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QJEEM22Ch91b9: JEE Main 2022 Chemistry - Class 11 JEEM22Ch91b9 Solution: If the solubility product of PbS is 8 10 28 , then the…

If the solubility product of PbS is 8 ×\times 10 −- 28, then the solubility of PbS in pure water at 298 K is x ×\times 10 −- 16 mol L −- 1. The value of x is __________. (Nearest Integer)

[Given : 2\sqrt2 = 1.41]

Step-by-Step Verified Solution

Ksp=S2\mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}

S=Ksp=8×10−28=22×10−14\mathrm{S}=\sqrt{K_{s p}}=\sqrt{8 \times 10^{-28}}=2 \sqrt{2} \times 10^{-14}

=2.82×10−14=2.82 \times 10^{-14}

=282×10−16=282 \times 10^{-16}

∴\therefore Ans. 282

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