Question

calculate the equilibrium molarity of aqueous Ag+ ion. round to two sig figs

For the aqueous [ Ag(NH;)]* complex K; = 1.7x 10' at 25 °C. Suppose equal volumes of 0.0082 M AgNO, solution and 0.20 M NH, solution are mixed. Calculate the equilibrium molarity of aqueous Aglon. Round your answer to 2 significant digits. OM x 5 ?

Answer 1

Let V be the volume of each solution. Number of moles of Agt = 0.0082 M * V mL = 0.0082V mmol Number of moles of NH3 = 0.20 M * V mL = 0.2V mmol The balanced equation for the formation of the complex (Ag(NH3)2]* is Ag+ + 2 NH3 --------> [Ag(NH3)2]* Since the concentration of Agt is less than that of NH3, assume that all Agt is converted to the complex. mmol Ag+ + 2NH3 — (Ag(NH3)21+ Initial 0.0082V 0.2v Change -0.0082V -0.0082V + 0.0082V Final 0.1836V 0.0082V Concentration of the complex= moles/total volume = 0.0082V mmol/(2V)mL = 0.0041 M The conc. of NH3 remaining the solution = 0.1836 mmol/(2V)mL = 0.0918 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+ + 2NH3 — > [Ag(NH3)21* Initial 0.0918 0.0041 0 Change + x + 2x -X Final 0.0918+ 2x 0.0041 - x 2 =1.7x107 [Ag(NH3)2]* 0.0041-x [Ag+ ][NH,] x.(0.0918+2x)? x is small, so 0.0041-x=0.0041 and 0.0918+2x=0.0918 0.0041_=1.7x10 x.(0.0918) 0.0041 ** 1.7x10?x(0.0918) x= = 2.9x10-8 Concentration of Ag+ = 2.9x10 SM