MHT CET202526 Apr 2025Morning ShiftChemistryIonic EquilibriumActual
The solubility product of AgBr is 4.9 10⁻¹³ at a certain temperature. Calculate the solubility .
Options
- A4 10⁻⁶ ~mol~ dm ⁻³
- B4 10⁻⁷ ~mol~ dm ⁻³
- C7 10⁻⁷ ~mol~ dm ⁻³
- D3 10⁻⁸ ~mol~ dm ⁻³
Correct answer
C. 7 10⁻⁷ ~mol~ dm ⁻³
Step-by-step solution
The dissolution equilibrium for silver bromide is: AgBr(s) Ag^+(aq) + Br^-(aq) For this equilibrium, the solubility product expression gives K_ sp = [ Ag^+ ][ Br^- ] . If s represents the molar solubility, then [ Ag^+ ] = s and [ Br^- ] = s , yielding K_ sp = s^2 . Given K_ sp = 4.9 10⁻¹³ , we solve s^2 = 4.9 10⁻¹³ . Converting to 49 10⁻¹⁴ and taking square roots: s = 49 10⁻¹⁴ = 7 10⁻⁷~ mol~dm⁻³ This value corresponds to option C among the given choices. Answer: C