Question

无机化学求大神解答！！！27 28 29题！！！！

Answer 1

27、HAc(aq)<=>H+(aq)+Ac-(aq)
△G0= △fG0[Ac-(aq)] +△fG0[H+(aq)] -△fG0[HAc(aq)]
=(-372.46)-(-399.61)=27.15kJ/mol
△G0=-RTLnKa
27.15*10^3 J/mol=-8.314J/mol*K *298.15K *LnKa
Ka=1.75*10^-5

AgCl(s)<=>Ag+(aq)+Cl-(aq)
△G0= △fG0[Ag+(aq)] +△fG0[Cl-(aq)] -△fG0[AgCl(s)]
=(+77.11)+(-131.17)-(-109.72)=55.66kJ/mol
△G0=-RTLnKsp
55.66*10^3 J/mol=-8.314J/mol*K *298.15K *LnKsp
Ka=1.76*10^-10

28、Ka越大，酸度强
Ka（NH4+ ）=Kw/Kb（NH3）=10^-14/1.76*10^-5=5.6*10^-10
因此H3PO4 >H2PO4- >NH4+ >HCN

29、Kb越大，碱度强
Kb（Ac-）=Kw/KaHAc=10^-14/1.76*10^-5=5.6*10^-10
Kb(PO43-)=Kw/Ka3(H3PO4)=10^-14 /2.2*10^-13 =4.5*10^-2
Kb(H2PO4-)=Kw/Ka1(H3PO4)=10^-14 /7.52*10^-3 =1.33*10^-12
由于 Kb(PO43-)>KbNH3 >Kb（Ac-）>Kb(H2PO4-)
因此: PO43->NH3 >Ac->H2PO4-