ºþÄÏÀí¹¤Ñ§ÔºÎÞ»ú»¯Ñ§¿Î¼þÖ®¸´Ï°ÌâµÚ°ËÕÂÅäλ»¯ºÏÎïÓëÅäλµÎ¶¨1 ÁªÏµ¿Í·þ

·¢²¼Ê±¼ä : ÐÇÆÚÎå ÎÄÕºþÄÏÀí¹¤Ñ§ÔºÎÞ»ú»¯Ñ§¿Î¼þÖ®¸´Ï°ÌâµÚ°ËÕÂÅäλ»¯ºÏÎïÓëÅäλµÎ¶¨1¸üÐÂÍê±Ï¿ªÊ¼ÔĶÁ01d1dc1a964bcf84b9d57b9b

µÚ°ËÕ Åäλ»¯ºÏÎïÓëÅäλµÎ¶¨

Ï°Ìâ1£ºÅäλ»¯ºÏÎﲿ·Ö

1 ÎÞË®CrCl3ºÍ°±×÷ÓÃÄÜÐγÉÁ½ÖÖÅäºÏÎïAºÍB£¬×é³É·Ö±ðΪCrCl3¡¤6NH3ºÍCrCl3¡¤5NH3¡£¼ÓÈëAgNO3,AÈÜÒºÖм¸ºõÈ«²¿ÂȳÁµíΪAgCl£¬¶øBÈÜÒºÖÐÖ»ÓÐ2/3µÄÂȳÁµí³öÀ´¡£¼ÓÈëNaOH²¢¼ÓÈÈ£¬Á½ÖÖÈÜÒº¾ùÎÞ°±Î¶¡£ÊÔд³öÕâÁ½ÖÖÅäºÏÎïµÄ»¯Ñ§Ê½²¢ÃüÃû¡£

-½â£ºÒò¼ÓÈëAgNO3£¬AÈÜÒºÖм¸ºõÈ«²¿ÂȳÁµíΪAgCl£¬¿ÉÖªAÖеÄÈý¸öClÈ«²¿ÎªÍâ

--½çÀë×Ó£¬BÈÜÒºÖÐÖ»ÓÐ2/3µÄÂȳÁµí³öÀ´£¬ËµÃ÷BÖÐÓÐÁ½¸öClΪÍâ½ç£¬Ò»¸öClÊôÄڽ硣¼ÓÈëNaOH£¬Á½ÖÖÈÜÒºÎÞ°±Î¶£¬¿ÉÖª°±ÎªÄڽ硣Òò´ËA¡¢BµÄ»¯Ñ§Ê½ºÍÃüÃûӦΪ£¬A£º[Cr(NH3)6]Cl3 ÈýÂÈ»¯Áù°±ºÏ¸õ£¨¢ó£© B£º[Cr(NH3)5Cl]Cl2 ¶þÂÈ»¯Ò»ÂÈÎå°±ºÏ¸õ£¨¢ó£©

2 Ö¸³öÏÂÁÐÅäºÏÎïµÄÖÐÐÄÀë×Ó¡¢ÅäÌå¡¢ÅäλÊý¡¢ÅäÀë×ÓµçºÉÊýºÍÅäºÏÎïÃû³Æ¡£ ½â£º ÅäºÏÎï K2[HgI4] [CrCl2(H2O)4]Cl [Co(NH3)2(en)2](NO3)2 Fe3[Fe(CN)6]2 K[Co(NO2)4(NH3)2] Fe(CO)5 ÖÐÐÄÀë×Ó Hg2+ Cr3+ Co2+ Fe3+ Co3+ Fe ÅäÌå I- Cl-,H2O NH3, en CN- NO2-,NH3 CO ÅäλÊý 4 6 6 6 6 5 ÅäÀë×ÓµçºÉ -2 +1 +2 -3 -1 0 ÅäºÏÎïÃû³Æ ËĵâºÏ¹¯£¨¢ò£©Ëá¼Ø ÂÈ»¯¶þÂÈËÄË®ºÏ¸õ£¨¢ó£© ÏõËá¶þ°±¶þÒÒ¶þ°·ºÏîÜ(¢ò) ÁùÇèºÏÌú£¨¢ó£©ËáÑÇÌú ËÄÏõ»ù¶þ°±ºÏîÜ£¨¢ó£©Ëá¼Ø ÎåôÊ»ùºÏÌú

