AgBr(s) ⇄ Ag+(aq) + Br-(aq) Ksp = 7.7 x 10-13Ag+(aq) + 2NH3(aq) ⇄ Ag(NH3)2+(aq) Kf = 1.5 x 107When NH3 is added to AgBr(s), a combination of these reactions result:AgBr(s) + 2NH3(aq) ⇄ Ag(NH3)2+(aq) + Br-(aq)What mass of AgBr will dissolve in a 5.0 L container of 2.7 M NH3 solution?* 8.6×100 g* 2.8×10-2 g* 5.2×100 g* 9.2×10-3 g* 4.6×10-2 g

The reactions given are, 1) AgBr(s) ⇄ Ag+(aq) + Br-(aq) Ksp = 7.7 × 10-13 2) Ag+(aq) + 2 NH3(aq) ⇄ Ag(NH3)2+(aq) Kf = 1.5 × 107 Adding the above two reactions to get the net reaction as (on adding the reactions, the K will get multiplied), => AgBr(s) + 2 NH3(aq) ⇄ Ag(NH3)2+(aq) + Br-(aq) K = 7.7 × 10-13 × 1.5 × 107 = 1.155 × 10-5 Given: Concentration of NH3 = 2.7 M And volume of solution = 5.0 L Assuming y concentration of AgBr dissolves in the reaction. Since the overall reaction has K value << 1. Hence negligible reaction will take place and y <<<< 2.7. Therefore the equilibrium concentration of speices are, => [NH3 ] = 2.7 - y = 2.7 M approx. => [Ag(NH3)2]+ = y = [Br- ] The equilibrium constant expression for the reaction is given by, => K = [Ag(NH3)2+] [Br-][NH3]2Hence substituting the values we get,=> 1.155 × 10-5 = y × y2.72=> y = 0.009176 M approx.Hence moles of AgBr dissolved = concentration × volume of solution in L = 0.009176 × 5.0 = 0.04588 mol.Molar mass of AgBr = 107.9 + 79.9 = 187.8 g/mol.Hence mass of AgBr dissolved = moles × molar mass = 0.009176 × 187.8 = 8.6 × 100 g approx. Therefore the answer is option 8.6 × 100 g.