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The solubility constant of agbr is 5.0 x 10-13. the formation constant of the silver(i)

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.
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endreu2005
endreu2005
4,4(59 marks)

S AgBr = 2.82 E-3 mol/L

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

dh29229
dh29229
4,4(56 marks)

We must, therefore, acquiesce in the necessity, which denounces our Separation, and hold them, as we hold the rest of mankind, Enemies in War, in Peace Friends.

Explanation:

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