15. The complex ion Ag(NH,) is formed by mixing 5.00 mL of AgNO, and with 5.00 m
ID: 694645 • Letter: 1
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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