`H_(2)S` gas is passed through an acidic solution of `K_(2)Cr_(2)O_(7)` .The solution turns milky, why?

Answer

To understand why the solution turns milky when `H₂S` gas is passed through an acidic solution of `K₂Cr₂O₇`, we can break down the process step by step.

### Step 1: Identify the Reactants
We have two main reactants in this scenario:
- `H₂S` (hydrogen sulfide gas)
- `K₂Cr₂O₇` (potassium dichromate) in an acidic solution, typically with `H₂SO₄` (sulfuric acid).

### Step 2: Reaction Conditions
The `K₂Cr₂O₇` is in an acidic medium, which is necessary for the reaction to proceed. The acidic environment helps facilitate the oxidation-reduction reaction.

### Step 3: Reaction Overview
When `H₂S` is introduced into the acidic solution of `K₂Cr₂O₇`, a redox reaction occurs. In this reaction, `H₂S` acts as a reducing agent and reduces the chromium from its +6 oxidation state in `K₂Cr₂O₇` to a lower oxidation state, while itself being oxidized to sulfur.

### Step 4: Write the Balanced Chemical Equation
The balanced chemical equation for the reaction can be written as follows:

\[ 3 H₂S + K₂Cr₂O₇ + 4 H₂SO₄ \rightarrow 3 S + K₂SO₄ + Cr₂(SO₄)₃ + 7 H₂O \]

### Step 5: Formation of Colloidal Sulfur
In the above reaction, sulfur (`S`) is produced. The sulfur formed is in a colloidal state, which means it is dispersed in the solution but not dissolved. This colloidal sulfur gives the solution a milky appearance.

### Step 6: Conclusion
The milky appearance of the solution is due to the formation of colloidal sulfur when `H₂S` is passed through the acidic solution of `K₂Cr₂O₇`.

### Summary
To summarize, the solution turns milky because the reaction between `H₂S` and `K₂Cr₂O₇` in an acidic medium produces colloidal sulfur.

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