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If the solubility product of PbS is 8 × 10^-28, then the solubility of PbS in pure water at 298 K is x × 10^-16 mol L^-1. The value of x is ________. (Nearest Integer) [Given √2 = 1.41]
- 282
- 283
- 284
- 285
Correct answer: 282
Solution
The solubility product (Ksp) of PbS can be expressed as Ksp = [Pb^2+][S^2-]. Since the solubility of PbS in pure water is 's', we have Ksp = s^2, leading to s = √(Ksp). Substituting Ksp = 8 × 10^-28 gives s = √(8 × 10^-28) = 2.83 × 10^-14 mol L^-1, which corresponds to x = 282 when expressed in the required format.
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