Question

7. Silver ion forms a stable 1:1 complex with a chelating amine molecule. Calculate the silver ion concentration at equilibrium when 25mL of 0.10 M silver nitrate is added to 50 mL of 0.15 Mamine. Given that Kf = 5.0 x 10'.

Answer 1

The balanced equation for the formation of the complex [Ag(amine)]* is Ag+ + amine --------> [Ag(amine)] Number of moles of Ag+ = M*V = 0.10 M * 25.0 mL = 2.5 mmol Number of moles of amine = M*V = 0.15 M * 50.0 mL = 7.5 mmol Since the concentration of Agt is less than that of amine, assume that all Agt is converted to the complex. mmol Ag+ + amine — → [Ag(amine)]* Initial 2.5 7.5 Change -2.5 -2.5 +2.5 Final 0 5.0 2.5 Total volume of the solution = 25.0 + 50.0 = 75.0 mL The conc. of amine remaining the solution = net moles/total volume = 5.0 mmol/75.0 mL = 0.066667 M Conc. of the complex formed = 2.5 mmol/75.0 mL = 0.033333 M At this point, there is no Agt left in the solution. The concentration of Agt is determined by the dissociation of the complex. Ag+ + amine → → [Ag(amine)]* Initial 0 0.066667 0.033333 Change +X -X Final 0.066667 + 0.033333 - X 0.033333 K, = -= 5.0x10 x.(0.066667+x) x is small, so 0.066667+x=0.066667 0.033333 x.(0.066667) 0.033333 =5.0x10? 3x=500107 x(0.066667) 5.0x107x(0.066667) =1.00x10-8 Concentration of Agt = 1.00 x 10-M