3.酸效应一增大溶解度例5.1计算CaC,O.在不同情况下的溶解度在纯水中KO(CaC,0,)=10-8.6(1=0)SDH2C204(/=0pKa1=1.3pKa2=4.3Ca?++C,O.2-CaC,04sS个=5X10-5mol-L-1S=[Ca2+]=[C2O42]=sp6
6 3.酸效应—增大溶解度 例5.1 计算CaC2O4在不同情况下的溶解度 CaC2O4 Ca2++C2O4 2- S S Kθ sp(CaC2O4 )=10-8.6 (I＝0) H2C2O4 pKa1=1.3 pKa2=4.3 (I＝0) 在纯水中 S=[Ca2+]=[C2O4 2- ]= =5×10-5 mol·L-1 sp K
在pH=1.0的酸性溶液中用10.1的常数Ca2++C2042-CaC204 H+K'sp=[Ca?+][C2O4'] =S2HC,0411[C2042] α H2C204C,02-(H)SSα c,02(H) =1+[H+1β +[H+J?β2=1+10-1.0+4.0 +10-2.0+5.1 =103.4C,03()10-7.8+3.4 =10-4.4Ks=KspoαspK!S一= 10-2.2 = 6 X 10-3 (mol-L-1)sp7
7 在pH=1.0的酸性溶液中 CaC2O4 Ca2++C2O4 2- S S H+ HC2O4 - H2C2O4 用I＝0.1的常数 K´ sp =[Ca2+][C2O4 ] =S2 = [C2O4 2- ] (H) 2- 2 4 C O (H) =1+[H+ ]b1+[H+ ] 2b2 2- 2 4 C O -1.0+4.0 -2.0+5.1 3.4 =1+10 +10 =10 Ksp ′= Ksp =10-7.8+3.4 =10-4.4 2- 2 4 C O (H) S = = 10-2.2 = 6×10-3 (mol·L-1 ) sp K
若pH=4.0，过量H2C204（c=0.10mol-L-1）αc,0 (H) = 100.3C,02(H) 10-7.8+0.3=10-7.5Ksp= KspαsoK'sp= 10-7.5+1.0mol.L-1= 10-6.5 = 3 ×10-7,S=[C,04 ']Ca2+沉淀完全KMnO.法测Ca2+，MO指示剂8
8 若pH=4.0，过量H2C2O4 ( c=0.10mol·L-1 ) 2 4 0.3 C O (H) = 10 sp 2 4 -7.5+1.0 -6.5 -7 -1 = = 10 = 10 = 3 10 mol L [C O '] K S Ca2＋沉淀完全 KMnO4法测Ca2＋ ，MO指示剂 Ksp ′= Ksp = 10-7.8+0.3=10-7.5 2- 2 4 C O (H)
例5.2MnSKO=10-12.6H,SpKal=7.1 pK,2=12.9spKs,较大,S2-定量成为HS-,产生同量OH：假设MnS+H,0=Mn2++HS-+OHsSS[Mn?*][S?-][H*][OH"]K = [Mn?+][HS' ][OH' ] =Ka2010-12.6-14.0K= 10-13.7spW10-12.9Ka2S = 3/K = 10-13.7/3 = 10-4.6 mol-L-19
9 ∵ Ksp较大, S2-定量成为HS- ,产生同量OH- ∴假设 MnS+H2O=Mn2++HS－+OH－ S S S 2+ 2- s + - 2+ - 2 p w - a a 2 -12.6-14.0 -13.7 -12.9 [H ][OH ] = [Mn ][HS ][OH ] = 10 = = = 10 10 [Mn ][S ] K K K K K 3 -13.7/3 - 4.6 -1 S K = = 10 = 10 mol L 例5.2 MnS Kθ sp ＝10-12.6, H2S pKa1=7.1 pKa2=12.9
检验：[OH]=S=10-4.6mol-L-1,[H+]=10-9.4 mol-L-1s(H) =1 + 10-9.4+12.9+10-18.8+20.0=1 +103.5+101.2=103.5S2-HS-H2S主要以HS-存在故假设合理。亦可不必计算，由优势区域图可知：H2SHS-S2-pH7.19.412.9pKa1pKa210
10 检验：[OH- ]=S=10-4.6 mol·L-1 , [H+ ]=10-9.4 mol·L-1 S(H)＝1＋10-9.4+12.9+10-18.8+20.0 ＝1 + 103.5 + 101.2＝103.5 S2- HS- H2S 主要以HS-存在 故假设合理。 7.1 pKa1 12.9 pKa2 pH H2S HS－ S2－ 9.4 亦可不必计算，由优势区域图可知：