ÎÞɫ͸Ã÷ÈÜÒºÖпÉÄܺ¬ÓÐÏÂÁÐÀë×Ó:K+¡¢Al3+¡¢Fe3+¡¢Ba2+¡¢NO3¡ª¡¢SO42¡ª¡¢HCO32¡ª¡¢Cl-,È¡¸ÃÈÜÒº½øÐÐÈçÏÂʵÑé:
¢ÙÓÃÀ¶É«Ê¯ÈïÊÔÖ½¼ì²â¸ÃÈÜÒº,ÊÔÖ½ÏÔºìÉ«;
¢ÚÈ¡Ô­ÈÜÒºÉÙÐí,¼ÓÈëͭƬºÍÏ¡ÁòËá¹²ÈÈ,²úÉúÎÞÉ«ÆøÌå,¸ÃÆøÌåÓö¿ÕÆøÁ¢¼´±äΪºì×ØÉ«;
¢ÛÈ¡Ô­ÈÜÒºÉÙÐí,¼ÓÈ백ˮÓа×É«³ÁµíÉú³É,¼ÌÐø¼ÓÈë¹ýÁ¿°±Ë®,³Áµí²»Ïûʧ;
¢ÜÈ¡Ô­ÈÜÒºÉÙÐí,µÎÈëÂÈ»¯±µÈÜÒº²úÉú°×É«³Áµí;
¢ÝȡʵÑé¢ÜºóµÄ³ÎÇåÈÜÒº,µÎÈëÏõËáÒøÈÜÒº²úÉú°×É«³Áµí,ÔÙ¼ÓÈë¹ýÁ¿µÄÏ¡ÏõËá,³Áµí²»Ïûʧ¡£
Çë»Ø´ðÏÂÁÐÎÊÌâ:
(1)¸ù¾ÝÉÏÊöʵÑéÅжÏÔ­ÈÜÒºÖÐÉÏÊöÀë×ӿ϶¨´æÔÚµÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡,¿Ï¶¨²»´æÔÚµÄÊÇ¡¡        ¡£ 
(2)д³öÓë¢Ú¢ÛÁ½¸öʵÑéÓйصÄÀë×Ó·½³Ìʽ:
¢Ú                        , ¢Û¡¡                         ¡£ 
(1)Al3+¡¢NO3¡ª¡¢SO42¡ª¡¡Fe3+¡¢Ba2+¡¢HCO3¡ª
(2)¢Ú3Cu+8H++2NO3¡ª3Cu2++2NO¡ü+4H2O
¢ÛAl3++3NH3¡¤H2OAl(OH)3¡ý+3NH4+
ÓÉÓÚÈÜÒºÎÞɫ͸Ã÷ÅųýÁËFe3+µÄ´æÔÚ;ÓÉ¢Ù¿ÉÖªÈÜÒºÏÔËáÐÔ,ÅųýÁËHCO3¡ªµÄ´æÔÚ;ÓÉ¢Ú¿ÉÖªNO3¡ªÒ»¶¨´æÔÚ;ÓÉ¢Û¿ÉÖªAl3+´æÔÚ;ÓɢܿÉÖªS´æÔÚ,ÄÇôÅųýÁËBa2+µÄ´æÔÚ;ÓÉÓÚ¢ÜÒýÈëÁËCl-,ËùÒÔ¢ÝÎÞ·¨Ö¤Ã÷Ô­ÈÜÒºÖÐÊÇ·ñº¬Cl-;Òò´ËÒ»¶¨´æÔÚµÄÀë×ÓÊÇAl3+¡¢NO3¡ª¡¢SO42¡ª,Ò»¶¨²»´æÔÚµÄÀë×ÓÊÇFe3+¡¢Ba2+¡¢HCO3¡ª¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ij´ý²âÈÜÒºÖпÉÄܺ¬ÓÐSO42-¡¢SO32-¡¢CO32-¡¢HCO3-¡¢NO3-¡¢Cl£­¡¢Br£­ÖеÄÈô¸ÉÖÖ¼°Ò»ÖÖ³£¼û½ðÊôÑôÀë×Ó(Mn£«)£¬ÏÖ½øÐÐÈçÏÂʵÑé(ÿ´ÎʵÑéËùÓÃÊÔ¼Á¾ùÊÇ×ãÁ¿µÄ£¬¼ø¶¨ÖÐijЩ³É·Ö¿ÉÄÜûÓиø³ö)¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¸ù¾ÝÉÏÊö¿òͼÐÅÏ¢Ìîдϱí(²»ÄÜÈ·¶¨µÄ²»Ìî)¡£
 
