< Question 12 of 19 > A volume of 500.0 mL of 0.110 M NaOH is added to 525 mL of 0.200 M weak acid (K₁ = 5.64 x 10-5). What is the pH of the resulting buffer? HA(aq) + OH(aq) -> → H₂O(1) + A¯(aq) pH =

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**Buffer Solution pH Calculation**

**Problem Statement:**

A volume of 500.0 mL of 0.110 M NaOH is added to 525 mL of 0.200 M weak acid (\( K_a = 5.64 \times 10^{-5} \)). What is the pH of the resulting buffer?

**Reaction:**

\[ \text{HA (aq)} + \text{OH}^- \text{(aq)} \rightarrow \text{H}_2\text{O (l)} + \text{A}^- \text{(aq)} \]

**To Find:**

The pH of the resulting buffer.

**Detailed Solution:**

To determine the pH of the resulting buffer solution after the addition of NaOH to the weak acid, a step-by-step calculation is required using the Henderson-Hasselbalch equation:

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

**Step 1: Calculate the moles of \(\text{HA}\) and \(\text{OH}^-\)**

- Moles of \(\text{HA} = \text{concentration} \times \text{volume}\)

  \[ \text{Moles of HA} = 0.200 \text{ M} \times 0.525 \text{ L} = 0.105 \text{ mol} \]

- Moles of \(\text{OH}^- = \text{concentration} \times \text{volume}\)

  \[ \text{Moles of OH}^- = 0.110 \text{ M} \times 0.500 \text{ L} = 0.055 \text{ mol} \]

**Step 2: Determine the limiting reagent and calculate the remaining moles of \(\text{HA}\) and \(\text{A}^- \)**

- Since \(\text{OH}^-\) is the limiting reagent:
  
  \[ \text{Remaining moles of HA} = 0.105 \text{ mol} - 0.055 \text{ mol} = 0.050 \text{ mol} \]
  
  \[ \text{Moles
Transcribed Image Text:**Buffer Solution pH Calculation** **Problem Statement:** A volume of 500.0 mL of 0.110 M NaOH is added to 525 mL of 0.200 M weak acid (\( K_a = 5.64 \times 10^{-5} \)). What is the pH of the resulting buffer? **Reaction:** \[ \text{HA (aq)} + \text{OH}^- \text{(aq)} \rightarrow \text{H}_2\text{O (l)} + \text{A}^- \text{(aq)} \] **To Find:** The pH of the resulting buffer. **Detailed Solution:** To determine the pH of the resulting buffer solution after the addition of NaOH to the weak acid, a step-by-step calculation is required using the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log \left(\frac{\text{[A}^- \text{]}}{\text{[HA]}}\right) \] **Step 1: Calculate the moles of \(\text{HA}\) and \(\text{OH}^-\)** - Moles of \(\text{HA} = \text{concentration} \times \text{volume}\) \[ \text{Moles of HA} = 0.200 \text{ M} \times 0.525 \text{ L} = 0.105 \text{ mol} \] - Moles of \(\text{OH}^- = \text{concentration} \times \text{volume}\) \[ \text{Moles of OH}^- = 0.110 \text{ M} \times 0.500 \text{ L} = 0.055 \text{ mol} \] **Step 2: Determine the limiting reagent and calculate the remaining moles of \(\text{HA}\) and \(\text{A}^- \)** - Since \(\text{OH}^-\) is the limiting reagent: \[ \text{Remaining moles of HA} = 0.105 \text{ mol} - 0.055 \text{ mol} = 0.050 \text{ mol} \] \[ \text{Moles
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