Which of the following pair can't exist in solution

Options

A. `NaHCO_(3)` and `NaOH`
B. `Na_(2)CO_(3)` and `NaOH`
C. `Na_(2)CO_(3)` and `NaCI`
D. `NaHCO_(3)` and `NaCI`

Explanation

To determine which pair cannot exist in solution, we will analyze each option provided in the question.

### Step-by-Step Solution:

1. **Identify the Compounds in Each Option:**
- Option A: NaHCO3 (sodium bicarbonate) and NaOH (sodium hydroxide)
- Option B: Na2CO3 (sodium carbonate) and NaOH
- Option C: Na2CO3 and NaCl (sodium chloride)
- Option D: NaHCO3 and NaCl

2. **Analyze Option A (NaHCO3 and NaOH):**
- NaHCO3 is a weak acid (bicarbonate) and NaOH is a strong base. When NaOH is added to NaHCO3, a reaction occurs:
\[ \text{NaHCO}_3 + \text{NaOH} \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \]
- This reaction shows that NaHCO3 will decompose into Na2CO3 and water when mixed with NaOH. Therefore, they cannot exist together in solution without undergoing a reaction.

3. **Analyze Option B (Na2CO3 and NaOH):**
- Both Na2CO3 and NaOH are basic. They can coexist in solution without any reaction occurring between them.

4. **Analyze Option C (Na2CO3 and NaCl):**
- Na2CO3 (sodium carbonate) and NaCl (sodium chloride) are both soluble in water and can exist together in solution without any reaction.

5. **Analyze Option D (NaHCO3 and NaCl):**
- NaHCO3 and NaCl can coexist in solution as they do not react with each other. NaHCO3 remains as a bicarbonate in the presence of NaCl.

6. **Conclusion:**
- The pair that cannot exist in solution is **Option A: NaHCO3 and NaOH** because they react with each other, leading to the decomposition of NaHCO3.

### Final Answer:
The pair that cannot exist in solution is **Option A: NaHCO3 and NaOH**.