ÓÐÒ»ÖÖ°×É«·ÛÄ©£¬º¬ÓÐÏÂÁÐÒõÀë×ÓºÍÑôÀë×ÓÖеļ¸ÖÖ¡£
ÒõÀë×Ó£ºS2£­¡¢Cl£­¡¢NO3¡ª¡¢SO42¡ª¡¢CO32¡ª¡¢HCO3¡ª¡¢MnO4¡ª¡£
ÑôÀë×Ó£ºNa£«¡¢Mg2£«¡¢Al3£«¡¢Ba2£«¡¢Fe2£«¡¢Fe3£«¡¢Cu2£«¡¢NH4+¡£
½«¸Ã°×É«·ÛÄ©½øÐÐÏÂÁÐʵÑ飬¹Û²ìµ½µÄÏÖÏóÈçÏ£º
ʵÑé²Ù×÷
ÏÖÏó
a.È¡ÉÙÁ¿·ÛÄ©£¬¼ÓË®¡¢Õñµ´
È«²¿Èܽ⡢
ÈÜÒºÎÞɫ͸Ã÷
 
b.ÏòËùµÃÈÜÒºÖÐÂýÂýµÎÈë¿ÁÐÔÄÆÈÜÒº£¬²¢¼ÓÈÈ
ÎÞÃ÷ÏÔÏÖÏó
c.È¡ÉÙÁ¿·ÛÄ©£¬¼ÓÑÎËá
ÎÞÃ÷ÏÔÏÖÏó
d.È¡ÉÙÁ¿·ÛÄ©£¬¼ÓÏ¡H2SO4ºÍÏ¡HNO3µÄ»ìºÏÒº
Óа×É«³ÁµíÉú³É
 
¸ù¾ÝʵÑéÍƶϣº
(1)´ÓaʵÑéÖУ¬¿ÉÍƶϷÛÄ©Öв»¿ÉÄÜÓÐ______________(ÌîÀë×Ó·ûºÅ£¬ÏÂͬ)¡£
(2)´ÓbʵÑéÖУ¬¿ÉÍƶϷÛÄ©Öв»¿ÉÄÜÓÐ_____________________________________¡£
(3)´ÓcʵÑéÖУ¬¿ÉÍƶϷÛÄ©Öв»¿ÉÄÜÓÐ________________________________¡£
(4)´ÓdʵÑéÖУ¬¿ÉÍƶϷÛÄ©Öв»¿ÉÄÜÓÐ________£¬Ò»¶¨º¬ÓÐ________¡£
(5)ÒÔÉϸ÷ʵÑéÈÔÎÞ·¨È·¶¨ÊÇ·ñ´æÔÚµÄÀë×ÓÊÇ____________¡£
(1)Fe2£«¡¢Fe3£«¡¢Cu2£«¡¢MnO4¡ª
(2)NH4+¡¢Mg2£«¡¢Al3£«
(3)CO32¡ª¡¢HCO3¡ª¡¢S2£­
(4)SO42¡ª¡¡Ba2£«
(5)Cl£­¡¢NO3¡ª¡¢Na£«
(1)Fe3£«Îª×Ø»ÆÉ«£¬Fe2£«ÎªÇ³ÂÌÉ«£¬Cu2£«ÎªÀ¶É«£¬MnO4¡ªÎª×ÏÉ«¡£
(2)NH4+ÓëOH£­¿É·´Ó¦Éú³ÉNH3£¬Mg2£«¡¢Al3£«¿ÉÓëOH£­·´Ó¦Éú³É³Áµí¡£
(3)CO32¡ª¡¢HCO3¡ª¿É·Ö±ðÓëH£«·´Ó¦Éú³ÉCO2ÆøÌå¡£
(4)¼ÓÈëÁòËáºÍÏõËáÉú³ÉµÄ³ÁµíÖ»ÄÜÊÇBaSO4£¬Òò´Ë¿ÉÍƳöº¬ÓÐBa2£«£¬Ôò±Ø²»º¬ÓÐSO42¡ª
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚº¬ÓдóÁ¿K+¡¢OH£­¡¢CO32-µÄÈÜÒºÖл¹¿ÉÄÜ´óÁ¿´æÔÚµÄÀë×ÓÊÇ
A£®NH4+B£®Al3+C£®Ca2+D£®SO42-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÒºÌå¾ù´¦ÓÚ25¡æ£¬ÓйØÐðÊöÕýÈ·µÄÊÇ
A£®Ä³ÎïÖʵÄÈÜÒºpH<7£¬Ôò¸ÃÎïÖÊÒ»¶¨ÊÇËá»òÇ¿ËáÈõ¼îÑÎ
B£®pH=4£®5µÄ·¬ÇÑÖ­ÖÐc(H£«£©ÊÇpH=6£®5µÄÅ£ÄÌÖÐc£¨H£«)µÄ102±¶
C£®³£ÎÂʱ£¬Ä³ÈÜÒºÖÐÓÉË®µçÀë³öÀ´µÄc(H£«)ºÍc(OH£­)µÄ³Ë»ýΪ1¡Á10£­24£¬¸ÃÈÜÒºÖÐÒ»¶¨¿ÉÒÔ´óÁ¿´æÔÚK£«¡¢Na£«¡¢AlO2£­¡¢SO42£­
D£®³£ÎÂʱ£¬0£®l mol/L HAÈÜÒºµÄpH>l£¬0£®1 mol/L BOHÈÜÒºÖÐc(OH£­)/c(H£«)=l012£¬½«ÕâÁ½ÖÖÈÜÒºµÈÌå»ý»ìºÏ£¬»ìºÏºóÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС¹ØϵΪ£ºc£¨B£«£©>c£¨OH£­£©>c£¨H£«£©>c£¨A£­£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