¿Ï¶¨´æÔÚµÄÀë×Ó
¿Ï¶¨Ã»ÓеÄÀë×Ó
³ÁµíD
»¯Ñ§Ê½»òÀë×Ó·ûºÅ
 
 
 
 
(2)´ý²âÈÜÒºÖÐÊÇ·ñÓÐSO32-¡¢SO42-             ¡£ÈôÆøÌåDÓö¿ÕÆø±äºìÉ«£¬ÔòÉú³É³ÁµíDʱ¿Ï¶¨·¢ÉúµÄ·´Ó¦µÄÀë×Ó·½³ÌʽΪ             £¬ÐγɳÁµíBʱ·´Ó¦µÄÀë×Ó·½³ÌʽΪ             ¡£
(3)ÈôMn£«Îª³£¼û½ðÊôÑôÀë×ÓÇÒÔ­×ÓÐòÊý²»´óÓÚ20£¬ÔòҪȷ¶¨Ëü¾ßÌåÊǺÎÖÖÀë×ӵķ½·¨ÊÇ                                                     ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚ¸ø¶¨µÄËÄÖÖÈÜÒºÖУ¬º¬ÓÐÒÔϸ÷ÖÖ΢Á££¬Ò»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ  (   )
A£®ÓÉË®µçÀëµÄc(OH£­)=1¡Á10£­12mol/LµÄÈÜÒºÖУºBa2£«¡¢K£«¡¢Br£­¡¢SiO32£­
B£®³£ÎÂÏÂÆÏÌÑÌÇÈÜÒºÖУºSCN¡ª¡¢Cl£­¡¢K£«¡¢NH4£«
C£®äåË®ÖУºNa+¡¢CO32£­¡¢NH4£«¡¢SO42¡ª
D£®PH´óÓÚ7µÄÈÜÒº£º Na+¡¢Ba2+¡¢SO32£­¡¢ClO¡ª

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÄÜÕýÈ·±íʾÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£º£¨  £©
A£®ÂÈÆøͨÈëË®ÖУ¬ÈÜÒº³ÊËáÐÔ£ºCl2+H2O 2H++Cl£­+ClO£­
B£®½ðÊôÂÁÈÜÓÚÇâÑõ»¯ÄÆÈÜÒº£ºAl+2OH-£½AlO2-+H2¡ü
C£®ÏòÃ÷·¯ÈÜÒºÖеμÓBa(OH)2ÈÜÒº£¬´ï³ÁµíÎïÖʵÄÁ¿×î´ó£º2Al3++3SO42-+3Ba2++6OH£­=" 2" Al(OH)3¡ý+3BaSO4¡ý
D£®¹ýÁ¿ÌúÈÜÓÚÏ¡ÏõË᣺Fe+4H++NO3£­=Fe3++NO¡ü+2H2O

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ij»¯Ñ§Ñо¿ÐÔѧϰС×é¶Ôµç½âÖÊÈÜÒº×÷ÈçϵĹéÄÉ×ܽᣨ¾ùÔÚ³£ÎÂÏ£©£¬ÆäÖÐÕýÈ·µÄÊÇ
¢ÙpH=lµÄÇ¿ËáÈÜÒº¼ÓˮϡÊͺó£¬ÈÜÒºÖи÷Àë×ÓŨ¶ÈÒ»¶¨¼õС
¢ÚpH=2µÄÑÎËáºÍpH=lµÄÑÎËᣬc£¨H£«£©Ö®±ÈΪ2:1
¢ÛpHÏàµÈµÄËÄÖÖÈÜÒº£ºa£®CH3 COONa¡¢b£®C6H5 ONa¡¢c£®NaHCO3¡¢d£®NaOH£¬ÆäÈÜÒºÎïÖʵÄÁ¿Å¨¶ÈÓÉСµ½´óµÄ˳ÐòΪd<b<c<a
¢ÜNH4HSO4ÈÜÒºÖеμÓNaOHÈÜÒºÖÁÈÜÒºpH=7£¬Ôòc£¨Na£«£©=2c£¨SO42£­£©
¢ÝÒÑÖª´×ËáµçÀëƽºâ³£ÊýΪKa£»´×Ëá¸ùË®½â³£ÊýΪKh£»Ë®µÄÀë×Ó»ýΪKw£»ÔòÈýÕß¹ØϵΪKa¡¤Kh=Kw
¢Þ¼×¡¢ÒÒÁ½ÈÜÒº¶¼ÊÇÇ¿µç½âÖÊÈÜÒº£¬ÒÑÖª¼×ÈÜÒºµÄpHÊÇÒÒÈÜÒºpHµÄÁ½±¶£¬Ôò¼×¡¢ÒÒÁ½ÈÜÒºµÈÌå»ý»ìºÏ£¬»ìºÏÒºpH¿ÉÄܵÈÓÚ7
A£®¢Û¢Ý¢ÞB£®¢Û¢Ü¢ÞC£®¢Ü¢Ý¢ÞD£®¢Ù¢Ú¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

