Calculate the pH of a 0.33 M nitrous acid (HNO2, Ka = 4.5 x10-4) solution.
0.48
3.82
1.92
None is correct
The Correct Answer and Explanation is :
To calculate the pH of a 0.33 M nitrous acid (HNO₂) solution, we use the following steps:
Step 1: Write the Dissociation Equation
Nitrous acid (HNO₂) is a weak acid and partially dissociates in water as follows:
[
HNO_2 \rightleftharpoons H^+ + NO_2^-
]
Given that the acid dissociation constant (( K_a )) for nitrous acid is:
[
K_a = 4.5 \times 10^{-4}
]
Step 2: Set Up the ICE Table
Let’s define the initial concentration and the changes due to dissociation:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HNO₂ | 0.33 | -x | 0.33 – x |
| H⁺ | 0 | +x | x |
| NO₂⁻ | 0 | +x | x |
Step 3: Apply the ( K_a ) Expression
[
K_a = \frac{[H^+][NO_2^-]}{[HNO_2]}
]
Substituting the equilibrium values:
[
4.5 \times 10^{-4} = \frac{x^2}{0.33 – x}
]
Since ( K_a ) is small, we assume ( x ) is small compared to 0.33, so we approximate:
[
0.33 – x \approx 0.33
]
Thus, the equation simplifies to:
[
4.5 \times 10^{-4} = \frac{x^2}{0.33}
]
Step 4: Solve for ( x )
[
x^2 = (4.5 \times 10^{-4}) \times (0.33)
]
[
x^2 = 1.485 \times 10^{-4}
]
[
x = \sqrt{1.485 \times 10^{-4}}
]
[
x \approx 0.0122
]
Since ( x = [H^+] ), we find:
[
pH = -\log(0.0122)
]
[
pH \approx 3.92
]
Step 5: Compare with Given Answers
The closest option is none of the given answers because 3.92 is not listed.
Explanation (300 words)
The pH of a 0.33 M nitrous acid solution was determined using the acid dissociation constant (( K_a )) and the ICE table method. Since nitrous acid is a weak acid, it does not completely dissociate in solution. We set up an equilibrium expression using the ( K_a ) value and approximated the dissociation to simplify calculations.
By solving for ( [H^+] ), we found it to be approximately 0.0122 M. Taking the negative logarithm of the hydrogen ion concentration gave us a pH of 3.92, which was not listed among the answer choices.
This calculation highlights an important principle in acid-base chemistry: weak acids partially ionize, and their pH must be determined using equilibrium expressions rather than assuming full dissociation. Approximations are often used in these calculations when ( K_a ) is small compared to the initial concentration, as in this case.
This result confirms that none of the provided answer choices are correct. If an answer choice of 3.92 had been available, it would have been the correct selection.