ÔÚÊÒÎÂÏ£¬ÏÂÁÐÎåÖÖÈÜÒº¢Ù0.1mol/LNH4Cl ¢Ú0.1mol/LCH3COONH¢Û0.1mol/ L NH4HSO4    ¢Ü0.1mol/LNH3¡¤H2OºÍ0.1mol/LNH4ClµÄ»ìºÏÈÜÒº  ¢Ý0.1mol/LNH3¡¤H2O
(1)ÈÜÒº¢Ù³Ê___ÐÔ£¨Ìî¡°Ëᡱ¡°¼î¡±»ò¡°ÖС±£©ÆäÔ­ÒòÊÇ____________________________(ÓÃÀë×Ó·½³Ìʽ±íʾ)
(2)±È½ÏÈÜÒº¢Ú¡¢¢ÛÖÐC(NH4+)µÄ´óС¹ØϵÊÇ____________________(Ìî>,<»ò=)
(3)ÔÚÈÜÒº¢ÜÖУ¬_____________Àë×ÓµÄŨ¶ÈΪ0.1mol/L
NH3¡¤H2OºÍ_________Àë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÖ®ºÍΪ0.2mol/L
(4) ÊÒÎÂϲâµÃÈÜÒº¢ÚµÄPH=7£¬Ôò˵Ã÷CH3COO-µÄË®½â³Ì¶È_______(Ìî>,<»ò=")" NH4+µÄË®½â³Ì¶È,C(CH3COO-)_________C(NH4+)(Ìî>,<»ò=)
1)Ëá , NH4£«£«H2ONH3¡¤H2O£«H£« £¬NH4£«ºÍË®µçÀë³öÀ´µÄ½áºÏ³ÉNH3¡¤H2O£¬´Ù½øË®µÄµçÀ룬ƽºâʱʹC(H+)£¾C(OH-)£¬ËùÒÔÈÜÒº³ÊËáÐÔ¡£(2)¢Ú<¢Û   (3)Cl£­ ,  NH4£«   (4)   ="  " £¬ =
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨6·Ö£©ÒÑÖª£º
£¨1£©Cu2+¡¢Fe2+ÔÚpHΪ4~5µÄÌõ¼þϲ»Ë®½â£¬¶øÕâÒ»Ìõ¼þÏÂFe3+¼¸ºõÈ«²¿Ë®½â¡£
£¨2£©Ë«ÑõË®£¨H2O2£©ÊÇÇ¿Ñõ»¯¼Á£¬ÔÚËáÐÔÌõ¼þÏ£¬ËüµÄ»¹Ô­²úÎïÊÇH2O¡£ÏÖÓôÖÑõ»¯Í­£¨º¬ÉÙÁ¿FeÔÓÖÊ£©ÖÆÈ¡CuCl2ÈÜÒºµÄ¹ý³ÌÈçÏ£º
¢ÙÈ¡50mL´¿¾»µÄÑÎËᣬ¼ÓÈëÒ»¶¨Á¿µÄ´ÖCuO¼ÓÈȽÁ°è¡¢³ä·Ö·´Ó¦ºó¹ýÂË£¬²âÖªÂËÒºµÄpH=3¡£¢ÚÏòÂËÒºÖмÓÈëË«ÑõË®¡¢½Á°è¡£
¢Ûµ÷½Ú¢ÚÖÐÈÜÒºµÄpHÖÁ4£¬¹ýÂË¡£
¢Ü°Ñ¢ÛËùµÃÂËҺŨËõµÃµ½CuCl2ÈÜÒº¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¢ÚÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌÊÇ                         ¡£
£¨2£©¢ÛÖÐʹpHÉý¸ßµ½4£¬²ÉÈ¡µÄ´ëÊ©ÊÇ£º¼ÓÈë¹ýÁ¿µÄ    £¨Ìî×ÖĸÐòºÅ£©²¢Î¢ÈÈ¡¢½Á°è¡£
A£®NaOHB£®°±Ë®C£®CuCl2D£®CuO
£¨3£©¢ÛÖйýÂ˺óÂËÔüµÄ³É·ÖÊÇ                ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

