MHT CET202520 Apr 2025Morning ShiftChemistryIonic EquilibriumActual
The solubility of AgBr is 7.1 10⁻⁷ ~mol~ dm ⁻³ . Calculate it's solubility product at the same temperature.
Options
- A7.08 10⁻¹³
- B3.67 10⁻¹³
- C5.89 10⁻¹³
- D5.04 10⁻¹³
Correct answer
D. 5.04 10⁻¹³
Step-by-step solution
The dissolution equilibrium for silver bromide is given by AgBr(s) Ag^+(aq) + Br^-(aq) At equilibrium, the concentrations are [ Ag^+ ] = [ Br^- ] = s where s = 7.1 10⁻⁷ mol ,dm⁻³ is the molar solubility. The solubility product expression is K_ sp = [ Ag^+ ][ Br^- ] = s^2 Substituting the given solubility: K_ sp = (7.1 10⁻⁷)^2 = 50.41 10⁻¹⁴ = 5.041 10⁻¹³ This value matches option D among the given choices. The correct answer is D .