¢Ù ×°Óд¿Ë®µÄÉÕ±ÖеĴ¿Ë®Öð½¥Ôö¼Ó ¢Ú ×°Óд¿Ë®µÄÉÕ±ÖеĴ¿Ë®Öð½¥¼õÉÙ
¢Û ×°Óб¥ºÍÌÇË®µÄÉÕ±ÖеÄË®Öð½¥Ôö¼Ó ¢Ü ×°Óб¥ºÍÌÇË®µÄÉÕ±ÖеÄË®Öð½¥¼õÉÙ ÕýÈ·´ð°¸Îª¢ÚºÍ¢Û¡£
3£®ÊÔÓÃÈܶȻý¹æÔò½âÊÍÏÂÁÐÊÂʵ: £¨1£© Mg(OH)2ÈÜÓÚNH4ClÈÜÒºÖС£
-´ð£ºNH4ClΪǿËáÈõ¼îÑΣ¬ÏÔËáÐÔ£¬¿ÉÖкͲ¿·ÖOH £¬[ Mg(OH)2
2+-= Mg + 2OH]ʹMg(OH)2µÄÀë×Ó»ýСÓÚÈܶȻý¶øÈܽ⡣
£¨2£© ZnSÄÜÈÜÓÚÑÎËáºÍÏ¡ÁòËᣬ¶øCuSÈ´²»ÄÜÈÜÓÚÑÎËáºÍÏ¡ÁòËᣬµ«ÄÜÈÜÓÚÏõËáÖС£
´ð£ºÓÉʽc(H+)2Óйأ¬µ±ÈÜ
¡ª
ÒºÖÐc(H+)Ôö¼Óʱ£¬c(S2)c¼õС£¬ÔÚÑÎËáºÍÏ¡ÁòËáÈÜÒºÖеģ¨cH+£©
¡ª
×ãÒÔʹc (Zn2+)Óëc (S2) µÄÀë×Ó»ýСÓÚZnSµÄÈܶȻý £¨KspZnS = 1.6¡Á10-24£©£¬¶øKspCuS = 6.3¡Á10-37, ·ÇÑõ»¯ÐÔËá²»×ãÒÔʹ c
¡ª
(Cu2+) Óëc (S2)µÄÀë×Ó»ýСÓÚCuSµÄÈܶȻý£¬ Ö»ÓÐÑõ»¯ÐÔËá
¡ª
£¨HNO3µÈ£©¿ÉÒÔ½«S2-Ñõ»¯£¬¶ø´ó´ó½µµÍc (S2) ,ʹCuSÈܽâ. £¨3£© BaSO4²»ÈÜÓÚÏ¡ÑÎËáÖС£
´ð£ºÑÎËáÖеÄH+ »òCl- ²»ÄÜÓëBaSO4ÖнâÀë³öÀ´µÄÀë×Ó
¡ª
£¨Ba2+ »ò SO42£©·´Ó¦£¬¼´ÑÎËá²»ÄܽµµÍÈÜÒºÖÐc(Ba2+) »òc(SO42¡ª
)£¬ËùÒÔBaSO4²»ÈÜÓÚÏ¡ÑÎËá¡£
4£®Ìî³äÌâ: £¨1£© Èȵ糧·ÏË®ÅÅ·ÅÔì³ÉÈÈÎÛȾµÄÔÒòÊÇ___ÈȵķÏˮʹÌìȻˮÌåµÄÎÂ¶È Éý¸ß£¬ÈܽâÑõµÄŨ¶È½µµÍ£¬ÑáÑõ¾úºÍË®ÔåÔö¼Ó£¬Ë®ÌåÖÊÁ¿¶ñ»¯___¡£ £¨2£© Ë®ÖмÓÒÒ¶þ´¼¿ÉÒÔ·À¶³µÄÔÒòÊÇË®ÖÐÈܽâÈÜÖʺóʹˮµÄÄý¹Ìµã½µµÍ¡£ £¨3£© ÂÈ»¯¸ÆºÍÎåÑõ»¯¶þÁ׿ÉÓÃ×÷¸ÉÔï¼ÁµÄÔÒòÊÇ_ÔÚËüÃǵıíÃæËùÐγɵÄ
ÈÜÒºµÄÕôÆøѹÏÔÖøϽµ£¬¿É²»¶ÏµØÎüÊÕË®ÕôÆø£¬Ê¹ÌåϵÖеÄË®·Ö½µµÍ¡£ £¨4£© ÈËÌåÊäÒºÓõÄÉúÀíÑÎË®¼°ÆÏÌÑÌÇÈÜÒºµÄŨ¶È²»ÄÜËæÒâ¸ÄµÄÔÒòÊÇ _
c(S2?)?K1K2c(H2S)c2(H?)¿ÉÖª£¬c(S2 - )ÓëÈÜÒºÖÐ
±£Ö¤ËùÊäÈëµÄÒºÌåÓëÈËÌåѪҺµÄÉø͸ѹÏà½ü»òÏàµÈ__¡£
3