´óÆøÖÐÁò¡¢µªµÄÑõ»¯ÎïÊÇÐγÉËáÓêµÄÖ÷ÒªÎïÖÊ¡£Ä³µØËáÓêÖпÉÄܺ¬ÓÐÏÂÁÐÀë×Ó£ºNa£«¡¢Ba2£«¡¢NH4+¡¢Al3£«¡¢Cl£­¡¢SO32-¡¢SO42-¡¢NO3-µÈ¡£Ä³Ñо¿Ð¡×éÈ¡¸ÃµØÒ»¶¨Á¿µÄËáÓ꣬ŨËõºó½«ËùµÃ³ÎÇåÊÔÒº·Ö³ÉÈý·Ý£¬½øÐÐÈçÏÂʵÑ飺
ÊÔÑù
Ëù¼ÓÊÔ¼Á
ʵÑéÏÖÏó
µÚÒ»·ÝÊÔÒº
µÎ¼ÓÊÊÁ¿µÄµí·Û­KIÈÜÒº
ÈÜÒº³ÊÀ¶É«
µÚ¶þ·ÝÊÔÒº
µÎ¼ÓÓÃÑÎËáËữµÄBaCl2ÈÜÒº
Óа×É«³Áµí²úÉú
µÚÈý·ÝÊÔÒº
µÎ¼ÓNaOHÈÜÒº£¬¼ÓÈÈ£¬¼ÓÈëµÄNaOHÈÜÒºÌå»ý(V)ÓëÉú³ÉµÄ³Áµí¡¢²úÉúµÄÆøÌåµÄÎïÖʵÄÁ¿(n)µÄ¹ØϵÈçÓÒͼ

 
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¸ù¾ÝʵÑé½á¹ûÅжϸÃËáÓêÖп϶¨²»´æÔÚµÄÀë×ÓÊÇ______________£¬²»ÄÜÈ·¶¨µÄÀë×ÓÓÐ________________¡£
(2)д³öµÚÒ»·ÝÊÔÒºµÎ¼Óµí·Û­KIÈÜҺʱ·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ£º__________________¡£
(3)µÚÈý·ÝÊÔÒºµÎ¼ÓNaOHÈÜÒº£¬¼ÓÈÈ£¬Õû¸ö¹ý³ÌÖз¢ÉúÁ˶à¸ö·´Ó¦£¬Ð´³öÆäÖÐÁ½¸ö·´Ó¦µÄÀë×Ó·½³Ìʽ£º__________________________________¡¢__________________________¡£
(4)Éè¼ÆʵÑé·½°¸£¬¼ìÑé¸ÃËáÓêÖÐÊÇ·ñ´æÔÚCl£­£º___________________________________
______________________________¡£
(5)¸ÃС×éΪÁË̽¾¿NO²ÎÓëÁòËáÐÍËáÓêµÄÐγɹý³Ì£¬ÔÚÉÕÆ¿ÖгäÈ뺬ÓÐÉÙÁ¿NOµÄSO2ÆøÌ壬ÔÙÂýÂýͨÈëO2£¬·¢Éú»¯Ñ§·´Ó¦ºó£¬ÔÙÅçÈ÷ÊÊÁ¿ÕôÁóË®¼´µÃÁòËáÐÍËáÓ꣬ÔòNOÔÚÉÏÊö·´Ó¦ÖеÄ×÷ÓÃÊÇ________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÔÚÔÖºóÖؽ¨ÖУ¬ÒûÓÃË®°²È«Õ¼Óм«ÎªÖØÒªµÄµØλ£¬Ä³Ñо¿Ð¡×éÌáÈ¡Èý´¦±»ÎÛȾµÄˮԴ½øÐзÖÎö£¬²¢¸ø³öÁËÈçÏÂʵÑéÐÅÏ¢£ºÆäÖÐÒ»´¦±»ÎÛȾµÄˮԴº¬ÓÐA¡¢BÁ½ÖÖÎïÖÊ£¬Ò»´¦º¬ÓÐC¡¢DÁ½ÖÖÎïÖÊ£¬Ò»´¦º¬ÓÐEÎïÖÊ£¬A¡¢B¡¢C¡¢D¡¢EÎåÖÖ³£¼û»¯ºÏÎﶼÊÇÓÉϱíÖеÄÀë×ÓÐγɵģº
ÑôÀë×Ó
K£«¡¢Na£«¡¢Cu2£«¡¢Al3£«
ÒõÀë×Ó
SO42¡ª¡¢HCO3¡ª¡¢NO3¡ª¡¢OH£­
 
