Verified Solution • 2025–26 Session moderate

QAIPMT-2003-Ch-2680: NEET 2003 Chemistry - Class 11 AIPMT-2003-Ch-2680 Solution: The solubility product of AgI at \(25^ C\) is \(1.0 10^{-16}\…

The solubility product of AgI at 25∘C25^\circ C is 1.0×10−16 mol2 L−21.0\times10^{-16}\ \text{mol}^2\ \text{L}^{-2}. The solubility of AgI in 10−410^{-4} N solution of KI at 25∘C25^\circ C is approximately (in mol L−1^{-1})

Multiple Choice Options

  • A 1.0×10−161.0\times10^{-16}
  • B 1.0×10−121.0\times10^{-12}
  • C 1.0×10−101.0\times10^{-10}
  • D 1.0×10−81.0\times10^{-8}

Step-by-Step Verified Solution

AgI | _ | Ag + | + | I - s | s | s K sp = s 2 _ 1.0 _ 10 �16 = s 2 _ s = 1.0 _ 10 �8 mol L �1 _ [Ag + ] = 1.0 _ 10 �8 mol L �1 Also, in 10 �4 N KI solution, [I �1 ] = (10 �4 + 1.0 _ 10 �8 ) mol L �1 _ [I �1 ] = (10 �4 ) mol L �1 [As 1.0 _ 10 �8 mol L �1 << 1.0 _ 10 �4 mol L �1 ] _ K sp of AgI = [Ag + ][I � ] = (1.0 _ 10 �8 )(10 �4 ) = 1.0 _ 10 �12 mol L �1

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