`Na_(2)CO_(3)+H_(2)O+CO_(2)rarr(A).` White crystalline powder (A) on reaction with phenolphthalein gives

Options

A. Pink colour
B. Yellow colour
C. Orange colour
D. No colour

Answer

To solve the question step by step, we will analyze the reaction and the properties of the products involved.

### Step 1: Identify the Reactants
The reactants given in the question are:
- Sodium carbonate (Na₂CO₃)
- Water (H₂O)
- Carbon dioxide (CO₂)

### Step 2: Write the Reaction
When sodium carbonate reacts with water and carbon dioxide, the reaction can be represented as:
\[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \rightarrow \text{A} \]

### Step 3: Determine the Product (A)
In this reaction, sodium carbonate reacts with water and carbon dioxide to form sodium bicarbonate (NaHCO₃). Thus, we can write:
\[ \text{A} = \text{NaHCO}_3 \]

### Step 4: Identify the Properties of Product A
Sodium bicarbonate (NaHCO₃) is a white crystalline powder.

### Step 5: Reaction with Phenolphthalein
Phenolphthalein is an acid-base indicator that changes color depending on the pH of the solution:
- In acidic solutions, phenolphthalein is colorless.
- In neutral solutions, it remains colorless.
- In basic solutions, it turns pink.

Since sodium bicarbonate is a weak base, when it is dissolved in water, it will create a slightly basic solution. Therefore, when sodium bicarbonate reacts with phenolphthalein, it will turn pink.

### Final Answer
The white crystalline powder (A) on reaction with phenolphthalein gives a **pink color**.