1. Determine the pH when 5.89 g of sodium acetate is dissolved in 190 mL of water. Ka of acetic Na + OOCCH, (NaOOCCH₂) acid is 5.56 10-¹⁰ NaOOCCH₂ OOCCH3 + H₂O HOOCCH₂ + OH-

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**Weak Base Equilibria**

1. **Determine the pH when 5.89 g of sodium acetate (NaOOCCH₃) is dissolved in 190 mL of water.** Ka of acetic acid is \(5.56 \times 10^{-10}\).

   \[ \text{NaOOCCH}_3 \rightarrow \text{Na}^+ + \text{OOCCH}_3^- \]

   \[ \text{OOCCH}_3^- + \text{H}_2\text{O} \rightleftharpoons \text{HOOCCH}_3 + \text{OH}^- \]

**Explanation:**

- **First Reaction:** Sodium acetate (\(\text{NaOOCCH}_3\)) dissociates in water to produce sodium ions (\(\text{Na}^+\)) and acetate ions (\(\text{OOCCH}_3^-\)).

- **Second Reaction:** The acetate ion (\(\text{OOCCH}_3^-\)) acts as a weak base, reacting with water to form acetic acid (\(\text{HOOCCH}_3\)) and hydroxide ions (\(\text{OH}^-\)). This establishes an equilibrium state for the weak base.
Transcribed Image Text:**Weak Base Equilibria** 1. **Determine the pH when 5.89 g of sodium acetate (NaOOCCH₃) is dissolved in 190 mL of water.** Ka of acetic acid is \(5.56 \times 10^{-10}\). \[ \text{NaOOCCH}_3 \rightarrow \text{Na}^+ + \text{OOCCH}_3^- \] \[ \text{OOCCH}_3^- + \text{H}_2\text{O} \rightleftharpoons \text{HOOCCH}_3 + \text{OH}^- \] **Explanation:** - **First Reaction:** Sodium acetate (\(\text{NaOOCCH}_3\)) dissociates in water to produce sodium ions (\(\text{Na}^+\)) and acetate ions (\(\text{OOCCH}_3^-\)). - **Second Reaction:** The acetate ion (\(\text{OOCCH}_3^-\)) acts as a weak base, reacting with water to form acetic acid (\(\text{HOOCCH}_3\)) and hydroxide ions (\(\text{OH}^-\)). This establishes an equilibrium state for the weak base.
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