1.) What is the molar concentration of Ag+(aq) in a 1.00 M solution of Ag(NH3)2
ID: 966269 • Letter: 1
Question
1.) What is the molar concentration of Ag+(aq) in a 1.00 M solution of Ag(NH3)2 with no excess ammonia? Given: Kf= 1.7x10^7 for Ag(NH3)22.) What is the equilibrium concentration of Cl^-(aq) in a solution that is .0200 M solution in CuNO3 and .450 M HCl(aq)? Given: Cu^+(aq) + 3Cl^-(aq) -----> CuCl3^-2 Kf= 5x 10^5 1.) What is the molar concentration of Ag+(aq) in a 1.00 M solution of Ag(NH3)2 with no excess ammonia? Given: Kf= 1.7x10^7 for Ag(NH3)2
2.) What is the equilibrium concentration of Cl^-(aq) in a solution that is .0200 M solution in CuNO3 and .450 M HCl(aq)? Given: Cu^+(aq) + 3Cl^-(aq) -----> CuCl3^-2 Kf= 5x 10^5 Given: Kf= 1.7x10^7 for Ag(NH3)2
2.) What is the equilibrium concentration of Cl^-(aq) in a solution that is .0200 M solution in CuNO3 and .450 M HCl(aq)? Given: Cu^+(aq) + 3Cl^-(aq) -----> CuCl3^-2 Kf= 5x 10^5
Explanation / Answer
1) Chemical equation:
Ag+ + 2NH3 <--> Ag(NH3)2+
Formation constant (Kf):
Kf = 1.7x107
[Ag(NH3)2+] = 1.00 M
Equilibrium equation:
Kf = [Ag(NH3)2]/([Ag+].[NH3]2)
In this case, [Ag+] = [NH3]
Kf = [Ag(NH3)2]/([Ag+]3)
Clear [Ag+]:
[Ag+] = ([Ag(NH3)2]/Kf)1/3
[Ag+] = (1.00 M/1.7x107)1/3
[Ag+] = 3.9x10-3 M
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.