½«0.2 mol/LµÄHXÈÜÒººÍ0.1mol/LµÄNaOHÈÜÒºµÈÌå»ý»ìºÏ£¨ºöÊÓ»ìºÏÇ°ºóÈÜÒºÌå»ýµÄ±ä»¯£©£¬ÏÂÁйØϵʽÖÐÒ»¶¨²»ÕýÈ·µÄÊÇ
A£®c(HX) >c(X¡ª)B£®c(Na+)<c(X¡ª)
C£®c(H+)+c(Na+) = c(OH¡ª)+c(X¡ª)D£®c(HX)+c(X¡ª)="0.2" mol/L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

(6·Ö)³£ÎÂÏ£¬ÓÐ0.1mol/LµÄÑÎËáºÍ0.1mol/LµÄ´×ËáÈÜÒº£¬»Ø´ðÏÂÁÐÎÊÌ⣺
¢Ù±È½ÏÁ½ÈÜÒºµÄpH £¬ÑÎËá__£¨Ìî¡°<¡± ¡¢¡°="¡±" »ò ¡°>¡±£©´×Ëá,д³ö´×ËáµçÀëµÄ·½³Ìʽ__________¡£
¢ÚÁ½ÖÖËá¾ùÄÜÓëÇâÑõ»¯ÄÆ·´Ó¦Éú³ÉÑΣ¬ÆäÖд×ËáÓëÇâÑõ»¯ÄÆ·´Ó¦ÄÜÉú³É´×ËáÄÆ£¬ÊµÑéÊÒÏÖÓв¿·Ö´×ËáÄƹÌÌ壬ȡÉÙÁ¿ÈÜÓÚË®£¬ÈÜÒº³Ê____£¨Ìî¡°ËáÐÔ¡± ¡¢¡°ÖÐÐÔ¡± »ò ¡°¼îÐÔ¡±£©£¬ÆäÔ­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©__________¡£
¢ÛÏò´×ËáÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄNaOHÈÜÒº£¬µ±²âµÃÈÜÒºµÄpH=7ʱ£¬ÈÜÒºÖÐÀë×ÓµÄŨ¶È´óСΪ
     £¨Ìî×Öĸ£¬ÏÂͬ£©£¬µ±²âµÃÈÜÒºµÄpH<7ʱ£¬ÈÜÒºÖÐÀë×ÓµÄŨ¶È´óСΪ     
a£®c(Na+)£¾c(CH3COO¡ª)£¾c(OH¡ª)£¾c(H+)
b£®c(Na+) = c(CH3COO¡ª)£¾c(OH¡ª) =c(H+
c£®c(CH3COO¡ª)£¾c(Na+)£¾c(H+)£¾c(OH¡ª)
d£®c(Na+)£¾c(CH3COO¡ª)£¾c(H+)£¾c(OH¡ª)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁи÷ÏîËù¸øµÄÁ½¸öÁ¿£¬Ç°Õß¡ª¶¨´óÓÚºóÕßµÄÊÇ
¢Ù´¿Ë®ÔÚ25¡æºÍ80¡æµÄpH
¢Ú1moINaHSO4ºÍ1mo1 Na2SO4ÔÚÈÛ»¯×´Ì¬ÏµÄÀë×ÓÊý
¢Û25¡æʱ£¬µÈÌå»ýÇÒpH¶¼µÈÓÚ3µÄÑÎËáºÍAlCl3µÄÈÜÒºÖУ¬ÒѵçÀëµÄË®·Ö×ÓÊý
¢Üº¬1mol FeCl3µÄ±¥ºÍÈÜÒºÓëË®ÍêÈ«·´Ó¦×ª»¯ÎªÇâÑõ»¯Ìú½ºÌåºóÆäÖнºÌåÁ£×ÓµÄÊýÄ¿Óë°¢·ü¼ÓµÂÂÞ³£Êý
A£®¢Ú¢ÜB£®¢Ù¢ÛC£®Ö»ÓТÙD£®¢Ú¢Û

