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Date Posted: 4/7/2008 8:33:32 PM  Status: Live
molar solubility
Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
What is the molar solubility of AgBr(s) in 1.0 M NH3?
Kf of Ag(NH3)2+ = 1.7e7 and Ksp of AgBr = 5.0e-13.
 
 
 
 
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Date Posted: 5/16/2008 9:45:03 AM  Status: Live
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Response:
 
 Given that
  The molarity of NH3 = 1.0 M
  Kf of Ag(NH3)2+ = 1.7*107
  Ksp of AgBr = 5.0 *10-13.
The given reactions are
   AgBr (s) ------->Ag+    +   Br-        Ksp=5.0*10-13
  Ag+   +   2NH3----->Ag(NH3)2+        Kf=1.7*107
--------------------------------------------------------
 AgBr (s) +2NH3 ------> Br-   +Ag(NH3)2+
initial:          1.0M            0          0
change:           -2x              x           x
equilibrium:   (1.0-2x)        x          x
The equilibrium constant K for this reaction can be obtained from Ksp and Kf values.
K= Ksp* Kf = (5.0*10-13)*(1.7*107) = 8.5*10-6
∴K=[ Br-][Ag(NH3)2+ ]/[NH3 ]2=(x*x)/(1.0-2x)2=8.5*10-6
==>x2= 8.5*10-6 *(1.0-2x)2
on solving above equation we get only one positive value.
  x=0.0029M
   Hence [Br-] = x
                      = 0.0029 M
  ∴ The molar solubility of AgBr = [Br-]
                                                  = 0.0029 M



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