When phenolphythalein is added to an aqueous solution containing one of the following solutes the solution turns pink. Which solute is present?

${ \textbf{(A)}\ \text{NaCl} \qquad\textbf{(B)}\ \text{KC}_2\text{H}_3\text{O}_2 \qquad\textbf{(C)}\ \text{LiBr} \qquad\textbf{(D)}\ \text{NH}_4\text{NO}_3 } $
1. **Understanding the Role of Phenolphthalein**:
   Phenolphthalein is a pH indicator that turns pink in basic solutions (pH > 8.2) and remains colorless in acidic or neutral solutions (pH < 8.2).

2. **Analyzing Each Solute**:
   - **(A) NaCl**:
     - NaCl dissociates into Na\(^+\) and Cl\(^-\) ions in water.
     - Na\(^+\) is the conjugate acid of a strong base (NaOH), and Cl\(^-\) is the conjugate base of a strong acid (HCl).
     - Both ions do not hydrolyze in water, so the solution remains neutral.
     - Therefore, NaCl does not turn phenolphthalein pink.

   - **(B) KC\(_2\)H\(_3\)O\(_2\)**:
     - KC\(_2\)H\(_3\)O\(_2\) dissociates into K\(^+\) and C\(_2\)H\(_3\)O\(_2\)^-\) ions in water.
     - K\(^+\) is the conjugate acid of a strong base (KOH), and C\(_2\)H\(_3\)O\(_2\)^-\) is the conjugate base of a weak acid (acetic acid, HC\(_2\)H\(_3\)O\(_2\)).
     - The acetate ion (C\(_2\)H\(_3\)O\(_2\)^-\)) hydrolyzes in water to form OH\(^-\) ions, making the solution basic:
       \[
       \text{C}_2\text{H}_3\text{O}_2^- + \text{H}_2\text{O} \rightarrow \text{HC}_2\text{H}_3\text{O}_2 + \text{OH}^-
       \]
     - Therefore, KC\(_2\)H\(_3\)O\(_2\) turns phenolphthalein pink.

   - **(C) LiBr**:
     - LiBr dissociates into Li\(^+\) and Br\(^-\) ions in water.
     - Li\(^+\) is the conjugate acid of a strong base (LiOH), and Br\(^-\) is the conjugate base of a strong acid (HBr).
     - Both ions do not hydrolyze in water, so the solution remains neutral.
     - Therefore, LiBr does not turn phenolphthalein pink.

   - **(D) NH\(_4\)NO\(_3\)**:
     - NH\(_4\)NO\(_3\) dissociates into NH\(_4^+\) and NO\(_3^-\) ions in water.
     - NH\(_4^+\) is the conjugate acid of a weak base (NH\(_3\)), and NO\(_3^-\) is the conjugate base of a strong acid (HNO\(_3\)).
     - NH\(_4^+\) hydrolyzes in water to form H\(^+\) ions, making the solution acidic:
       \[
       \text{NH}_4^+ + \text{H}_2\text{O} \rightarrow \text{NH}_3 + \text{H}_3\text{O}^+
       \]
     - Therefore, NH\(_4\)NO\(_3\) does not turn phenolphthalein pink.

3. **Conclusion**:
   Based on the analysis, the only solute that makes the solution basic and turns phenolphthalein pink is KC\(_2\)H\(_3\)O\(_2\).

The final answer is $\boxed{\text{B}}$.