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50
HCOOH
H+ + HCOO?
¿ªÊ¼Å¨¶È/ mol¡¤dm-3 0.050 0 0 ƽºâŨ¶È/ mol¡¤dm-3 0.050 - x x x
x2 K(HCOOH) == 1.8 ? 10-4 x = 2.9 ? 10-3 mol¡¤dm-3
0050?x ¹Ê [H+] = [HCOO? ] = 2.9 ? 10-3 mol¡¤dm-3 [CN? ] =
K(HCN)?[HCN]?
[CN] = 1.7 ? 10-8 mol¡¤dm-3 +[H] 51
(4.20?4.00)50.0= 0.100 (mol¡¤dm-3)
100.04.00?50.0? [NH4]== 2.00 (mol¡¤dm-3)
100.0 (1)»ìºÍºóNH3¡¤H2O¹ýÁ¿£º[NH3¡¤H2O]= ¸ÃÈÜÒºÖÐ[OH?] = x NH3¡¤H2O
?NH4 + OH?
ƽºâŨ¶È£º 0.10-x 2.00 + x x
+[NH4][OH?]18.?10?5(010.?x)?
Kb= ËùÒÔ[OH]==9.0 ? 10-7(mol¡¤dm-3)
[NH3?H2O]2.00?x (2) pH = 14 - pOH = 14 - (-lg 9.0 ? 10-7) = 7.95 (3) [Fe2+] =
Ksp(Fe(OH)2)[OH?]210.?10?15?= 1.2 ? 10-3 mol¡¤dm-3 ?72(9.0?10) 52
ÓÉÓÚ pH = 9.0 ¼´ pOH = 5.0 [OH?] = 1 ? 10-5 mol¡¤dm-3
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¸ù¾Ý X- + H2OHX + OH?
ƽºâŨ¶È/ mol¡¤dm-3 0.20 - 1.0 ? 10-5 1 ? 10-5 1 ? 10-5
(1?10?5)21?10?10 ¿ÉµÃ Kb =?= 5 ? 10-10 ?50.200.20?10.?10Kw1?10?14 ËùÒÔKa ? ?? 2 ? 10-5 ?10Kb5?10 53
ÉèÐè¼Ó KHSO4 W g
[H+]2 = K2£¬ ¼´??c(HSO)?[H]4 ËùÒÔ W = 0.155 g 54
(2.20?10?3)2=1.26 ? 10-2
W?2.20?10?3136?0.440 ÒòΪK1 >> K2 >> K3 ÒòΪÈÜÒºÖеÄH+ Ö÷ÒªÀ´×ÔH3AsO4µÄÒ»¼¶½âÀë¡£
[H+]2[H+]2? ¼´ K1 =
[H3AsO4]c(H3AsO4)?[H?]?[H+]3[AsO3?4]ÓÉ H3AsO4 µÄ½âÀëÆ½ºâÓУº K1 K2 K3== K1 [H+] [AsO34] [H3AsO4] ËùÒÔ [H+] =
5.6?10?8?3.0?10?13?= 4.2 ? 10-3 (mol¡¤dm-3) 3??18[AsO4]4.0?10K2K3[H?]2(4.2?10?3)2+
ËùÒÔc(H3AsO4) =+ [H] = + 4.2 ? 10-3 = 7.5 ? 10-2 (mol¡¤dm-3) ?4K12.5?10 55
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K1K2K3510.?10?22? (1) K == 50 Ksp2.0?10?24 KÖµ²»ÊǺܴó£¬Ò²²»ÊǺÜС£¬·´Ó¦·½Ïò²»ÄÜÈ·¶¨¡£·´Ó¦·½Ïò¾ö¶¨Êµ¼ÊŨ¶È±ÈÖµ
Q¡£
ÒÔQ =>KÀ´ÅжÏ
<(10.?10?16)2 (2) K == 1.0 ? 1017 ??49Ksp(Ag2S)10.?10??×Ô·¢½øÐС£ ·´Ó¦Æ½ºâ³£Êý >> 107£¬´Ë³Áµíת»¯·´Ó¦? 56
Ksp(AgI)250= 0.15 (mol¡¤dm-3) 10050 [HAc] = 0.10 ? = 0.050 (mol¡¤dm-3)
100[HAc]0.050 pH = pKa -lg= 4.74 - lg= 5.22
[NaAc]0.15 (1) »ìºÏºó [NaAc] = 0.30 ?
(2) »ìºÏºó [NH4Cl] = 0.10 mol¡¤dm-3
[H+] =(Kw/Kb)?c?(1?10?14?010.)/18.?10?5= 7.5 ? 10-6 (mol¡¤dm-3) pH = 5.13
(3) [H+] =(K1?K2?59.?10?2?6.4?10?5= 1.9 ? 10-3 (mol¡¤dm-3) pH = 2.71 57
(1) ÒòΪHCl ΪǿËᣬËùÒÔ[H3O+]= [HCl]=
010.?2.0= 0.0040 (mol¡¤dm-3)
48.0?2.0 ËùÒÔHClÈÜÒºµÄ [H+] = 0.0040 mol¡¤dm-3 ÉècA= x£¬Ôò°´HAµÄµçÀëÆ½ºâÓУº
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[H3O?][A?]0.00402? Ka == 4.0 ? 10-5
[HA]x?0.0040 x = 0.40 mol¡¤dm-3
(2) NaAΪHAµÄ¹²éî¼î£¬ÆäË®½â³£ÊýΪ£º Kh =
Kw= 2.5 ? 10-10 Ka ÒòΪpH = 9.00 ËùÒÔpOH = 14.00 - 9.00 = 5.00 ËùÒÔ[OH?] = 1.0 ? 10-5 mol¡¤dm-3
[OH?]2 Kh == 2.5 ? 10-10
?[A] ËùÒÔ[NaA]0 = 0.40 mol¡¤dm-3 58
??rGm= -2.30 ? 8.31 ? 10-3 ? 298
6.4?10?5lg= -32.8 (kJ¡¤mol-1) ?10112.?10 ¦¤rG m = -32.8 + 2.30 ? 8.31 ? 10-3 ? 298 lg ËùÒÔ·´Ó¦ÏòÕýÏò½øÐС£ 59
HNO2 ƽºâŨ¶È/ mol¡¤dm-3
010.?010.= -21.4 (kJ¡¤mol-1)
0.010?0.010?H+ + NO2
0.25?300- x x + y x
500H+ + Ac-
HAc ƽºâŨ¶È/ mol¡¤dm-3 Ôò 4.5 ? 10-4 =
0.50?200- y x + y y
500(x?y)x
015.?x.. ..