Question bank/JEE Main/Chemistry

Determine solubility of PbI2 in a 0.1 M solution of Pb(NO3)2.is
( ___ × 10-4). (Ksp of PbI2 = 7.1 × 10-9)
Pb(NO3)2 ds 0.1 M PbI2 ( ___ × 10-4).(PbI2

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x + 0.1 2x
Ksp = (x + 0.1)(2x)2 = 8 × 10-9
$\Rightarrow$ 4x2 = 8 × 10-8
$\Rightarrow$ x = 1.4 × 10-4 M

Source: Grass-G / CH-27 Chemistry Paper 2 15 Jan.docx