Calculating the Concentrations of Complex Ions
Assume that all of the Ag(H2O)2+ is converted Ag(NH3)2+; set
up the table with x = [Ag(H2O)2+] at equilibrium.
Ammonia reacted = [NH3]reacted = 2(1.3 x 10-3 M) = 2.6 x 10-3 M
Concentration (M) Ag(H2O)2+(aq) 2NH3 (aq) Ag(NH3)2+ 2 H2O(aq)
Initial 1.3 x 10-3 5.0 x 10-2 0 ----
Change ~(-1.3 x 10-3) ~(-2.6 x 10-3) ~(+1.3 x 10-3) ----
Equilibrium x 4.7 x 10-2 1.3 x 10-3 ----
Kf = = = 1.7 x 107
[Ag(H2O)2+][NH3]2
[Ag(NH3)2+]
1.3 x 10-3
x(4.7 x 10-2)2
x = 3.46 x 10 -8 M = [Ag(H2O)2+]
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