1 Lij»ìºÏÈÜÒº£¬¿ÉÄܺ¬ÓеÄÀë×ÓÈçÏÂ±í¡£
¿ÉÄÜ´óÁ¿º¬ÓеÄÑôÀë×Ó
H£«¡¢K£«¡¢Mg2£«¡¢Al3£«¡¢NH4+¡¢Fe2£«¡¢Fe3£«
¿ÉÄÜ´óÁ¿º¬ÓеÄÒõÀë×Ó
Cl£­¡¢Br£­¡¢I£­¡¢CO32¡ª¡¢AlO2¡ª
 
(1)Íù¸ÃÈÜÒºÖÐÖðµÎ¼ÓÈëNaOHÈÜÒº£¬²úÉú³ÁµíµÄÎïÖʵÄÁ¿(n)Óë¼ÓÈëNaOHÈÜÒºµÄÌå»ý(V)µÄ¹ØϵÈçͼËùʾ¡£Ôò¸ÃÈÜÒºÖÐÒ»¶¨²»º¬ÓеÄÀë×ÓÊÇ_________________________¡£

(2)BC¶ÎÀë×Ó·½³ÌʽΪ_______________________________________________¡£
(3)V1¡¢V2¡¢V3¡¢V4Ö®¼äµÄ¹ØϵΪ__________________________________________¡£
(4)¾­¼ì²â£¬¸ÃÈÜÒºÖл¹º¬ÓдóÁ¿µÄCl£­¡¢Br£­¡¢I£­£¬ÈôÏò1 L¸Ã»ìºÏÈÜÒºÖÐͨÈëÒ»¶¨Á¿µÄCl2£¬ÈÜÒºÖÐCl£­¡¢Br£­¡¢I£­µÄÎïÖʵÄÁ¿ÓëͨÈëCl2µÄÌå»ý(±ê×¼×´¿ö)µÄ¹ØϵÈç±íËùʾ£¬·ÖÎöºó»Ø´ðÏÂÁÐÎÊÌâ¡£
Cl2µÄÌå»ý(±ê×¼×´¿ö)
2.8 L
5.6 L
11.2 L
n(Cl£­)
1.25 mol
1.5 mol
2 mol
n(Br£­)
1.5 mol
1.4 mol
0.9 mol
n(I£­)
a mol
0
0
 
¢Ùµ±Í¨ÈëCl2µÄÌå»ýΪ2.8 Lʱ£¬ÈÜÒºÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_________________¡£
¢ÚÔ­ÈÜÒºÖÐCl£­¡¢Br£­¡¢I£­µÄÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ_____________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚÏÂÁи÷ÈÜÒºÖУ¬Àë×ÓÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ
A£®Ç¿¼îÐÔÈÜÒºÖУºK£«¡¢Al3£«¡¢Cl£­¡¢SO42£­
B£®º¬ÓÐ0.1 mol¡¤L-1 Fe3£«µÄÈÜÒºÖУºK£«¡¢Mg2£«¡¢I£­¡¢NO3£­
C£®º¬ÓÐ0.1 mol¡¤L-1 Ca2£«ÈÜÒºÔÚÖУºNa£«¡¢K£«¡¢CO32£­¡¢Cl£­
D£®ÊÒÎÂÏ£¬pH=1µÄÈÜÒºÖУºNa£«¡¢Fe3£«¡¢NO3£­¡¢SO42£­

