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 can follow these steps. The given data includes the concentration of the acid (HA) and the pH of the solution. We will use the relationship between the concentration of the acid and its dissociation to find the Ka.
Step 1: Determine the concentration of hydrogen ions [H⁺]
The pH of the solution is given as 2.53. To find the concentration of hydrogen ions, we use the following formula:
[
\text{pH} = -\log[\text{H}^+]
]
Rearranging this formula to solve for [H⁺]:
[
[\text{H}^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
]
Step 2: Set up an ICE table
Let’s assume that the monoprotic acid dissociates as follows:
[
\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-
]
We know the initial concentration of HA is 0.0192 M. Since the acid dissociates, we will use an ICE (Initial, Change, Equilibrium) table to organize the calculations:
| Species | HA | H⁺ | A⁻ |
|---|---|---|---|
| Initial | 0.0192 M | 0 M | 0 M |
| Change | -x | +x | +x |
| Equilibrium | 0.0192 – x | x | x |
Since we know that the concentration of hydrogen ions [H⁺] at equilibrium is approximately 2.95 × 10⁻³ M, we can substitute this value for x in the ICE table:
[
x = 2.95 \times 10^{-3} \, \text{M}
]
Step 3: Apply the equilibrium expression for Ka
The acid dissociation constant (Ka) for a monoprotic acid is given by:
[
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
]
At equilibrium, the concentration of HA is:
[
[\text{HA}] = 0.0192 – x = 0.0192 – 2.95 \times 10^{-3} \approx 0.01625 \, \text{M}
]
Now, we can substitute the equilibrium concentrations into the Ka expression:
[
K_a = \frac{(2.95 \times 10^{-3})(2.95 \times 10^{-3})}{0.01625}
]
[
K_a \approx \frac{8.70 \times 10^{-6}}{0.01625} \approx 5.35 \times 10^{-4}
]
Final Answer:
The acid dissociation constant (Ka) for the monoprotic acid is approximately 5.35 × 10⁻⁴.
Explanation:
In this problem, we used the pH of the solution to calculate the concentration of hydrogen ions at equilibrium. From there, we used an ICE table to track the dissociation of the acid and determine the equilibrium concentrations of the species involved. Finally, we applied the definition of the acid dissociation constant, Ka, to calculate its value.