Question
Calculate the solubility in mol dm ^ -3 of sparingly soluble salt BA at 293 K if its solubility product is 8.56 10^ -5 at same temperature.
MHT CET
Chemistry
Ionic Equilibrium
2026
MHT CET 2026 (17 April Shift 1)
MHT CET Chemistry Question (2026) — Solution
Options
- A. 8.123 10^ -3
- B. 8.780 10^ -3
- C. 9.252 10^ -3
- D. 7.756 10^ -3
Answer
C. 9.252 10^ -3
Step-by-step solution
The dissociation of the sparingly soluble salt BA is given by: BA B^+ + A^- Let the solubility of the salt be s mol dm ^ -3 . The solubility product K_ sp is given by: K_ sp = [B^+][A^-] = s s = s^2 Given K_ sp = 8.56 10^ -5 , we have: s^2 = 8.56 10^ -5 = 85.6 10^ -6 s = 85.6 10^ -6 = 9.252 10^ -3 mol dm ^ -3 Answer: 9.252 10^ -3
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