Question

At T(K), the solubility products (K\(_{sp}\)) of AgCl, AgBr and AgI are \(1.8 \times 10^{-10}\), \(5 \times 10^{-13}\) and \(8.3 \times 10^{-17}\) respectively. A 1L solution contains \(10^{-5}\) moles each of NaCl, NaBr and NaI. This solution is titrated with \(10^{-6}\) M AgNO\(_3\) until all halides (Cl\(^{-}\), Br\(^{-}\), I\(^{-}\)) are precipitated as silver salts.
The order of precipitation of these respectively is:

• A. AgCl, AgBr, AgI
• B. AgI, AgBr, AgCl
• C. AgI, AgCl, AgBr
• D. All the three salts are precipitated simultaneously

Hint:

Lower \(K_{sp}\) means lower solubility — such salts precipitate first upon reaction with AgNO\(_3\).

Answer 1

The Correct Option is B

Solubility product (K\(_{sp}\)) indicates the ease of dissolution. Lower K\(_{sp}\) → Less soluble → Precipitates first.
Given:
- AgI: \(K_{sp} = 8.3 \times 10^{-17}\) → Least soluble
- AgBr: \(K_{sp} = 5 \times 10^{-13}\)
- AgCl: \(K_{sp} = 1.8 \times 10^{-10}\) → Most soluble
Therefore, the order of precipitation: \[ \boxed{\text{AgI} \rightarrow \text{AgBr} \rightarrow \text{AgCl}} \]

Answer 2

At T(K), the solubility products (K_sp) of AgCl, AgBr and AgI are:
- AgCl: \(1.8 \times 10^{-10}\)
- AgBr: \(5 \times 10^{-13}\)
- AgI: \(8.3 \times 10^{-17}\)

A 1L solution contains \(10^{-5}\) moles each of NaCl, NaBr and NaI (i.e., \(10^{-5}\) M of Cl⁻, Br⁻, and I⁻). The solution is titrated with \(10^{-6}\) M AgNO₃.

Step 1: Concept – Precipitation and Ionic Product:
A precipitate forms when the ionic product (IP = [Ag⁺][X⁻]) exceeds the solubility product (K_sp) for that salt.

To compare when each halide starts precipitating, we calculate the minimum [Ag⁺] required to start precipitation using:
\[ [Ag^+] = \frac{K_{sp}}{[X^-]} \]

Step 2: Calculate [Ag⁺] required for precipitation of each halide:
- For AgCl: \[ [Ag^+] = \frac{1.8 \times 10^{-10}}{10^{-5}} = 1.8 \times 10^{-5} \]
- For AgBr: \[ [Ag^+] = \frac{5 \times 10^{-13}}{10^{-5}} = 5 \times 10^{-8} \]
- For AgI: \[ [Ag^+] = \frac{8.3 \times 10^{-17}}{10^{-5}} = 8.3 \times 10^{-12} \]

Step 3: Order of precipitation:
The halide that requires the least [Ag⁺] will precipitate first. So we compare:
- AgI precipitates first (requires \(8.3 \times 10^{-12}\) M Ag⁺)
- Then AgBr (requires \(5 \times 10^{-8}\) M Ag⁺)
- Finally AgCl (requires \(1.8 \times 10^{-5}\) M Ag⁺)

Final Answer:
\[ \boxed{\text{AgI, AgBr, AgCl}} \]