What is the Ka of HA if 0.10 M solution has pH = 3.50?
The correct answer and explanation is :
To determine the acid dissociation constant ( K_a ) of the weak acid ( HA ), we use the given information:
- Initial concentration of HA: ( 0.10 ) M
- pH of the solution: ( 3.50 )
Step 1: Determine ( [H^+] )
The pH is related to the hydrogen ion concentration by the formula:
[
\text{pH} = -\log [H^+]
]
Solving for ( [H^+] ):
[
[H^+] = 10^{-\text{pH}} = 10^{-3.50}
]
[
[H^+] = 3.16 \times 10^{-4} \text{ M}
]
Step 2: Set Up the ICE Table
For the dissociation of the weak acid:
[
HA \rightleftharpoons H^+ + A^-
]
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HA | 0.10 | (-x) | (0.10 – x) |
| (H^+) | 0 | (+x) | (x) |
| (A^-) | 0 | (+x) | (x) |
From Step 1, we found ( [H^+] = x = 3.16 \times 10^{-4} ) M.
Step 3: Calculate ( K_a )
The acid dissociation constant is given by:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Substituting the values:
[
K_a = \frac{(3.16 \times 10^{-4}) (3.16 \times 10^{-4})}{0.10 – 3.16 \times 10^{-4}}
]
Approximating ( 0.10 – 3.16 \times 10^{-4} \approx 0.10 ):
[
K_a \approx \frac{(3.16 \times 10^{-4})^2}{0.10}
]
[
K_a \approx \frac{9.98 \times 10^{-8}}{0.10}
]
[
K_a \approx 1.00 \times 10^{-6}
]
Final Answer:
[
K_a = 1.0 \times 10^{-6}
]
Explanation (Summary)
The problem involves a weak acid ( HA ) that partially ionizes in solution. The pH provides the hydrogen ion concentration, which we use in the equilibrium expression. By assuming ( x ) is small compared to 0.10 M, we simplify calculations and determine ( K_a ) as ( 1.0 \times 10^{-6} ), characterizing the acid’s strength.