3 ÊÔÓüۼüÀíÂÛ˵Ã÷ÏÂÁÐÅäÀë×ÓµÄÀàÐÍ¡¢¿Õ¼ä¹¹ÐͺʹÅÐÔ¡£

3--½â£º£¨1£©CoF6ºÍCo(CN)63

CoF63-ΪÍâ¹ìÐÍ£¬¿Õ¼ä¹¹ÐÍΪÕý°ËÃæÌ壬˳´ÅÐÔ£¬´Å¾Ø=4(4?2)?4.90?B¡£ Co(CN)63-ΪÄÚ¹ìÐÍ£¬¿Õ¼ä¹¹ÐÍΪÕý°ËÃæÌ壬¿¹´ÅÐÔ£¬´Å¾ØΪÁã¡£

2-£¨2£©Ni(NH3)42+ºÍNi(CN)4

Ni(NH3)42+ΪÍâ¹ìÐÍ£¬ÕýËÄÃæÌåÐÍ£¬Ë³´ÅÐÔ£¬´Å¾Ø=2(2?2)?2.83?B¡£

Ni(CN)4ΪÄÚ¹ìÐÍ£¬Æ½ÃæÕý·½ÐÍ£¬¿¹´ÅÐÔ£¬´Å¾ØΪÁã¡£ 4 ½«0.1mol¡¤L-1ZnCl2ÈÜÒºÓë1.0 mol¡¤L-1NH3ÈÜÒºµÈÌå»ý»ìºÏ£¬Çó´ËÈÜÒºÖÐZn(NH3)42+

ºÍZn2+µÄŨ¶È¡£

½â£ºµÈÌå»ý»ìºÏºó£¬ZnCl2ºÍNH3¸÷×ÔµÄŨ¶È¼õ°ë£¬Éú³ÉµÄZn(NH3)42+µÄŨ¶ÈΪ0.05 mol¡¤L-1£¬Ê£ÓàµÄNH3µÄŨ¶ÈΪ0.3 mol¡¤L-1£¬ÉèZn2+µÄŨ¶ÈΪx mol¡¤L-1¡£

Zn2+ + 4NH3 = Zn(NH3)42+ ƽºâʱ x 0.3+4x 0.05 ¨C x

Kf?2-

0.05?x9 ?2.9?104x(0.3?4x)-9

0.05 ¨C x ¡Ö 0.05 0.3 + 4x ¡Ö 0.3 x = 2.1¡Á10 (mol¡¤L-1) 5 ÔÚ100ml 0.05mol¡¤L-1Ag(NH3)2+ÈÜÒºÖмÓÈë1ml 1 mol¡¤L-1NaClÈÜÒº£¬ÈÜÒºÖÐNH3µÄŨ¶ÈÖÁÉÙÐè¶à´ó²ÅÄÜ×èÖ¹AgCl³ÁµíÉú³É£¿

-½â£º»ìºÏºóc[Ag(NH3)2+] = 0.1¡Á100/101 = 0.099(mol¡¤L1)

1

c(Cl) = 1/101= 0.0099 (mol¡¤L1)

·½·¨Ò». Ag(NH3)2+ + Cl- = AgCl + 2NH3 0.099 0.0099 x

--

2-x1 Kj?L1) ??5.14?102 x = 0.71 (mol¡¤0.099?0.0099Kf?Ksp·½·¨¶þ c(Ag+)c(Cl) ¡Ü Ksp,AgCl c(Ag+) ¡Ü Ksp,AgCl /0.0099 = 1.79¡Á10-8(mol¡¤L1) Ag+ + 2NH3 = Ag(NH3)2+ 1.79¡Á10-8 x 0.099

-0.0997 Kf? x = 0.71(mol¡¤L1) ?1.1?10(1.79?10?8)x2--

6 ¼ÆËãAgClÔÚ0.1mol¡¤L°±Ë®ÖеÄÈܽâ¶È¡£

-½â£º ÉèAgClÔÚ0.1mol¡¤L°±Ë®ÖеÄÈܽâ¶ÈΪx mol¡¤L1.