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(¡¡ ¡¡)
A£®100 ¡æʱ,½«pH=2µÄÑÎËáÓëpH=12µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ,ÈÜÒºÏÔÖÐÐÔ
B£®ÊÒÎÂÏÂ,ÏòpH=3µÄ´×ËáÈÜÒº¼ÓˮϡÊͺó,ÈÜÒºÖÐËùÓÐÀë×ÓŨ¶È¾ù
¼õС
C£®Ïòº¬ÓÐBaSO4³ÁµíµÄÈÜÒºÖмÓÈëNa2SO4¹ÌÌå,c(Ba2+)Ôö´ó
D£®ÏòCH3COONaÈÜÒºÖмÓÈëÊÊÁ¿CH3COOH,¿Éʹc(Na+)=c(CH3COO-)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁи÷×éÀë×ÓÖУ¬ÔÚ¸ø¶¨Ìõ¼þÏÂÄÜ´óÁ¿¹²´æµÄÊÇ£¨¡¡¡¡£©¡£
A£®ÔÚpH£½1µÄÈÜÒºÖУ¬NH4+£¬K£«£¬ClO£­£¬Cl£­
B£®ÓÐSO42-´æÔÚµÄÈÜÒºÖУºNa£«£¬Mg2£«£¬Ba2£«£¬I£­
C£®ÓÐNO3-´æÔÚµÄÇ¿ËáÐÔÈÜÒºÖУºNH4+£¬Ba2£«£¬Fe2£«£¬Br£­
D£®ÔÚc£¨H£«£©£½1.0¡Á10£­13 mol¡¤L£­1µÄÈÜÒºÖУºNa£«£¬S2£­£¬AlO2-£¬SO32-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖ³£¼û»¯ºÏÎ¶¼ÊÇÓÉϱíÖеÄÀë×ÓÐγɵģº
ÑôÀë×Ó
 K+¡¢Na+¡¢Al3+¡¢Cu2+
ÒõÀë×Ó
OH£­¡¢HCO3£­¡¢NO3£­¡¢SO42£­
 
ΪÁ˼ø±ðÉÏÊö»¯ºÏÎï¡£·Ö±ðÍê³ÉÒÔÏÂʵÑ飬Æä½á¹ûÊÇ£º
¢Ù½«ËüÃÇÈÜÓÚË®ºó£¬DΪÀ¶É«ÈÜÒº£¬ÆäËû¾ùΪÎÞÉ«ÈÜÒº£»
¢Ú½«EÈÜÒºµÎÈëµ½CÈÜÒºÖгöÏÖ°×É«³Áµí£¬¼ÌÐøµÎ¼Ó£¬³ÁµíÍêÈ«Èܽ⣻
¢Û½øÐÐÑæÉ«·´Ó¦£¬B¡¢CΪ×ÏÉ«(͸¹ýÀ¶É«îܲ£Á§)£¬A¡¢EΪ»ÆÉ«£»
¢ÜÔÚ¸÷ÈÜÒºÖмÓÈëÏõËá±µÈÜÒº£¬ÔÙ¼Ó¹ýÁ¿Ï¡ÏõËᣬAÖзųöÎÞÉ«ÆøÌ壬C¡¢DÖвúÉú°×É«³Áµí£¬BÖÐÎÞÃ÷ÏÔÏÖÏó¡£
¢Ý½«B¡¢DÁ½ÈÜÒº»ìºÏ£¬Î´¼û³Áµí»òÆøÌåÉú³É£®
¸ù¾ÝÉÏÊöʵÑéÌî¿Õ£º
£¨1£©Ð´³öB¡¢CµÄ»¯Ñ§Ê½£ºB               £»  C             ¡£
£¨2£©Ð´³ö¹ýÁ¿EµÎÈ˵½CÈÜÒºÖеÄÀë×Ó·´Ó¦·½³Ìʽ                        ¡£
£¨3£©½«º¬1 mol AµÄÈÜÒºÓ뺬1 mol EµÄÈÜÒº·´Ó¦ºóÕô¸É£¬½öµÃµ½Ò»ÖÖ»¯ºÏÎ¸Ã»¯ºÏÎïµÄ»¯Ñ§Ê½Îª              ¡£
£¨4£©ÔÚAÈÜÒºÖмӳÎÇåʯ»ÒË®£¬ÆäÀë×Ó·½³ÌʽΪ                      ¡£
£¨5£©ÉÏÊöÎåÖÖ»¯ºÏÎïÖÐÓÐÒ»ÖÖÊdz£ÓÃ×÷¾»Ë®¼Á£¬Æ侻ˮԭÀíÊÇ£º                           £¨Çë½áºÏ·½³Ìʽ¼°±ØÒªµÄÎÄ×Ö½øÐÐÐðÊö£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ
A£®ÂÈ»¯ÌúÈÜÒºÖмÓÈëÍ­·Û£º2Fe3+£«Cu=== 2Fe2+£«Cu2+
B£®ÄÆͶÈëË®ÖУºNa£«H2O===Na+£«2OH-£«H2¡ü
C£®ÂÈÆøºÍÀäµÄÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£º2Cl2£«2OH¡¥ £½ 3Cl¡¥£«ClO¡¥+H2O
D£®Ê¯»Òʯ¼ÓÈëÑÎËáÈÜÒºÖУº2H+ + CO32¨C£½ CO2¡ü+H2O

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