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 (Ka), we can use the following steps:
Step 1: Write the dissociation equation of the acid.
For a weak acid HA dissociating in water:
[ \text{HA (aq)} \rightleftharpoons \text{H}^+ (aq) + \text{A}^- (aq) ]
Step 2: Set up the expression for the equilibrium constant (Ka).
The expression for ( K_a ) is:
[
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
]
where:
- ( [\text{H}^+] ) is the concentration of hydrogen ions,
- ( [\text{A}^-] ) is the concentration of the conjugate base,
- ( [\text{HA}] ) is the concentration of the weak acid at equilibrium.
Step 3: Use the ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentrations.
We start with 0.2 M of HA, and initially, no ( \text{H}^+ ) or ( \text{A}^- ) ions are present.
| Species | Initial Concentration | Change in Concentration | Equilibrium Concentration |
|---|---|---|---|
| HA | 0.2 M | -x | 0.2 – x |
| H⁺ | 0 M | +x | x |
| A⁻ | 0 M | +x | x |
At equilibrium, the concentration of hydrogen ions ( [\text{H}^+] = x ), and the concentration of HA is ( [\text{HA}] = 0.2 – x ).
Step 4: Substitute the equilibrium concentrations into the expression for Ka.
[
K_a = \frac{x \cdot x}{0.2 – x}
]
Given that ( K_a = 8.80 \times 10^{-6} ), we can substitute this value:
[
8.80 \times 10^{-6} = \frac{x^2}{0.2 – x}
]
Step 5: Make an assumption for small ( x ).
Since the acid dissociates very weakly (due to the small ( K_a )), we can assume that ( x ) will be very small compared to 0.2 M, so we approximate ( 0.2 – x \approx 0.2 ). This simplifies the equation to:
[
8.80 \times 10^{-6} = \frac{x^2}{0.2}
]
Step 6: Solve for ( x ), the concentration of H⁺.
[
x^2 = (8.80 \times 10^{-6}) \times 0.2
]
[
x^2 = 1.76 \times 10^{-6}
]
[
x = \sqrt{1.76 \times 10^{-6}} = 1.33 \times 10^{-3} \, \text{M}
]
Step 7: Calculate the pH.
The pH is given by:
[
\text{pH} = -\log [\text{H}^+]
]
Substitute the value of ( x ) (the concentration of H⁺):
[
\text{pH} = -\log (1.33 \times 10^{-3}) = 2.876
]
Final Answer:
The pH of the 0.2 M weak acid solution is approximately 2.88.
Explanation:
To find the pH of a weak acid solution, we used the acid dissociation constant ( K_a ) and an equilibrium approach. The dissociation of the acid is minimal, so we made the assumption that ( x ) is very small, which simplifies the calculations. The result shows that the solution is acidic, with a pH of 2.88, indicating that the acid does not fully dissociate in water.