K_sp of PbF2 is 3.6 x 10^-8. Its solubility in 0.1 M Pb(NO3)2 solution (in mol L^-1), neglecting s compared with 0.1, is

JEE Main levelSingle correct optionDifficulty 2 of 3ie185

$K_{s p}$ of $PbF X_{2}$ is $3.6 \times 1 0^{- 8}$ . Its solubility in $0.1 M$ $Pb (NO X_{3}) X_{2}$ solution (in $mol L^{- 1}$ ), neglecting $s$ compared with $0.1$ , is

[Pb X^{2 +}] \approx 0.1 M

and

[F X^{-}] = 2 s

, so

K_{s p} = 0.1 (2 s)^{2} = 0.4 s^{2}

. Hence

s = 3.6 \times 1 0^{- 8} /0.4 = 3.0 \times 1 0^{- 4} M