- AgCl + 2NH3 = Ag(NH3)2+ + Cl

- ƽºâʱ 0.1 x x mol¡¤L1

Kj?cAg(NH)?cCl?32c2NH3?Kf,Ag(NH?3)2?Ksp,AgCl?1.1?107?1.77?10?10?1.95?10?3

x2-1?3?1.95?10 x = 0.0044 (mol¡¤L) 20.17 ÔÚ100ml 0.15 mol¡¤L1Ag(CN)2ÖмÓÈë+50ml 0.1mol¡¤L1 KIÈÜÒº£¬ÊÇ·ñÓÐAgI³ÁµíÉú

-³É£¿ÔÚÉÏÊöÈÜÒºÖÐÔÙ¼ÓÈë50ml 0.2mol¡¤L1KCNÈÜÒº£¬ÓÖÊÇ·ñ²úÉúAgI³Áµí £¿

---½â£º£¨1£©100ml0.15mol¡¤L1 Ag(CN)2ÖмÓÈë50mL 0.1mol¡¤L1 KIºó£¬

- c[Ag(CN)2-] = 0.15¡Á100/150 = 0.10(mol¡¤L1),

- - c(I ) = 0.1¡Á50/150 = 0.033 (mol¡¤L1)

-- Ag+ + 2CN = Ag(CN)2 ƽºâʱ x 2x 0.10 - x

---

Kf?0.1?x-L1) ?1.3?1021 x = 2.7¡Á10-8(mol¡¤2x(2x)-8

-10

c(Ag+)c(I-) = 2.7¡Á10¡Á0.033 = 8.9¡Á10> Ksp,AgI = 8.51¡Á10-17, ÓÐAgI³ÁµíÉú³É¡£

- £¨2£©ÔÙ¼ÓÈë50ml 0.2mol¡¤L1KCNºó£¬c(CN-) = 0.2¡Á50 /200 = 0.05(mol¡¤L-1) c[Ag(CN)2-] = 0.15¡Á100/200 = 0.075(mol¡¤L),

--c(I) = 0.1¡Á50/200 = 0.025 (mol¡¤L1)

-- Ag+ + 2CN = Ag(CN)2 ƽºâʱ y 0.05 0.075 ¨C y

- Kf?0.075?y?1.3?1021 y = 2.3¡Á10-20(mol¡¤L1)

(0.05)2y c(Ag+)c(I-) = 2.3¡Á10¡Á0.025 = 5.8¡Á10< Ksp,AgI = 8.51¡Á10, ÎÞAgI³ÁµíÉú³É¡£

2-8 0.08molAgNO3ÈܽâÔÚ1LNa2S2O3ÈÜÒºÖÐÐγÉAg(S2O3)23-£¬¹ýÁ¿µÄS2O3Ũ¶ÈΪ0.2 -mol¡¤L1¡£ÓûµÃµ½Â±»¯Òø³Áµí £¬ËùÐèI-ºÍCl-µÄŨ¶È¸÷Ϊ¶àÉÙ£¿ÄÜ·ñµÃµ½AgI£¬AgCl³Áµí£¿

-½â£º0.08molAgNO3ÈܽâÔÚ1LNa2S2O3ÈÜÒººó£¬c[Ag(S2O3)23-] = 0.08(mol¡¤L1)

-20

-22

-17

2

Ag+ + 2S2O3 = Ag(S2O3)2 x 0.2 0.08 ¨C x

2-3-

Kf?0.08?x-14-113 x = 6.9¡Á10(mol¡¤L) ?2.9?1020.2x ÓûµÃAgCl³Áµí£¬c(Ag+)c(Cl-) >Ksp,AgCl

