JEE Main202229 Jul 2022Morning ShiftChemistryIonic EquilibriumActual
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 ]
Correct answer
0
Step-by-step solution
PbS s ⇌ Pb 2 + aq + S 2 - aq Let us consider the solubility of PbS is " S " M . Thus, Solubility product = Pb 2 + aq S 2 - aq Ksp = Pb 2 + aq S 2 - aq       ⋯   i We know K sp = S 2 S = K sp = 8 × 10 - 28 = 2 2 × 10 - 14 = 2 . 82 × 10 - 14 = 282 × 10 - 16