4. Calculate the molarity of H3O* and the molarity of OH (at 37°C) in a solution having a pH equal to... a) pH = 1.30 b) pH = 5.73 c) pH = 7.80 d) pH = 10.74 e) pH = 12.61 %3D

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### Problem 4: Calculating Molarity of H₃O⁺ and OH⁻

Calculate the molarity of H₃O⁺ (hydronium ion) and the molarity of OH⁻ (hydroxide ion) at 37°C in a solution having a pH equal to:

- **a)** pH = 1.30
- **b)** pH = 5.73
- **c)** pH = 7.80
- **d)** pH = 10.74
- **e)** pH = 12.61

### Explanation:

#### Steps to Calculate Molarity:
1. **For H₃O⁺ (hydronium ion):**
   \[ \text{[H₃O⁺]} = 10^{-\text{pH}} \]

2. **For OH⁻ (hydroxide ion):**
   - Calculate pOH using the relation:
     \[ \text{pH} + \text{pOH} = 14 \]
   - Then find [OH⁻] using:
     \[ \text{[OH⁻]} = 10^{-\text{pOH}} \]

### Examples:
1. **For pH = 1.30:**
   - \(\text{[H₃O⁺]} = 10^{-1.30}\)
   - \(\text{pOH} = 14 - 1.30 = 12.70\)
   - \(\text{[OH⁻]} = 10^{-12.70}\)

2. **For pH = 5.73:**
   - \(\text{[H₃O⁺]} = 10^{-5.73}\)
   - \(\text{pOH} = 14 - 5.73 = 8.27\)
   - \(\text{[OH⁻]} = 10^{-8.27}\)

3. **For pH = 7.80:**
   - \(\text{[H₃O⁺]} = 10^{-7.80}\)
   - \(\text{pOH} = 14 - 7.80 = 6.20\)
   - \(\text{[OH⁻]} = 10^{-6.20
Transcribed Image Text:### Problem 4: Calculating Molarity of H₃O⁺ and OH⁻ Calculate the molarity of H₃O⁺ (hydronium ion) and the molarity of OH⁻ (hydroxide ion) at 37°C in a solution having a pH equal to: - **a)** pH = 1.30 - **b)** pH = 5.73 - **c)** pH = 7.80 - **d)** pH = 10.74 - **e)** pH = 12.61 ### Explanation: #### Steps to Calculate Molarity: 1. **For H₃O⁺ (hydronium ion):** \[ \text{[H₃O⁺]} = 10^{-\text{pH}} \] 2. **For OH⁻ (hydroxide ion):** - Calculate pOH using the relation: \[ \text{pH} + \text{pOH} = 14 \] - Then find [OH⁻] using: \[ \text{[OH⁻]} = 10^{-\text{pOH}} \] ### Examples: 1. **For pH = 1.30:** - \(\text{[H₃O⁺]} = 10^{-1.30}\) - \(\text{pOH} = 14 - 1.30 = 12.70\) - \(\text{[OH⁻]} = 10^{-12.70}\) 2. **For pH = 5.73:** - \(\text{[H₃O⁺]} = 10^{-5.73}\) - \(\text{pOH} = 14 - 5.73 = 8.27\) - \(\text{[OH⁻]} = 10^{-8.27}\) 3. **For pH = 7.80:** - \(\text{[H₃O⁺]} = 10^{-7.80}\) - \(\text{pOH} = 14 - 7.80 = 6.20\) - \(\text{[OH⁻]} = 10^{-6.20
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