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 find the Ka of a monoprotic acid from its concentration and pH, follow these steps:
Step-by-step Solution:
We are given:
- Acid concentration: ([HA] = 0.0192) M
- pH of the solution: 2.53
Step 1: Calculate ([H^+]) from pH
[
[H^+] = 10^{-pH} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
]
Step 2: Write the dissociation equation for a monoprotic acid:
[
HA \rightleftharpoons H^+ + A^-
]
Since it’s monoprotic, the amount of (H^+) produced is equal to the amount of (A^-) formed.
Let the initial concentration of (HA) be 0.0192 M. At equilibrium:
- ([H^+] = [A^-] = 2.95 \times 10^{-3}) M
- ([HA] = 0.0192 – 2.95 \times 10^{-3} = 0.01625) M
Step 3: Use the Ka expression
[
Ka = \frac{[H^+][A^-]}{[HA]} = \frac{(2.95 \times 10^{-3})^2}{0.01625}
]
[
Ka = \frac{8.70 \times 10^{-6}}{0.01625} \approx 5.35 \times 10^{-4}
]
✅ Final Answer:
[
\boxed{Ka = 5.35 \times 10^{-4}}
]
💬 Explanation (Approx. 300 words):
To calculate the acid dissociation constant ((Ka)) of a monoprotic acid, we need two main values: the concentration of the acid and the pH of the solution. The pH tells us how much hydrogen ion (([H^+])) has dissociated into the solution.
Since the acid is monoprotic, it donates only one hydrogen ion per molecule. This means that for each mole of acid that dissociates, one mole of hydrogen ions and one mole of conjugate base ((A^-)) are produced.
We first calculate ([H^+]) using the pH formula:
[
[H^+] = 10^{-\text{pH}} = 10^{-2.53}
]
This gives us the concentration of hydrogen ions at equilibrium.
Next, we assume the acid started with an initial concentration of 0.0192 M. Since some portion of the acid dissociates, we subtract the amount that dissociated (equal to ([H^+])) from the initial amount to get the equilibrium concentration of the undissociated acid.
Using the formula for the acid dissociation constant:
[
Ka = \frac{[H^+][A^-]}{[HA]}
]
we plug in our equilibrium concentrations. The numerator reflects the products of dissociation, while the denominator reflects the remaining reactant. Solving the expression gives the value of (Ka), which indicates the strength of the acid. A (Ka) value of (5.35 \times 10^{-4}) shows the acid is weak but not extremely weak—it partially dissociates in water.