-10-14- c(Cl-) > 1.77¡Á10 /6.9¡Á10 = 2565 (mol¡¤L1) £¬²»¿ÉÄܵõ½AgCl³Áµí¡£

ÓûµÃAgI³Áµí£¬c(Ag+)c(I-) >Ksp,AgI

-17-14-3- c(I-) > 8.51¡Á10 /6.9¡Á10 = 1.23¡Á10 (mol¡¤L1) £¬¿ÉÒԵõ½AgI³Áµí¡£ --9 50ml 0.1 mol¡¤L1 AgNO3ÈÜÒºÓëµÈÁ¿µÄ6 mol¡¤L1°±Ë®»ìºÏºó£¬Ïò´ËÈÜÒºÖмÓÈë0.119g KBr¹ÌÌ壬ÓÐÎÞAgBr³ÁµíÎö³ö£¿ÈçÓû×èÖ¹AgBrÎö³ö£¬Ô­»ìºÏÈÜÒºÖа±µÄ³õŨ¶ÈÖÁÉÙӦΪ¶àÉÙ£¿

--½â£º50ml 0.1 mol¡¤L1 AgNO3ÈÜÒºÓëµÈÁ¿µÄ6 mol¡¤L1°±Ë®»ìºÏºó£¬¸÷×ÔµÄŨ¶È¼õ°ë£¬

---AgNO3ÈÜÒººÍ°±Ë®µÄŨ¶È·Ö±ðΪ0.05 mol¡¤L1ºÍ3 mol¡¤L1£¬·´Ó¦Éú³É0.05 mol¡¤L1 Ag(NH3)2+£¬

-ƽºâʱAg+Ũ¶ÈΪxmol¡¤L1¡£

Ag+ + 2NH3 = Ag(NH3)2+

- ƽºâʱ x 2.9 + 2x 0.05 ¨C x mol¡¤L1

0.05?x?Kf?1.1?107 2x(2.9?2x)0.05 ¨Cx ¡Ö 0.05 2.9 + 2x ¡Ö 2.9 x = 5.4¡Á10-10 (mol¡¤L1)

-

¼ÓÈë0.119g KBr¹ÌÌåºó£¬Br-Ũ¶È=

0.119?0.01(mol?L?1)

119?0.1c(Ag+)c(Br-) = 5.4¡Á10-10¡Á0.01 = 5.4¡Á10-12 > KSP, AgBr = 5.35¡Á10-13

-Óû×èÖ¹Éú³ÉAgBr³Áµí£¬c(Ag+) ¡Ü5.35¡Á10-13/0.01 = 5.35¡Á10-11(mol¡¤L1)

Ag+ + 2NH3 = Ag(NH3)2+

- ƽºâʱ 5.35¡Á10-11 x +2¡Á5.35¡Á10-11 0.05 ¨C5.35¡Á10-11 mol¡¤L1

x +2¡Á5.35¡Á10-11 ¡Ö x 0.05 ¨C5.35¡Á10-11 ¡Ö 0.05

0.05-?1.1?107 x = 9.2 (mol¡¤L1) -1125.35?10?x°±Ë®µÄ³õŨ¶È = 9.2 + 0.1 = 9.3 (mol¡¤L)

-10 ·Ö±ð¼ÆËãZn(OH)2ÈÜÓÚ°±Ë®Éú³ÉZn(NH3)42+ºÍÉú³ÉZn(OH)42ʱµÄƽºâ³£Êý¡£ÈôÈÜ

-ÒºÖÐNH3ºÍNH4+µÄŨ¶È¾ùΪ0.1 mol¡¤L1£¬ÔòZn(OH)2ÈÜÓÚ¸ÃÈÜÒºÖÐÖ÷ÒªÉú³ÉÄÄÒ»ÖÖÅäÀë×Ó£¿

-½â£º Zn(OH)2 + 4NH3 = Zn(NH3)42+ + 2OH

Kj?c Zn(NHc2?3) 42cOH?4NH3?Kf,Zn(NH2?3) 4?Ksp,Zn(OH)2

?2.9?109?6.68?10?17?1.9?10?7Zn(OH)2 + 2OH = Zn(OH)42

--

Kj? cZn(OH)2- 4c2OH-?2cZn2??cOH?cZn2??c2OH??Kf,Zn(OH)42-?Ksp,Zn(OH)2

?4.6?1017?6.68?10?17?30.7 µ±ÈÜÒºÖÐNH3ºÍNH4+µÄŨ¶È¾ùΪ0.1 mol¡¤L1ʱ£¬OH Ũ¶ÈΪx mol¡¤L1£¬

