JEE Main202225 Jun 2022Evening ShiftChemistryIonic EquilibriumActual
The K sp for bismuth sulphide Bi 2 S 3 is 1 . 08 × 10 - 73 . The solubility of Bi 2 S 3 in molL - 1 at 298 K is
Options
- A1 . 0 × 10 - 15
- B2 . 7 × 10 - 12
- C3 . 2 × 10 - 10
- D4 . 2 × 10 - 8
Correct answer
A. 1 . 0 × 10 - 15
Step-by-step solution
Bi 2 S 3 will get dissociated as: Bi 2 S 3 ⇌ 2 Bi + 3 + 3 S - 2                             2 s                   3 s Here, K sp = 2 s 2 3 s 3 = 108   s 5 s = K sp 108 1 5 s = 1 . 08 × 10 - 73 108 1 5 s = 10 - 15   mol   L - 1