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 determine the acid dissociation constant ( K_a ) for the monoprotic acid, we follow these steps:
Step 1: Define the Dissociation Reaction
Let the monoprotic acid be represented as ( HA ):
[
HA \rightleftharpoons H^+ + A^-
]
The equilibrium expression for the dissociation constant ( K_a ) is:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Step 2: Determine ( [H^+] ) from pH
The given pH of the solution is 2.53. We calculate ( [H^+] ) using:
[
[H^+] = 10^{-\text{pH}} = 10^{-2.53}
]
Using a calculator:
[
[H^+] = 2.95 \times 10^{-3} \text{ M}
]
Step 3: Set Up an ICE Table
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| ( HA ) | 0.0192 | (-x) | ( 0.0192 – x ) |
| ( H^+ ) | 0 | ( +x ) | ( x ) |
| ( A^- ) | 0 | ( +x ) | ( x ) |
Since the concentration of ( H^+ ) at equilibrium is ( 2.95 \times 10^{-3} ) M, we set ( x = 2.95 \times 10^{-3} ):
[
[HA]_{\text{eq}} = 0.0192 – 2.95 \times 10^{-3} = 0.01625 \text{ M}
]
Step 4: Calculate ( K_a )
[
K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(2.95 \times 10^{-3})^2}{0.01625}
]
[
K_a = \frac{8.70 \times 10^{-6}}{0.01625} = 5.36 \times 10^{-4}
]
Final Answer:
[
K_a = 5.36 \times 10^{-4}
]
Explanation:
This problem involves calculating the acid dissociation constant ( K_a ) for a weak monoprotic acid from its pH and initial concentration. We first use the pH to find ( [H^+] ), then construct an ICE table to determine equilibrium concentrations. Finally, we substitute the values into the equilibrium expression to solve for ( K_a ). Since ( [H^+] ) is relatively small compared to the initial acid concentration, we assume minimal dissociation. The result, ( K_a = 5.36 \times 10^{-4} ), confirms that this is a weak acid, as weak acids typically have ( K_a ) values less than 1.