ΪÁ˼ø±ðÉÏÊö»¯ºÏÎ·Ö±ðÍê³ÉÒÔÏÂʵÑ飬Æä½á¹ûÊÇ£º
¢Ù½«ËüÃÇÈÜÓÚË®ºó£¬DΪÀ¶É«ÈÜÒº£¬ÆäËû¾ùΪÎÞÉ«ÈÜÒº¡£
¢Ú½«EÈÜÒºµÎÈëµ½CÈÜÒºÖУ¬³öÏÖ°×É«³Áµí£¬¼ÌÐøµÎ¼Ó³ÁµíÈܽ⡣
¢Û½øÐÐÑæÉ«·´Ó¦ÊµÑ飬ֻÓÐB¡¢Cº¬ÓмØÀë×Ó¡£
¢ÜÔÚ¸÷ÈÜÒºÖмÓÈëBa(NO3)2ÈÜÒº£¬ÔÙ¼ÓÈë¹ýÁ¿Ï¡ÏõËᣬAÖзųöÎÞÉ«ÆøÌ壬C¡¢DÖвúÉú°×É«³Áµí¡£
¢Ý½«B¡¢DÁ½ÈÜÒº»ìºÏ£¬Î´¼û³Áµí»òÆøÌåÉú³É¡£
¸ù¾ÝÉÏÊöʵÑéÏÖÏóÌîдÏÂÁпհףº
(1)д³öB¡¢C¡¢DµÄ»¯Ñ§Ê½£ºB________¡¢C________¡¢D________¡£
(2)½«º¬1 mol AµÄÈÜÒºÓ뺬1 mol EµÄÈÜÒº·´Ó¦ºóÕô¸É£¬½öµÃµ½Ò»ÖÖ»¯ºÏÎ¸Ã»¯ºÏÎïΪ________¡£
(3)д³öʵÑé¢Ú·¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ________________________________________
(4)C³£ÓÃ×÷¾»Ë®¼Á£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆ侻ˮԭÀí_______________________________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂʱ,ÏÂÁи÷×éÀë×ÓÔÚÖ¸¶¨ÈÜÒºÖÐÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÊÇ(¡¡¡¡)
A£®pH=0µÄÈÜÒºÖÐ:Na+¡¢Fe2+¡¢N¡¢ClO-
B£®c(Fe3+)="0.1" mol/LµÄÈÜÒºÖÐ:K+¡¢Ba2+¡¢S¡¢SCN-
C£®=1012µÄÈÜÒºÖÐ:N¡¢Al3+¡¢S¡¢Cl-
D£®ÓÉË®µçÀëµÄc(H+)=1¡Á10-14 mol/LµÄÈÜÒºÖÐ:Ca2+¡¢K+¡¢Cl-¡¢HS

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬ÏÂÁÐÓйØÈÜÒºµÄpH»ò΢Á£µÄÎïÖʵÄÁ¿Å¨¶È¹ØϵµÄÅжϲ»ÕýÈ·µÄÊÇ(¡¡¡¡)¡£
A£®pH£½3µÄ´×ËáÈÜÒºÓëpH£½11µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºµÄpH<7
B£®pH£½3µÄ¶þÔªÈõËáH2RÈÜÒºÓëpH£½11µÄNaOHÈÜÒº»ìºÏºó£¬»ìºÏÈÜÒºµÄpH£½7£¬Ôò·´Ó¦ºóµÄ»ìºÏÈÜÒºÖУº2c(R2£­)£«c(HR£­)£½c(Na£«)
C£®½«0.2 mol¡¤L£­1µÄijһԪËáHAÈÜÒººÍ0.1 mol¡¤L£­1 NaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬»ìºÏÈÜÒºµÄpH´óÓÚ7£¬Ôò·´Ó¦ºóµÄ»ìºÏÈÜÒºÖУº2c(OH£­)£½2c(H£«)£«c(HA)£­c(A£­)
D£®Ä³ÎïÖʵÄÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H£«)£½1¡Á10£­a mol¡¤L£­1£¬Èôa>7£¬Ôò¸ÃÈÜÒºµÄpHÒ»¶¨Îª14£­a

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