Let's solve each part step by step using the **Henderson-Hasselbalch equation**: \[ \text{pH} = \text{p}K_\text{a} + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] Where: - \([\text{A}^-]\) = concentration of acetate ion (from sodium acetate) - \([\text{HA}]\) = concentration of acetic acid - For acetic acid: \(K_a = 1.8 \times 10^{-5}\), so \(\text{p}K_a = 4.74\) --- ### **1. 1.00 M Acetic Acid & 1.00 M Sodium Acetate** #### **Step 1: Write the equation and identify concentrations** - \([\text{A}^-] = 1.00~\text{M}\) - \([\text{HA}] = 1.00~\text{M}\) - \(\text{p}K_a = 4.74\) #### **Step 2: Plug values into the Henderson-Hasselbalch equation** \[ \text{pH} = 4.74 + \log\left(\frac{1.00}{1.00}\right) \] #### **Step 3: Calculate and explain** \[ \text{pH} = 4.74 + \log(1) = 4.74 + 0 = 4.74 \] **Explanation:** When the concentrations of acid and its conjugate base are equal, the pH equals the pKa. **Final Answer:** **pH = 4.74** --- ### **2. 0.800 M Acetic Acid & 1.00 M Sodium Acetate** #### **Step 1: Identify concentrations** - \([\text{A}^-] = 1.00~\text{M}\) - \([\text{HA}] = 0.800~\text{M}\) - \(\text{p}K_a = 4.74\) #### **Step 2: Plug into the equation** \[ \text{pH} = 4.74 + \log\left(\frac{1.00}{0.800}\right) \] #### **Step 3: Calculate and explain** \[ \frac{1.00}{0.800} = 1.25 \\ \log(1.25) \approx 0.10 \\ \text{pH} = 4.74 + 0.10 = 4.84 \] **Explanation:** The base (acetate) is in higher concentration than the acid, so the pH is slightly higher than the pKa. **Final Answer:** **pH = 4.84** --- ### **3. 1.00 M Acetic Acid & 0.800 M Sodium Acetate** #### **Step 1: Identify concentrations** - \([\text{A}^-] = 0.800~\text{M}\) - \([\text{HA}] = 1.00~\text{M}\) - \(\text{p}K_a = 4.74\) #### **Step 2: Plug into the equation** \[ \text{pH} = 4.74 + \log\left(\frac{0.800}{1.00}\right) \] #### **Step 3: Calculate and explain** \[ \frac{0.800}{1.00} = 0.80 \\ \log(0.80) \approx -0.10 \\ \text{pH} = 4.74 + (-0.10) = 4.64 \] **Explanation:** The acid is in higher concentration than the base, so the pH is slightly lower than the pKa. **Final Answer:** **pH = 4.64** --- ## **Summary Table** | Acetic Acid (M) | Sodium Acetate (M) | pH | |-----------------|--------------------|------| | 1.00 | 1.00 | 4.74 | | 0.800 | 1.00 | 4.84 | | 1.00 | 0.800 | 4.64 |

The concept used to determine the pH of each buffer solution is the Henderson-Hasselbalch equation, which relates the pH of a buffer to the pKa of the acid and the ratio of the concentrations of its conjugate base and acid. By knowing the pKa of acetic acid (approximately 4.74), we can calculate the pH based on the given molar concentrations of acetic acid (the weak acid) and sodium acetate (its conjugate base). When the concentrations of acid and base are equal, the pH equals the pKa. If the conjugate base concentration is higher, the pH increases; if lower, it decreases. This equation simplifies the process of calculating the pH for buffer solutions with varying acid/base ratios.