Colourless salt `(A)`+dil.`H_(2)SO_(4)` or `CH_(3)COOH + KI to` blue colour with starch. `(A)` can be

Options

A. `K_(2)SO_(3)`
B. `Na_(2)CO_(3)`
C. `NH_(4)NO_(3)`
D. `NH_(4)Cl`

Answer

To determine the identity of the colorless salt (A) that, when reacted with dilute H₂SO₄ or CH₃COOH in the presence of KI, produces a blue color with starch, we can follow these steps:

### Step 1: Understand the Reaction
When a colorless salt (A) reacts with dilute sulfuric acid (H₂SO₄) or acetic acid (CH₃COOH) and potassium iodide (KI), it is important to identify the products formed. The blue color with starch indicates the presence of iodine (I₂), which suggests that iodine is being generated in the reaction.

### Step 2: Identify Possible Reactions
The formation of iodine (I₂) from KI typically occurs when iodide ions (I⁻) are oxidized. This oxidation can happen in the presence of certain acids. Therefore, we need to identify a colorless salt (A) that can release iodide ions and facilitate this oxidation.

### Step 3: Analyze the Colorless Salt
The colorless salt (A) must be capable of reacting with the acid to produce iodine. One common salt that fits this description is ammonium nitrate (NH₄NO₃). When ammonium nitrate reacts with dilute sulfuric acid, it can produce ammonium ions (NH₄⁺) and nitrate ions (NO₃⁻), which can lead to the oxidation of iodide ions from KI to iodine.

### Step 4: Confirm the Reaction
The reaction can be summarized as follows:
1. NH₄NO₃ + H₂SO₄ → NH₄⁺ + NO₃⁻ (producing ammonium and nitrate ions)
2. KI + I⁻ (from KI) + oxidizing agent (from the acid) → I₂ (iodine)

The iodine (I₂) produced will then react with starch to give a blue color.

### Step 5: Conclusion
Based on the above analysis, the colorless salt (A) that can react with dilute H₂SO₄ or CH₃COOH and KI to produce a blue color with starch is ammonium nitrate (NH₄NO₃).

### Final Answer:
(A) can be ammonium nitrate (NH₄NO₃).

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