¨C

NH3¡¤H2O = NH4+ + OH

---

3

ƽºâ 0.1 0.1 x Kb?0.1x?1.77?10?5 x = 1.77¡Á10-5(mol¡¤L-1) 0.1--

Zn(NH3)42+ + 4OH = Zn(OH)42 + 4NH3

Kj?cZn(OH)cZn(NHcZn(NH42-42- ?c4NH34?cOH?2-?Kf,Zn(OH)Kf,Zn(NH42- 3)42? 3)42? 4.6?1017??1.6?108 92.9?10Kj?cZn(OH)3)42? 44(0.1) (1.77?10?5)48?1.6?108

cZn(OH)cZn(NH 3)42? (1.77?10?5)4?1.6?10??1.6?10?8 40.1-

µ±ÈÜÒºÖÐNH3ºÍNH4+µÄŨ¶È¾ùΪ0.1 mol¡¤L1ʱ£¬Zn(OH)2ÈÜÓÚ¸ÃÈÜÒºÖÐÖ÷ÒªÉú³ÉZn(NH3)42+¡£

---11 ½«º¬ÓÐ0.2 mol¡¤L1NH3ºÍ1.0 mol¡¤L1NH4+µÄ»º³åÈÜÒºÓë0.02 mol¡¤L1Cu(NH3)42+ÈÜÒºµÈÌå»ý»ìºÏ£¬ÓÐÎÞCu(OH)2³ÁµíÉú³É£¿[ÒÑÖªCu(OH)2µÄKsp=2.2¡Á10-20]

-½â£ºµÈÌå»ý»ìºÏºó£¬NH3¡¢NH4+¡¢Cu(NH3)42+µÄŨ¶È¸÷×Ô¼õ°ë£¬ÉèCu2+Ũ¶ÈΪx mol¡¤L1, Cu2+ + 4NH3 = Cu(NH3)42+ ƽºâʱ x 0.1+4x 0.01 - x

Kf?0.01?x13 ?2.1?104x(0.1?4x)-

0.01 ¨C x ¡Ö 0.01 0.1+4x ¡Ö 0.1 x = 4.8¡Á10-12 (mol¡¤L1)

¨C- Éè»ìºÏÒºÖÐOHŨ¶ÈΪy mol¡¤L1£¬

NH3¡¤H2O = NH4+ + OH¨C

ƽºâʱ 0.y 0.5+y y

Kb?(0.5?y)y?1.77?10?5

0.1?y-

0.y¡Ö0.1 0.5+y¡Ö0.5 y = 3.5¡Á10-6 (mol¡¤L1)

- c ( Cu2+) c2(OH ) = 4.8¡Á10-12¡Á(3.5¡Á10-6)2 = 5.9¡Á10-23 < Cu(OH)2µÄKsp£¬ÎÞCu(OH)2

³ÁµíÉú³É¡£

12 ³öÏÂÁз´Ó¦µÄ·½³Ìʽ²¢¼ÆËãƽºâ³£Êý£º

£¨1£© AgIÈÜÓÚKCNÈÜÒºÖУ»

£¨2£© AgBr΢ÈÜÓÚ°±Ë®ÖУ¬ÈÜÒºËữºóÓÖÎö³ö³Áµí£¨Á½¸ö·´Ó¦£©¡£

--

½â£º£¨1£© AgI + 2CN- = Ag(CN)2 + I

Kj?cAg(CN)- cI?2c2CN??Kf,Ag(CN)- ?Ksp,AgI?1.3?1021?8.51?10?17?1.1?105

2(2) AgBr + 2NH3 = Ag(NH3)2+ + Br-

4