3. What is the pH of a solution that is 0.75 M in sodium acetate and 0.50 M in acetic acid? (K for acetic acid is 1.8x10-5.)

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**Problem 3: Calculating pH of a Buffered Solution**

What is the pH of a solution that is 0.75 M in sodium acetate and 0.50 M in acetic acid? Given that the \( K_a \) for acetic acid is \( 1.8 \times 10^{-5} \).

**Explanation:**

This problem involves calculating the pH of a buffer solution using the Henderson-Hasselbalch equation, which is suitable for solutions comprising a weak acid and its conjugate base. The equation is given by:

\[
\text{pH} = \text{pK}_a + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]

Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium acetate),
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid),
- \(\text{pK}_a = -\log(K_a)\).

**Calculation Steps:**

1. Calculate \(\text{pK}_a\):
   \[
   \text{pK}_a = -\log(1.8 \times 10^{-5})
   \]

2. Apply concentrations to the Henderson-Hasselbalch equation:
   \[
   \text{pH} = \text{pK}_a + \log \left(\frac{0.75}{0.50}\right)
   \]

This method effectively calculates the pH of buffer solutions in which both the acid and base components are present in significant amounts.
Transcribed Image Text:**Problem 3: Calculating pH of a Buffered Solution** What is the pH of a solution that is 0.75 M in sodium acetate and 0.50 M in acetic acid? Given that the \( K_a \) for acetic acid is \( 1.8 \times 10^{-5} \). **Explanation:** This problem involves calculating the pH of a buffer solution using the Henderson-Hasselbalch equation, which is suitable for solutions comprising a weak acid and its conjugate base. The equation is given by: \[ \text{pH} = \text{pK}_a + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] Where: - \([\text{A}^-]\) is the concentration of the conjugate base (sodium acetate), - \([\text{HA}]\) is the concentration of the weak acid (acetic acid), - \(\text{pK}_a = -\log(K_a)\). **Calculation Steps:** 1. Calculate \(\text{pK}_a\): \[ \text{pK}_a = -\log(1.8 \times 10^{-5}) \] 2. Apply concentrations to the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pK}_a + \log \left(\frac{0.75}{0.50}\right) \] This method effectively calculates the pH of buffer solutions in which both the acid and base components are present in significant amounts.
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