Question

A \(1.0\ L\) aqueous solution contains \(1\times 10^{-8}\ M\ NaBr\), \(1\times 10^{-8}\ M\ NaCl\) and \(1\times 10^{-8}\ M\ NaI\). To this solution, \(1\times 10^{-10}\ M\) aqueous \(AgNO_3\) solution is added drop wise. The order of precipitation of \(AgX\), where \(X=Cl, Br, I\), is
\[ K_{sp}(AgCl)=1.8\times 10^{-10} \] \[ K_{sp}(AgBr)=5\times 10^{-13} \] \[ K_{sp}(AgI)=8.3\times 10^{-17} \]

• A. \(AgBr,\ AgCl,\ AgI\)
• B. \(AgCl,\ AgBr,\ AgI\)
• C. \(AgI,\ AgBr,\ AgCl\)
• D. \(AgBr,\ AgI,\ AgCl\)

Hint:

When the concentration of anions is the same, the salt with the lowest \(K_{sp}\) precipitates first.

Answer 1

The Correct Option is C

Step 1: Rank silver halides by K_sp.
Smaller K_sp precipitates first at equal halide concentration. K_sp: AgI (8.3×10⁻¹⁷)<AgBr (5×10⁻¹³)<AgCl (1.8×10⁻¹⁰).

Step 2: Final Answer:
AgI, AgBr, AgCl (option 3).