Using systematic equilibrium calculations, find the molar solubility of AgBr in
ID: 901495 • Letter: U
Question
Using systematic equilibrium calculations, find the molar solubility of AgBr in 4.0 M NH_3 (this is the concentration of the free, uncomplexed NH_3). Also calculate the concentrations of Ag^+, Br^- AgNH_3^+ and Ag(NH_3)_2^+. Verify any approximations. The solubility product constant for AgBr is 4.0 Times 10^-13. The formation constants for the successive addition of NH_3 to Ag^+ are 2.3 Times 10^3 (for adding NH^3 to Ag^+) and 6.0 Times 10^3 (for adding NH_3 to AgNH_3^+). Start by writing the: solubility product constant equation, formation constant equations for AgNH_3^+ and Ag(NH_3)_2^+, mass balance on bromide, mass balance on silver, and charge balance equation.Explanation / Answer
a.
AgBr = Ag+ + Br - Ksp = [Ag+ ][Br-]
b.
Ag+ + NH3 = [Ag(NH3)]+ K1 = [Ag(NH3)]+ / ( [Ag+] [NH3] )
[Ag(NH3)]+ + NH3 = [Ag(NH3)2]+ K2 = [Ag(NH3)2]+ / ([Ag(NH3)]+ [NH3] )
c.
[Br-]
d.
[Ag+ ] + [Ag(NH3)]+ + [Ag(NH3)2]+
e.
[Br-] = [Ag+ ] + [Ag(NH3)]+ + [Ag(NH3)2]+
Calculations:
Add these equations
AgBr = Ag+ + Br - Ksp = [Ag+ ][Br-]
Ag+ + NH3 = [Ag(NH3)]+ K1 = [Ag(NH3)]+ / ( [Ag+] [NH3] )
[Ag(NH3)]+ + NH3 = [Ag(NH3)2]+ K2 = [Ag(NH3)2]+ / ([Ag(NH3)]+ [NH3] )
……………………………………………………………………….
AgBr + 2NH3 = [Ag(NH3)2]+ + Br - K = Ksp·K1·K2
K = [Br-] [Ag(NH3)2]+ / [NH3]2
K = 4.0 x 10-13 x 2.3 x103 x 6.0x103 = 5.52x10-6
Looking to K1 and K2 values observe that
[Ag(NH3)]+ / [Ag+] = 4 x 2.3 x103
[Ag(NH3)2]+ / ([Ag(NH3)]+ = 4 x6x103
Then
[Ag(NH3)2]+ >> ([Ag(NH3)]+ >>[Ag+]
And approximate that:
[Br-] = [Ag(NH3)2]+ (see at d.)
The molar solubility S of AgBr in 4 m NH3 is
S = [Br-] = [Ag(NH3)2]+= K1/2· [NH3]
= (5.52x10-6)1/2 x 4.0 =
= 9.4x10-3 M
[Ag(NH3)2]+ / [Ag(NH3)]+ = 4 x6x103
[Ag(NH3)]+= 9.4x10-3 M / (4 x6x103) = 4.0 x 10-7 M
[Ag(NH3)]+ / [Ag+] = 4 x 2.3 x103
[Ag+] = 4.0 x 10-7 / 4 x 2.3 x103 = 4.3x10-11 M
Observe the values and that the approximation is confirmed.
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