Find the pH of 0.2 M a weak acid (HA) with Ka = 8.80 x 10-6.
The correct answer and explanation is :
To find the pH of a 0.2 M solution of a weak acid (HA) with a given acid dissociation constant ( K_a = 8.80 \times 10^{-6} ), we follow these steps:
Step 1: Write the Ionization Equation
The weak acid dissociates in water as follows:
[
HA \rightleftharpoons H^+ + A^-
]
Step 2: Set Up the ICE Table
Let’s define the initial concentration of HA as 0.2 M and assume that the concentration of ( H^+ ) and ( A^- ) starts at 0.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HA | 0.2 | -x | 0.2 – x |
| ( H^+ ) | 0 | +x | x |
| ( A^- ) | 0 | +x | x |
Step 3: Apply the Ka Expression
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Substituting the values:
[
8.80 \times 10^{-6} = \frac{x^2}{0.2 – x}
]
Since ( K_a ) is small, we assume ( x ) is much smaller than 0.2, so ( 0.2 – x \approx 0.2 ):
[
8.80 \times 10^{-6} = \frac{x^2}{0.2}
]
Step 4: Solve for ( x ) (Hydrogen Ion Concentration)
[
x^2 = (8.80 \times 10^{-6}) \times 0.2
]
[
x^2 = 1.76 \times 10^{-6}
]
[
x = \sqrt{1.76 \times 10^{-6}}
]
[
x = 1.33 \times 10^{-3} \text{ M}
]
Since ( x = [H^+] ), we find the pH:
[
\text{pH} = -\log(1.33 \times 10^{-3})
]
[
\text{pH} \approx 2.88
]
Explanation:
This calculation follows a systematic approach to finding the pH of a weak acid. By using the ICE table and assuming ( x ) is small compared to the initial concentration, we simplify the quadratic equation, making it easier to solve. The small ( K_a ) confirms that HA only partially dissociates, leading to a pH greater than that of a strong acid of the same concentration.