Question

Which is the correct observation when is treated with NaF, NaCℓ and NaI taken in three test tubes labelled as (X), (Y) and (Z)?

Answer 1

In test tube (X) with NaF, no precipitate is observed. In test tube (Y) with NaCl, a white precipitate is observed. In test tube (Z) with NaI, a yellow precipitate is observed.

Answer 2

The reaction of silver nitrate ($\text{AgNO}_3$) with sodium halides ($\text{NaF}$, $\text{NaCl}$, $\text{NaI}$) involves a double displacement reaction, leading to the formation of silver halides. The key to the observation lies in the solubility and color of these silver halides.

Let's analyze each reaction:

1. Test Tube X: $\text{AgNO}_3 + \text{NaF}$

   • Reaction: $\text{AgNO}_3(aq) + \text{NaF}(aq) \rightarrow \text{AgF}(aq) + \text{NaNO}_3(aq)$

   • Solubility of $\text{AgF}$: Unlike other silver halides, silver fluoride ($\text{AgF}$) is highly soluble in water.

   • Observation: No precipitate will form. The solution will remain clear and colorless.

2. Test Tube Y: $\text{AgNO}_3 + \text{NaCl}$

   • Reaction: $\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)$

   • Solubility of $\text{AgCl}$: Silver chloride ($\text{AgCl}$) is insoluble in water.

   • Observation: A white precipitate of $\text{AgCl}$ will form.

3. Test Tube Z: $\text{AgNO}_3 + \text{NaI}$

   • Reaction: $\text{AgNO}_3(aq) + \text{NaI}(aq) \rightarrow \text{AgI}(s) + \text{NaNO}_3(aq)$

   • Solubility of $\text{AgI}$: Silver iodide ($\text{AgI}$) is insoluble in water.

   • Observation: A yellow precipitate of $\text{AgI}$ will form.

Summary of Observations:

• Test Tube X (with NaF): No precipitate is formed; the solution remains clear.

• Test Tube Y (with NaCl): A white precipitate is formed.

• Test Tube Z (with NaI): A yellow precipitate is formed.

Explanation:

The observations are based on the solubility of silver halides. Silver fluoride ($\text{AgF}$) is soluble, while silver chloride ($\text{AgCl}$) and silver iodide ($\text{AgI}$) are insoluble. The characteristic colors of the precipitates help distinguish between $\text{AgCl}$ (white) and $\text{AgI}$ (yellow).

• Core Solution:

  • Test Tube X ($\text{AgNO}_3 + \text{NaF}$): No precipitate forms ($\text{AgF}$ is soluble).

  • Test Tube Y ($\text{AgNO}_3 + \text{NaCl}$): A white precipitate of $\text{AgCl}$ forms.

  • Test Tube Z ($\text{AgNO}_3 + \text{NaI}$): A yellow precipitate of $\text{AgI}$ forms.