5£®ÑÎËẬHCl 37.0%(ÖÊÁ¿·ÖÊý)£¬ÃܶÈΪ1.19g/cm¡£¼ÆËã: £¨1£© ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶È(c)¡£ £¨2£© ÑÎËáµÄÖÊÁ¿Ä¦¶ûŨ¶È(bB)¡£ £¨3£© HClºÍH2OµÄÎïÖʵÄÁ¿·ÖÊý(x2 ºÍx1)¡£ ½â£º(1)ÑÎËáµÄ·Ö×ÓÁ¿Îª
mol¡¤L-1 (2) bB =
1000?1.19?37.06.536.5, c = = 12.06
1000?37.06.5?(1?37.0%) = 16.09 mol¡¤kg-1
37.0%/36.5?0.22537.0%/36.5?(1?37.0%)/18 (3) x2 = £»x1 = 1-x2 =0.775
6£® ÓÉÓÚʳÑζԲݵØÓÐËðÉË£¬Òò´ËÓÐÈ˽¨ÒéÓû¯·ÊÈçÏõËá炙òÁò
Ëá林úÌæ
ʳÑÎÀ´ÈÚ»¯ÈËÐеÀÅԵıùÑ©¡£ÏÂÁл¯ºÏÎï¸÷100gÈÜÓÚ1kgË®ÖУ¬ÎÊÄÄÒ»ÖÖ±ùµãϽµµÄ¶à£¿Èô¸÷0.1molÈÜÓÚ1kgË®ÖУ¬ÓÖÎÊÄÄÒ»ÖÖ±ùµãϽµµÄ¶à£¿
£¨1£© NaCl (2) NH4NO3 (3) (NH4)2SO4
´ð: ¸ù¾ÝÒÀÊýÐÔ¶¨ÂÉ£¬´ð°¸·Ö±ðΪ£¨1£©£»£¨3£© NaCl NH4NO3 ¡Á1002/80 (NH4)2SO4 ¡Á100gÎïÖʺ¬100Àë×Ó¡Á1003/132 Êý2/58.5 £¨mol£© 0.1mol Îï0.2 Öʺ¬Àë×ÓÊý£¨mol£© 7£® 3.62gÄá¹Å¶¡ÈÜÓÚ73.4gË®ÖУ¬ÆäÄý¹Ìµã½µµÍÁË0.563¡æ¡£ÇóÄá¹Å¶¡µÄÏà¶Ô·Ö×ÓÖÊÁ¿¡£
½â£ºÓɹ«Ê½?t?Kf?bB0.2 0.3 ¿ÉµÃ bB = 0.563/ 1.86 = 0.303
mol¡¤kg
-1
M?3.62?1000?162.8 Äá¹Å¶¡µÄ·Ö×ÓÁ¿Îª
73.4?0.303
8£®Ê÷¸ÉÄÚ²¿Ê÷ÖÉÏÉýÊÇÉø͸ѹËùÖÂ.ÉèÊ÷ÖÊÇŨ¶ÈΪ
0.20mol?L-1
µÄÈÜÒº,ÔÚÊ÷ֵİë͸ĤÍⲿˮÖзǵç½âÖÊŨ¶ÈΪ
0.02mol?L-1
.ÊÔ¹À¼ÆÔÚ25¡æʱ,Ê÷ÖÄܹ»ÉÏÉý¶à¸ß¡£
½â£º¸ù¾ÝÇóÉø͸ѹ¹«Ê½¦°=¦¤cRT =(0.20-0.02)¡Á8.314¡Á298 =446kPa
48m
9. 25¡æʱ,0.1mol l-1
¼×°·£¨CH3NH2£©ÈÜÒºµÄ½âÀëΪ6.9%
CHH+
3NH2(aq) + 2O(l) = CH3NH3(aq) +
OH-(aq)
ÊÔÎÊ:ÏàͬŨ¶ÈµÄ¼×°·Ó백ˮÄĸö¼îÐÔÇ¿? ½â£º ¼×°·ÖÐ c(OH - )=c£¨CH3NH2£©??= 0.1 ¡Á 6.9% = 0.0069 mol¡¤L-1
0.1mol¡¤l1°±Ë®ÖÐ c(OH - ) =KbCb=1.74?10?5?0.1= 0.00133 mol L-1
½áÂÛ£ºÏàͬŨ¶Èʱ£¬¼×°·µÄ¼îÐÔ¸üÇ¿
10£®ÔÚ1 L 0.1mol L-1
