Asked by courtney
Find the solubility of AgBr in 3.9 M NH3 [Ksp of AgBr = 5.0 10-13 and Kf of Ag(NH3)2+ = 1.7 107]
Answers
Answered by
DrBob222
AgBr==> Ag^+ + Br^-
Ag^+ + 2NH3 ==> Ag(NH3^+
eqn 1..Ksp = (Ag^+)(Br^-)
Kf = [Ag(NH3)2^+]/(Ag^+)(NH3)^2
Let S = solubility, then
eqn 2..S = (Br^-) = (Ag^+) + [Ag(NH3)2^+]
eqn 3..3.9 = (NH3) + (NH4^+) + 2[Ag(MH3)2^+]
Three equations as above.
1..as is
2..(Ag^+)<< [Ag(NH3)2^+]
3..3.9 = (NH3) as (NH4^+) is small and 2[Ag(NH3)2^+] is much smaller than (NH3)
Three equation and three unknowns.
Ag^+ + 2NH3 ==> Ag(NH3^+
eqn 1..Ksp = (Ag^+)(Br^-)
Kf = [Ag(NH3)2^+]/(Ag^+)(NH3)^2
Let S = solubility, then
eqn 2..S = (Br^-) = (Ag^+) + [Ag(NH3)2^+]
eqn 3..3.9 = (NH3) + (NH4^+) + 2[Ag(MH3)2^+]
Three equations as above.
1..as is
2..(Ag^+)<< [Ag(NH3)2^+]
3..3.9 = (NH3) as (NH4^+) is small and 2[Ag(NH3)2^+] is much smaller than (NH3)
Three equation and three unknowns.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.