MHT CET Chemistry Question (2026) — Solution

Step-by-step solution

For a sparingly soluble salt BA, the dissociation equilibrium is given by BA B^+ + A^-. Let the solubility of the salt be s in mol dm ^ -3 . The solubility product K_ sp is given by: K_ sp = [B^+][A^-] = s s = s^2 Substituting the given value of K_ sp : s^2 = 6.4 10^ -13 = 64 10^ -14 s = 64 10^ -14 = 8 10^ -7 \ mol dm ^ -3 To convert the solubility from mol dm ^ -3 to g dm ^ -3 , multiply by the molar mass of the salt: Solubility in g dm ^ -3 = s Molar mass Solubility = 8 10^ -7 \ mol dm ^ -3 190\ g mol ^ -1 Solubility = 1520 10^ -7 \ g dm ^ -3 = 1.52 10^ -4 \ g dm ^ -3 Answer: 1.52 10^ -4