of the strength of the acid, and it reflects the extent to which the acid dissociates into ions in solution. The higher the value of (Ka), the stronger the acid. In this case, (Ka = 5.4 \times 10^{-4}) suggests that the acid is weak, as it only partially dissociates in solution.