0.05x=Kp(AgC1)=1.77x1010,x=3.54x109molL Ag" +2NH3 Ag(NH3)2 Initial 0.1 0 At equilibria 3.54 x 10 y-2(0.1-3.54x10%0.1-3.54x109 (0.1-3.54x10913.54x109y-2(0.1-3.54x10=K(Ag(NH2)=1.12x107 y=1.79 molL! the total concentration of NH3 is 1.79molL. (2)AgI Ag"I 3.54x109 0.001 3.54x10x0.001=3.54 x 102>Ksp(Agl)=8.52x107 So AgI precipitates. 2.(5 points) Anwser:p(SbCls)initial nRT/V=0.0125mol x8.314J-K*x 521K/3.50L=15.47KPa SbCls(g)=SbCl3 (g)+Cl2(g) Initial /KPa 15.47 0 0 At equilibria 15.47-x x (X/100)2/[(15.47-x)/100]=Km°=1.07， x=P(SbCl)=P(C2)=13.71Kpa,P(SbCs)=15.47-13.71=1.76kpa Poa=1.76+13.71x2=29.17kpa 3.(5 points) Anwser:Ag2SO(s)=2Ag*+SO2 0.01 X 0.012x=Kp(Ag2S0)=1.4x103X=0.14 CaSO4(s)Ca2++SOa2- 0.01y 0.01y=Kp(CaS04)=9.1x106y=9.1x10 So Caprecipitates first 第3页 0.05x = Ksp(AgCl) = 1.77 x 10-10 , x = 3.54 x 10-9 molL-1 Ag+ + 2NH3 Ag(NH3)2 Initial 0.1 y 0 At equilibria 3.54 x 10-9 y – 2 (0.1 – 3.54 x 10-9) 0.1 – 3.54 x 10-9 (0.1 – 3.54x10-9) /3.54 x 10-9 [y- 2 (0.1 – 3.54 x 10-9)]2 = Kf(Ag(NH3)2) = 1.12 x 107 y = 1.79 molL-1 the total concentration of NH3 is 1.79molL-1. (2) AgI Ag+ + I- 3.54x10-9 0.001 3.54x10-9 x 0.001 = 3.54 x 10-12 > Ksp(AgI) = 8.52 x 10-17, So AgI precipitates. 2．(5 points) Anwser: p(SbCl5)initial = nRT/V = 0.0125mol x8.314J⋅K-1 x 521K/3.50L = 15.47KPa SbCl5(g) SbCl3 (g) + Cl2 (g) Initial /KPa 15.47 0 0 At equilibria 15.47-x x x (X/100)2 /[(15.47-x)/100] = KP Θ = 1.07, x = P(SbCl3) = P(Cl2) =13.71Kpa, P(SbCl5) = 15.47-13.71=1.76kpa Ptotal = 1.76 + 13.71 x 2 = 29.17kpa 3. (5 points) Anwser: Ag2SO4(s) 2Ag+ + SO4 2- 0.01 x 0.012 x = Ksp(Ag2SO4) = 1.4 x 10-5 X = 0.14 CaSO4(s) Ca2+ + SO4 2- 0.01 y 0.01y = Ksp(CaSO4) = 9.1 x 10-6 y = 9.1 x 10-4 So Ca2+ precipitates first. 第 3 页