Question

Consider a solution made by mixing 500.0 mL of 4.0 M NH3 and 500.0 mL of 0.40 M AgNO3. Agt reacts with NH3 to form AgNH3* and Ag(NH3)2*: Ag* + NH3 <--> AgNH3* Ki = 2.1 x 103 AgNH3* + NH3 <--> Ag(NH3)2* K2 = 8.2 x 103 The concentration of Ag(NH3)2* at equilibrium is

Answer 1

Soli the objective is to calculate the concentration of Agt at Equilibrium Given data: 1 Agt reacts with NH₂ to from Aguilty + & Ag (N143), Consider the reactions is given Agt + NHz - Ag Netzt ko =2.1X103 Ag Nht + NH3 - Ag (WHz)," k, = 8.2 X163 Net reaction: Agt + 2NH3 = Ag (NH2), Ka ki xke =(2.1x103) (8.2x103) K = 1.72 X167 Now write ICE table for the given Net reaction Agt + 2NH3 ² AS Belose 0.2M { 207 1,6 tzn 1.6tza 0,20 After Equilísnum x 0.20- to calculate the eruilibrium Constant K= (Ag CNH3)+ ] [Ag+] CNA,y? 1.78107 = 0.20 - x X(116727)

in value of a we know that x=small number So, neglect the change 24 = Neglected 1.7X107 - 0.20-1 261.6) 197X107 = 0.20 2.62448 1.4x10+ = 0.076201895(3) P = 0.076 207895 1.7x107 x=4.482 x 10m 2~4.5x109an Concentration of Ag+, Cast) = 2 =4.581019mg Concentration of Ag (NH3)* = 0.20-4 =0.20 - (4.5x109) ng concentration of Ag (NH3)* = 0.1999999955 Ž0.1917