Question

The solubility constant of AgBr is 5.0 x 10-13. The formation constant of the silver(I) ammonia complex, Ag(NH3)2, is 1.6 x 107 Calculate the solubility, in moles/L, of AgBr in 1.00 Mammonia solution.

Answer 1

S AgBr = 2.82 E-3 mol/L

Step-by-step explanation:

• AgBr ↔ Ag+ + Br- .....(1)

∴ Ksp = [ Ag+ ] * [ Br- ] = 5.0 E-13

• Ag+ + 2NH3 ↔ [ Ag(NH3)2 ]+ ..........(2)

∴ Kf = 1.6 E7 = α[Ag(NH3)2]+ / ( αAg+ )*( αNH3 )²

∴ C NH3(sln) = 1.00 M

from (1) + (2):

• AgBr(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Br-(aq)

1 M 0 0

1 - 2x x x

∴ K = Ksp*Kf = ( 5.0 E-13 )*( 1.6 E7 ) = 8.0 E-6

⇒ K = ( [ Br- ] * [ Ag(NH3)2+] ) / [ NH3 ]² = 8.0 E-6

⇒ K = (( x )*( x )) / ( 1 - 2x ) = 8.0 E-6

⇒ x² = 8.0 E-6*( 1- 2x )

⇒ x² + 1.6 E-5x - 8.0 E-6 = 0

⇒ x = 2.82 E-3 M

⇒ [ Br- ] = [ AgBr ] = 2.82 E-3 M