Question

please answer both questions

QUESTION 26 Given the two equilibria below, Ag(NH3)2 (aq) = Ag (aq) + 2 NH3(aq) Kd-5.9x 10-8 Agl(s) Ag (aq)+I (aq) what is Ke for the following equilibrium? 10-17 Agl(s)+2NH3(aq) Ag(NH3)2"(aq)+ I (aq) а. 7.1 x 108 b. 1.4x 10-9 2.0 x 10-18 d. 4.9x 10-24 2.7 x 100

Answer 1

Ag(NH3)2^+ (aq) -------------> Ag^+ (aq) + 2NH3(aq)     Kd = 5.9*10^-8

Ag^+ (aq) + 2NH3(aq) -----------> Ag(NH3)2^+ (aq)     1/Kd   =

AgI(s) -------------> Ag^+ (aq) + I^- (aq)                         Ksp   = 8.3*10^-17

Ag^+ (aq) + 2NH3(aq) -----------> Ag(NH3)2^+ (aq)   

AgI(s) -------------> Ag^+ (aq) + I^- (aq)        

-----------------------------------------------------

AgI(s) + 2NH3(aq) -----------> Ag(NH3)2^+ (aq) + I^- (aq) KC = (1/Kd)*Ksp

Kc = 1/Kd * KSp

   = (1/5.9*10^-8 )*8.3*10^-17

         = 1.7*10^7 *8.3*10^-17

         = 1.4*10^-9

b.1.4*10^-9 >>>>.answer

Q27.

molarity of Zn(OH)2   = W/G.M.Wt * volume in L

                                  = 0.000222/99.424   = 2.23*10^-6M

Zn(OH)2(s) --------------> Zn^2+ (aq) + 2OH^- (aq)

                                            s                 2s

Ksp   = [Zn^2+][OH^-]^2

         = s*(2s)^2

          = 4s^3

          = 4(2.23*10^-6)^3

          = 4.5*10^-17

a.4.5*10^-17 >>>answer