Solubility product of the sparingly soluble salt AgBrO ₃ in aqueous medium is 9.3 10⁻¹⁰ Calculate the mass in gram of AgBrO ₃ present in 100 ml of its saturated solution. (Molar mass of AgBrO ₃ is 236 ~g / mol )

Step-by-step solution

The solubility equilibrium for AgBrO ₃ is given by: AgBrO ₃(s) Ag ⁺(aq) + BrO ₃⁻(aq) Let s be the solubility of AgBrO ₃ in mol/L. Then [ Ag ⁺] = s and [ BrO ₃⁻] = s . The solubility product constant K_ sp is defined as: K_ sp = [ Ag ⁺][ BrO ₃⁻] = s^2 Given K_ sp = 9.3 10⁻¹⁰ , we have: s = 9.3 10⁻¹⁰ = 93 10⁻¹¹ = 9.3 10⁻⁵ 3.04959 10⁻⁵ mol/L The volume of the solution is 100 ml = 0.1 L . The number of moles of AgBrO ₃ in 0.1 L is: n = s V = 3.04959 10⁻⁵ mol/L 0.1 L = 3.04959 10⁻⁶ mol The molar mass of AgBrO ₃ is 236 g