Aqueous solution of a salt 'A', when mixed with NaoH solution and warmed, a black precipitate is formed. Black ppt. is filtered and dissolved in dil. `H_(2)SO_(4)` solution. The resulting solution gives a chocolate coloured precipitate with potassium ferrocyanide solution. the filtrate obtained after filtering off the black precipitate, upon warming with Zn and NaOH evolves a pungent smelling gas. the resulting solution also responds to the ring test. the filtrate does not evolve any gas when it is boiled with urea in the presence of `H_(2)SO_(4)`.
Q. The filtrate obtained after filtering off the black precipitate consists of:

Options

A. `NO_(2)^(-)`
B. `NO_(3)^(-)`
C. `CO_(3)^(2-)`
D. `Cl^(-)`

Explanation

To solve the given question step by step, we will analyze the information provided and deduce the components of the filtrate obtained after filtering off the black precipitate.

### Step 1: Identify the Salt 'A'
The question states that an aqueous solution of a salt 'A' forms a black precipitate when mixed with NaOH and warmed. The black precipitate is likely to be copper oxide (CuO), which is formed from copper hydroxide (Cu(OH)₂) upon heating. Therefore, we can deduce that salt 'A' is likely copper nitrate (Cu(NO₃)₂).

**Hint:** Look for common salts that produce black precipitates with NaOH.

### Step 2: Formation of Black Precipitate
When copper nitrate (Cu(NO₃)₂) reacts with NaOH, it forms copper hydroxide (Cu(OH)₂), which is initially a blue precipitate. Upon warming, this precipitate dehydrates to form copper oxide (CuO), which is black.

**Hint:** Consider the thermal decomposition of hydroxides to identify the black precipitate.

### Step 3: Analyze the Filtrate
After filtering off the black precipitate (CuO), we are left with a filtrate that contains the remaining ions in solution. Since we started with copper nitrate, the filtrate will contain sodium nitrate (NaNO₃) as the soluble product.

**Hint:** The ions that remain in solution after filtering out the precipitate are usually the counter-ions from the original salt.

### Step 4: Reaction with Zinc and NaOH
The filtrate, which contains NaNO₃, is then warmed with zinc (Zn) and NaOH. This reaction produces ammonia (NH₃), which is a pungent-smelling gas. The presence of ammonia confirms the presence of nitrate ions (NO₃⁻) in the filtrate.

**Hint:** Look for reactions that produce gases, especially ammonia, when nitrates are involved.

### Step 5: Confirming the Presence of Nitrate Ions
The evolution of ammonia gas upon warming with zinc and NaOH indicates that nitrate ions are indeed present in the filtrate. This is a key point in confirming the composition of the filtrate.

**Hint:** Consider the characteristic reactions of nitrate ions when treated with reducing agents.

### Step 6: Conclusion
Based on the above deductions, the filtrate obtained after filtering off the black precipitate consists of sodium nitrate (NaNO₃), which contains the nitrate ions (NO₃⁻) that were originally part of the copper nitrate salt.

**Final Answer:** The filtrate obtained after filtering off the black precipitate consists of sodium nitrate (NaNO₃).