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 information: the concentration of the acid (0.0192 M) and the pH of the solution (2.53). The relationship between the pH and the concentration of hydrogen ions (([H^+])) is crucial to solve for Ka.
Step 1: Calculate ([H^+])
The pH of a solution is defined as:
[
\text{pH} = -\log [H^+]
]
Rearranging this equation to solve for ([H^+]):
[
[H^+] = 10^{-\text{pH}} = 10^{-2.53} \approx 2.95 \times 10^{-3} \, \text{M}
]
Step 2: Set up the dissociation equation
For a monoprotic acid (HA), the dissociation in water can be written as:
[
HA \rightleftharpoons H^+ + A^-
]
Initially, the concentration of the acid, ([HA]_0), is 0.0192 M, and the concentration of ([H^+]) and ([A^-]) are both initially 0 M.
After dissociation, let (x) represent the concentration of the dissociated hydrogen ions and conjugate base ions. From the pH calculation, we already know that ([H^+] = 2.95 \times 10^{-3} \, \text{M}), so:
[
x = [H^+] = 2.95 \times 10^{-3} \, \text{M}
]
Thus, at equilibrium:
[
[HA] = [HA]_0 – x = 0.0192 – 2.95 \times 10^{-3} \approx 0.01625 \, \text{M}
]
[
[A^-] = x = 2.95 \times 10^{-3} \, \text{M}
]
Step 3: Write the expression for (K_a)
The acid dissociation constant (K_a) is given by the equation:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Substitute the values:
[
K_a = \frac{(2.95 \times 10^{-3})(2.95 \times 10^{-3})}{0.01625} \approx \frac{8.70 \times 10^{-6}}{0.01625} \approx 5.36 \times 10^{-4}
]
Conclusion
The acid dissociation constant (Ka) for the monoprotic acid is approximately (5.36 \times 10^{-4}).
This process involves using the relationship between pH and the hydrogen ion concentration to find the degree of dissociation, and from there, we can calculate the (K_a) value. It demonstrates how pH provides valuable insight into the dissociation equilibrium of acids in solution.