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To find the pH of a solution with a weak monoprotic acid, we can use the acid dissociation constant, \( K_a \), and the initial concentration of the acid, \( [\text{HA}]_0 \).

The \( K_a \) expression for a weak acid \( \text{HA} \) dissociating into \( \text{H}^+ \) and \( \text{A}^- \) is:
\[
K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}
\]

Given:
- \( K_a = 1.90 \times 10^{-5} \)
- \( [\text{HA}]_0 = 0.0815 \, \text{M} \)

### Step-by-Step Solution

1. **Set up the \( K_a \) expression:**  
   Let \( x \) represent the concentration of \( \text{H}^+ \) (and \( \text{A}^- \)) ions at equilibrium.
   \[
   K_a = \frac{x^2}{[\text{HA}]_0 - x}
   \]

2. **Make the assumption that \( x \) is small compared to \( [\text{HA}]_0 \):**  
   Since the acid is weak, we can assume \( [\text{HA}]_0 - x \approx [\text{HA}]_0 \).
   So,
   \[
   K_a \approx \frac{x^2}{[\text{HA}]_0}
   \]

3. **Solve for \( x \):**
   Substitute \( K_a = 1.90 \times 10^{-5} \) and \( [\text{HA}]_0 = 0.0815 \):
   \[
   1.90 \times 10^{-5} = \frac{x^2}{0.0815}
   \]
   \[
   x^2 = (1.90 \times 10^{-5}) \times 0.0815
   \]
   \[
   x^2 = 1.5485 \times 10^{-6}
   \]
   \[
   x = \sqrt{1.5485 \times 10^{-6}}
   \]
   \[
   x \approx 0.001244 \, \text{M}
   \]

4. **Calculate the pH:**
   Since \( x \approx [\text{H}^+] \),
   \[
   \text{pH} = -\log [\text{H}^+]
   \]
   \[
   \text{pH} = -\log(0.001244)
   \]
   \[
   \text{pH} \approx 2.905
   \]

### Final Answer
\[
\text{pH} \approx 2.91
\]

Would you like more details or have any questions on this calculation?

---

1. How would the pH change if the initial concentration of the acid was increased?
2. What assumptions do we make when using the \( K_a \) approximation?
3. How would we calculate the pH if the \( K_a \) value was very high (stronger acid)?
4. How does temperature affect the \( K_a \) and pH of a solution?
5. Can this method be used for polyprotic acids?

**Tip:** For weak acids, if \( K_a \) is small, assume that the dissociation is minimal, which simplifies calculations.

The Ka of a weak monoprotic acid is 1.90 x 10^-5. What is the pH of a 0.0815 M solution of this acid?

Learn how to find the pH of a weak monoprotic acid solution with Ka = 1.90 x 10^-5 and a concentration of 0.0815 M. This detailed solution includes acid-base equilibrium concepts and step-by-step calculations. Suitable for advanced high school chemistry students.

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