In: Chemistry
Calculate [Ag1+] in solution at equilibrium when 2.39 x 10-2 M AgNO3 is added to 1.00 L of a 1.65 M NH3 solution, assuming no volume change? (Kf for Ag(NH3)21+ is 1.1 x 107)
a) 3.8 x 10-2
b) 2.39 x 10-2
c) 4.9 x 10-11
d) 8.5 x 10-10
e) 4.1 x 10-3
no of moles of AgNo3 = molarity * volume in L
= 2.39*10-2 *1 = 2.39*10-2 moles = 0.0239moles
no of moles of NH3 = molarity * volume in L
= 1.65*1 = 1.65 moles
Ag+ + 2NH3 ---------> [Ag(NH3)2]+
I 0.0239 1.65 0
C -0.0239 -2*0.0239 0.0239
E 0 1.6022 0.0239
Kf = [Ag(NH3)2]+/[Ag+][NH3]2
1.1*107 = 0.0239/[Ag+]*0.0239
[Ag+] = 0.0239/1.1*107 *0.0239
= 9.1*10-8 M >>>> answer