Chemistry, asked by abhishekjangir4716, 11 months ago

Silver nitrate is gradually added to an aqueous solution containing 0.01 m each of chloride bromide and iodide ions. The correct sequence (decreasing order) in which the halides will be precipitated is

Answers

Answered by shubhamjoshi033
29

The correct sequence will be AgI, AgBr and then AgCl.

The silver ions in the silver nitrate will react with the chloride, bromide and iodide ions to make the respective salt.

The equation can be given below:

Ag⁺  + Cl⁻ = AgCl

Ag⁺  + Br⁻ = AgBr

Ag⁺  + l⁻ = Agl

The solubility product constant of the three halides is given below:

Ksp of AgI = 8.3 x 10⁻¹⁷

Ksp of AgBr = 5.4 x 10⁻¹³

Ksp of AgCl = 1.77 x 10⁻¹⁰

Since the Ksp of AgI is least hence it will be precipitated first, then AgBr and then AgCl


Sarah2626: How have u calculated these values
Answered by Priyanka3353
26

Answer:

Explanation:

Ag+ is a soft cation or acid due to large size & small charge and I- is a soft anion or base, so, bonding between them is strong and AgI will be precipitated most readily whereas Cl- is a hard base and AgCl will be most soluble out of the three & will precipitated at last.

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