Question

Given:

AgBr_(s) <--> Ag⁺_(aq) + Br ^-_(aq)

and

Ag⁺_(aq) + 2NH_3(aq) <--> Ag(NH₃)₂⁺_(aq)

a) What would be the value of K for AgBr_(s) + 2NH₃(aq) <--> Ag(NH₃)₂⁺_(aq) + Br ^-_(aq) ?

b) What would be the molar solubility of silver bromide in 2.00 M aqueous ammonia?

Answer 1

Answer – We are given, reactions –

AgBr_(s) <---> Ag⁺_(aq) + Br ^-_{(aq) ,} Ksp = 5.0*10^-13

Ag⁺_(aq) + 2NH_3(aq) <-----> Ag(NH₃)₂⁺_(aq) , Kf = 1.6*10⁷

We need to calculate K for the reaction follow-

AgBr_(s) + 2NH₃(aq) <-----> Ag(NH₃)₂⁺_(aq) + Br ^-_(aq)

We know when we added two reactions then equilibrium constants get multiply to each other, so when we added these both reaction there is formed third reaction

AgBr_(s) <---> Ag⁺_(aq) + Br ^-_{(aq) ,} Ksp = 5.0*10^-13

Ag⁺_(aq) + 2NH_3(aq) <-----> Ag(NH₃)₂⁺_(aq) , Kf = 1.6*10⁷

AgBr_(s) + 2NH₃(aq) <-----> Ag(NH₃)₂⁺_(aq) + Br ^-_(aq) , K = 5.0*10^-13*1.6*10⁷

                                                                                       = 8.0*10^-6

b) We are given the , [NH₃] = 2.00 M ,

We calculated the K for the following reaction -

AgBr_(s) + 2NH₃(aq) <-----> Ag(NH₃)₂⁺_(aq) + Br ^-_(aq) , K = 8.0*10^-6

We need to put ICE chart

      AgBr_(s) + 2NH₃(aq) <-----> Ag(NH₃)₂⁺_(aq) + Br ^-_(aq)

I                    2.00                            0                 0

C                   -2x                             +x               +x

E                 2.00-2x                        +x                 +x

We know,

K = [Ag(NH₃)₂⁺_(aq) ] [Br ^-_(aq) ] / [NH₃]²

8.0*10^-6 = x *x / (2.0-2x)²

8.0*10^-6 = x² / (2.0-2x)²

Square root from both side

0.00283 = x / 2.0-2x

0.00283*(2-2x) = x

0.00566 -0.00566x = x

0.00566 = 1.00566x

So, x = 0.00562 M

So, molar solubility of silver bromide in 2.00 M aqueous ammonia is 0.00562 M