To solve the question regarding the color change of the indicator HIn in acid-base titrations, we will analyze the dissociation reaction and the implications of different pH environments on the indicator's color.
### Step-by-Step Solution:
1. **Understanding the Indicator Reaction**:
The indicator HIn in aqueous solution can be represented as:
\[
\text{HIn (aq)} \rightleftharpoons \text{H}^+ (aq) + \text{In}^- (aq)
\]
- Here, HIn has color X, and when it dissociates, it produces H⁺ ions and In⁻ ions, which have color Y.
2. **Analyzing the Color Changes**:
- In acidic solutions (high concentration of H⁺), the equilibrium will shift to the left, favoring the formation of HIn, resulting in color X being observed.
- In alkaline solutions (low concentration of H⁺), the equilibrium will shift to the right, favoring the formation of In⁻, resulting in color Y being observed.
3. **Evaluating the Statements**:
- **Statement 1**: In a strong alkaline solution, color Y will be observed.
- **True**: In strong alkaline conditions, the concentration of H⁺ ions is low, shifting the equilibrium to the right, thus color Y is observed.
- **Statement 2**: In a strong acidic solution, color Y will be observed.
- **False**: In strong acidic conditions, the concentration of H⁺ ions is high, shifting the equilibrium to the left, thus color X is observed.
- **Statement 3**: Concentration of In⁻ is higher than that of HIn at the equivalence point.
- **False**: At the equivalence point in a titration, the amounts of acid and base are equal, and the concentration of HIn and In⁻ will depend on the specific conditions of the titration.
- **Statement 4**: In a strong alkaline solution, color X is observed.
- **False**: As established, in strong alkaline conditions, color Y is observed.
4. **Conclusion**:
The only correct statement is that in a strong alkaline solution, color Y will be observed.
### Final Answer:
The correct statement is: **In a strong alkaline solution, color Y will be observed.**