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

(8·Ö)ÒÑÖª£ºAËáµÄÈÜÒºpH£½a£¬B¼îµÄÈÜÒºpH£½b¡£
(1)ÈôAΪÑÎËᣬBΪÇâÑõ»¯±µ£¬ÇÒa£«b£½14£¬Á½ÕßµÈÌå»ý»ìºÏ£¬ÈÜÒºµÄpH£½________¡£Ëá¼î°´Ìå»ý±ÈΪ1 ¡Ã10»ìºÏºóÈÜÒºÏÔÖÐÐÔ£¬Ôòa£«b£½________¡£
(2)ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa£½4£¬b£½12£¬ÄÇôAÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ________mol¡¤L£­1£¬BÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ________mol¡¤L£­1¡£
(3)ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa£«b£½14£¬ÓÃÌå»ýΪVAµÄ´×ËáºÍÌå»ýΪVBµÄÇâÑõ»¯ÄÆÈÜÒº»ìºÏºó£¬ÈÜÒºÏÔÖÐÐÔ£¬ÔòÆäÌå»ý¹ØϵΪVA________VB(Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±£¬ÏÂͬ)£¬»ìºÏºóÈÜÒºÖеÄÀë×ÓŨ¶È¹ØϵΪc(Na£«)________
c(CH3COO£­)¡£
(4)ÈôAµÄ»¯Ñ§Ê½ÎªHR£¬BµÄ»¯Ñ§Ê½ÎªMOH£¬ÇÒa£«b£½14£¬Á½ÕßµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ¡£Ôò»ìºÏÈÜÒºÖбض¨ÓÐÒ»ÖÖÀë×ÓÄÜ·¢ÉúË®½â£¬¸ÃË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ________________________________________________________________________
________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

±íʾ0.1mol/LNaHCO3ÈÜÒºÖÐÓйØ΢Á£Å¨¶È´óСµÄ¹Øϵʽ£¬ÕýÈ·µÄÊÇ£¨   £©
A£®£¨Na+£©£¾c£¨HCO3-£©£¾c£¨CO32-£©£¾c£¨H+£©£¾c£¨OH-£©
B£®£¨Na+£©+c£¨H+£©=c£¨HCO3-£©+2c£¨CO32-£©+c£¨OH-£©
C£®c£¨Na+£©£¾c£¨HCO3-£©£¾c£¨CO32-£©£¾c£¨H2CO3£©£¾c£¨OH-£©
D£®c£¨H+£©£¾c£¨H2CO3£©£¾c£¨OH-£©£¾c£¨HCO3-£©£¾c£¨CO32-£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÈÜÒºÕô¸Éºó£¬Äܵõ½ÈÜÖʹÌÌåµÄÊÇ
A£®AlCl3 B£®KHCO3 C£®Fe2(SO4)3D£®NaClO

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏÂÓÃ0.10 mol /L KOHÈÜÒº£¬µÎ¶¨0.10 mol /L CH3COOHÈÜÒº£¬Óйصζ¨¹ý³ÌÖеÄ˵·¨ÕýÈ·µÄÊÇ
A£®µÎ¶¨ÖÕµãʱ£ºÈÜÒºµÄpH=7
B£®µÎ¶¨ÖÕµãʱ£ºc(K+) = c(CH3COO£­)£¬c(H+) < c(OH£­)
C£®µ±ÈÜÒºpH=7ʱ£ºc(K+) = c(CH3COO£­) + c(CH3COOH)
D£®½Ó½üµÎ¶¨ÖÕµãʱ£ºc(K+) + c(H+) = c(CH3COO£­) + c(OH£­)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