HAc ÈÜÒºÖÐ,Ðè¼ÓÈë¶àÉٿ˵ÄNaAc¡¤3H2O²ÅÄÜʹÈÜÒºµÄpH Ϊ5.5£¿£¨¼ÙÉèNaAc¡¤3H2OµÄ¼ÓÈë²»¸Ä±äHAcµÄÌå»ý£©¡£
½â£º¸ÃϵͳΪ»º³åÈÜÒº c£¨H+£©=
Kcaacs £»
0.110-5.5 = 1.74¡Á10 -5 cs
£» cs = 0.055 mol?L-1¡£
Ðè¼Ó´×Ëá 0.55¡Á136 = 74.8g
11£®Ä³Ò»ÔªÈõ¼î£¨MOH£©µÄÏà¶Ô·Ö×ÓÁ¿Îª125£¬ÔÚ25¡æʱ½«1g´Ë¼îÈÜÓÚ0.1 LË®ÖÐ,ËùµÃÈÜÒºµÄpHΪ11.0,Çó¸ÃÈõ¼îµÄ
½âÀë³£ÊýK?b¡£
½â£ºÎªÈõ¼îµçÀëÌåϵͳ£¬´ËÈõ¼îµÄŨ¶È
c?1000b?1?0.08mol?L?1125?100£¬
¡Ö
??14?11.0(OH)?10/10?Kbcb?Kb?0.08£¬ cÇó³öKb?1.25¡Á10 -5¡£ 12£®Pb(NO3)2ÈÜÒºÓëBaCl2ÈÜÒº»ìºÏ£¬Éè»ìºÏÒºÖÐPb(NO3)2
µÄŨ¶ÈΪ0.20mol?L-1£¬ÎÊ
-4
(1) ÔÚ»ìºÏÈÜÒºÖÐCl- µÄŨ¶ÈµÈÓÚ5.0¡Á10 mol?L-1 ʱ£¬ÊÇ·ñÓгÁµíÉú³É?
(2) »ìºÏÈÜÒºÖÐCl-µÄŨ¶È¶à´óʱ£¬¿ªÊ¼Éú³É³Áµí?
-2
(3) »ìºÏÈÜÒºÖÐCl-µÄƽºâŨ¶ÈΪ6.0¡Á10 mol?L-1ʱ£¬²ÐÁôÓÚÈÜÒºÖеÄPb2+µÄŨ¶ÈΪ¶àÉÙ? ½â£º£¨1£©Àë×Ó»ýc(Pb2+)c(Cl-1)2 = 0.2¡Á(5.0¡Á10-4)2 =5¡Á10-8 £¨2£©¿ªÊ¼Éú³É³Áµíʱ 2 c£¨Cl£© = £¨3£©²ÐÁôµÄǦÀë×ÓŨ¶È c£¨Pb2+£©= ¡Á10 - 3¡£ ¨C Kb/[Pb2?] = 1.6?10?5/0.2?8.9?10?3£» KPbCl2/ c (Cl¨C)2 = 1.6¡Á10-5/(6.0¡Á10 ¨C2 ) 2 = 4.44 µÚÁùÕ Ô×ӽṹÓëÖÜÆÚϵ Ï°ÌâÓë½â´ð 1. ÏÂÁÐ˵·¨ÊÇ·ñÕýÈ·£¿Èç²»ÕýÈ·£¬Çë˵Ã÷ÔÒò¡£ £¨1£© Ô×Ó¹ìµÀÊÇÖ¸Ô×ÓºËÍâµç×Ó³öÏÖ¸ÅÂÊ×î´óµÄÇøÓò¡£ ´ð£º²»ÕýÈ·¡£Ô×Ó¹ìµÀÊǺËÍâµç×ÓµÄÒ»ÖÖ¿ÉÄܵÄÔ˶¯×´Ì¬£¬²»½ö½öÊÇÖ¸¸ÅÂÊ×î´óµÄÇøÓò¡£ £¨2£©ÒòΪ²¨º¯ÊýÓÐÒ»¶¨µÄÎïÀíÒâÒ壬Òò´Ën¡¢l¡¢mÈý¸öÁ¿×ÓÊýµÄÈ¡ÖµÓÐÒ»¶¨ÏÞÖÆ£¬Ö»ÓÐÂú×ãÒ»¶¨¹ØϵµÄÈ¡ÖµµÃµ½µÄº¯Êý²ÅÄÜÃèÊöµç×ÓµÄÔ˶¯×´Ì¬¡£ ´ð£ºÕýÈ·¡£ £¨3£©ÒòΪp¹ìµÀÊÇ8×ÖÐεģ¬ËùÒÔ´¦ÓڸùìµÀµÄµç×ÓÊÇÑØ×Å8×ÖÐεĹìµÀÔ˶¯¡£ ´ð£º´í¡£µç×ÓÔ˶¯Ã»Óй̶¨¹ì¼££¬Ö»Óм¸ÂÊ·Ö²¼¹æÂÉ£¬8×ÖÐÍÊÇp¹ìµÀµÄ½Ç¶È·Ö²¼º¯Êýͼ¡£ËüÖ»±íʾÔ×Ó¹ìµÀÔÚ²»Í¬·½ÏòÉÏ