15. The complex ion Ag(NH,) is formed by mixing 5.00 mL of AgNO, and with 5.00 m

Question

15. The complex ion Ag(NH,) is formed by mixing 5.00 mL of AgNO, and with 5.00 mL of aqueous NH,. The complex forms according the equation Ag. (aq) + 2 NH, (aq) Ag(NHih" (aq) . Initial concentrations of the solutions prior to mixing are shown in Table 1. Absorption data for the complex ion are shown in Table 2 Table 1. Initial Concentration Data Table 2. Absorption Data Solution Absorbance @254 nm 95 .24 Concentration (M) Ag(NHsh 1 (M) AgINH1standard 7.56 x 10 TAgINH 121 test tube #2 NH 6.00x10 a. what is the equilibrium concentration of Ag(Hah" in test tube #2? b. Set up an ICE chart to find the equilibrium concentrations of Ag' and ammonia. c Write the Kf expression. d. Calculate Kf.

Explanation / Answer

15. For the given reaction data,

Ag+ + 2NH3 <==> [Ag(NH3)2]+

a. Equilibrium concentration of [Ag(NH3)2]+ in test tube #2

= 0.24 x 7.56 x 10^-3 M/0.95

= 1.91 x 10^-3 M

b. initial [Ag+] = 0.06 M x 5 ml/10 ml = 0.03 M

initial [NH3] = 0.06 M x 5 ml/10 ml = 0.03 M

ICE chart

                    Ag+           +            2NH3 <=======> [Ag(NH3)2]+

I                  0.03                           0.03                              -

C          1.91 x 10^-3          -2 x 1.91 x 10^-3             +1.91 x 10^-3

E             0.02809                      0.02618                     1.91 x 10^-3

So at equilibrium

concentration [Ag+] = 0.02809 M

concentration [NH3] = 0.02618 M

c. Kf = [Ag(NH3)2]+/[Ag+][NH3]^2

d. Kf = (1.91 x 10^-3)/(0.02809)(0.02618)^2

        = 99.21

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