A certain indicator, HA, has a K₂ value of 4.0 x 10-7. The protonated form of the indicator is blue and the ionized form is red. What is the pKa of the indicator? pK₂ = What is the color of this indicator in a solution with pH = 4? red purple blue

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### Understanding Indicators and pKa

**Problem Statement:**

A certain indicator, HA, has a \( K_a \) value of \( 4.0 \times 10^{-7} \). The protonated form of the indicator is blue, and the ionized form is red.

1. **Question**: What is the \( pK_a \) of the indicator?
   - **Answer Box**: (space provided)

2. **Question**: What is the color of this indicator in a solution with \( \text{pH} = 4 \)?
   - **Options**:
     - ☐ red
     - ☐ purple
     - ☐ blue

---

### Explanation:

- **Calculating pKa**: 
  The \( pK_a \) is calculated using the formula \( pK_a = -\log(K_a) \).

- **Color Indication**:
  - The color of an indicator in solution is based on the pH compared to its \( pK_a \).
  - If the pH is lower than the \( pK_a \), the solution is likely dominated by the **protonated form (blue)**.
  - If the pH is higher, the **ionized form (red)** is predominant.
  
In this case, with \( pH = 4 \), you would compare it to the calculated \( pK_a \) to determine the color.
Transcribed Image Text:### Understanding Indicators and pKa **Problem Statement:** A certain indicator, HA, has a \( K_a \) value of \( 4.0 \times 10^{-7} \). The protonated form of the indicator is blue, and the ionized form is red. 1. **Question**: What is the \( pK_a \) of the indicator? - **Answer Box**: (space provided) 2. **Question**: What is the color of this indicator in a solution with \( \text{pH} = 4 \)? - **Options**: - ☐ red - ☐ purple - ☐ blue --- ### Explanation: - **Calculating pKa**: The \( pK_a \) is calculated using the formula \( pK_a = -\log(K_a) \). - **Color Indication**: - The color of an indicator in solution is based on the pH compared to its \( pK_a \). - If the pH is lower than the \( pK_a \), the solution is likely dominated by the **protonated form (blue)**. - If the pH is higher, the **ionized form (red)** is predominant. In this case, with \( pH = 4 \), you would compare it to the calculated \( pK_a \) to determine the color.
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