The solubility of AgBr at 298 K is 2 10⁻⁷ kg/dm ^3 . Find the solubility product of AgBr at same temperature. [Molar mass of AgBr =188 g/mol]

Step-by-step solution

Given solubility of AgBr = 2 10⁻⁷ kg/dm ^3 Since 1 kg = 10^3 g and 1 dm ^3 = 1 L , the solubility in g/L is: s = 2 10⁻⁷ 10^3 g/L = 2 10⁻⁴ g/L Molar mass of AgBr = 188 g/mol Molar solubility ( S ) in mol/L is: S = 2 10⁻⁴ 188 mol/L 1.06 10⁻⁶ mol/L The dissociation of AgBr is given by: AgBr(s) Ag ^+( aq ) + Br ^-( aq ) The solubility product K_ sp is: K_ sp = [ Ag ^+][ Br ^-] = S^2 K_ sp = (1.06 10⁻⁶)^2 = 1.1236 10⁻¹² 1.124 10⁻¹² Answer: 1.124 10⁻¹²