(5·Ö) ³£ÎÂÏ£¬ÓÐŨ¶È¾ùΪ0.5mol/LµÄËÄÖÖÈÜÒº£º¢ÙNa2CO3¢ÚNaHCO3¢ÛHCl ¢ÜNH3?H2O
£¨1£©ÉÏÊöÈÜÒºÖУ¬¿É·¢ÉúË®½âµÄÊÇ         £¨ÌîÐòºÅ£©¡£
£¨2£©ÏòÈÜÒº¢ÜÖмÓÈëÉÙÁ¿ÂÈ»¯ï§¹ÌÌ壬´Ëʱc(NH4+)/c(OH£­)µÄÖµ        £¨ÌîÔö´ó¡¢¼õС»ò²»±ä£©¡£
£¨3£©Èô½«¢ÛºÍ¢ÜµÄÈÜÒº»ìºÏºóÈÜҺǡºÃ³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°¢ÛµÄÌå»ý     ¢ÜµÄÌå»ý£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©£¬´ËʱÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ                   ¡£
£¨4£©È¡10mLÈÜÒº¢Û£¬¼ÓˮϡÊ͵½500mL,Ôò¸ÃÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)=        ¡£


£¨1£©¢Ù¢Ú  (©ѡ¡¢´íÑ¡²»µÃ·Ö)
£¨2£©Ôö´ó
£¨3£©£¼      c(NH4+)=c(Cl-)£¾c(H+)=c(OH-)
£¨4£©  10-12 mol/L

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ͼÖРA¡¢B¡¢C¡¢D¡¢E¾ùΪÓлú»¯ºÏÎÒÑÖª£ºCÄܸúNaHCO3·¢Éú·´Ó¦£¬CºÍDµÄÏà¶Ô·Ö×ÓÖÊÁ¿ÏàµÈ£¬ÇÒEΪÎÞÖ§Á´µÄ»¯ºÏÎ

¸ù¾Ýͼ»Ø´ðÎÊÌ⣺
£¨1£©C·Ö×ÓÖеĹÙÄÜÍÅÃû³ÆÊÇ£º
ôÈ»ù
ôÈ»ù
£»
ÏÂÁз´Ó¦ÖУ¬»¯ºÏÎïB²»ÄÜ·¢ÉúµÄ·´Ó¦ÊÇ
e
e
£¨Ìî×ÖĸÐòºÅ£©£º
a¡¢¼Ó³É·´Ó¦  b¡¢È¡´ú·´Ó¦  c¡¢ÏûÈ¥·´Ó¦    d¡¢õ¥»¯·´Ó¦  e¡¢Ë®½â·´Ó¦  f¡¢Öû»·´Ó¦
£¨2£©·´Ó¦¢ÚµÄ»¯Ñ§·½³ÌʽÊÇ
CH3COOH+CH3CH2CH2OH
ŨÁòËá
¼ÓÈÈ
CH3COOCH2CH2CH3+H2O
CH3COOH+CH3CH2CH2OH
ŨÁòËá
¼ÓÈÈ
CH3COOCH2CH2CH3+H2O
£®
£¨3£©AµÄ½á¹¹¼òʽÊÇ
£®
£¨4£©Í¬Ê±·ûºÏÏÂÁÐÈý¸öÌõ¼þµÄBµÄͬ·ÖÒì¹¹ÌåµÄÊýÄ¿ÓÐ
4
4
¸ö£®
¢ñ£®º¬Óмä¶þÈ¡´ú±½»·½á¹¹£»¢ò£®ÊôÓÚ·Ç·¼ÏãËáõ¥£»¢ó£®Óë FeCl3 ÈÜÒº·¢ÉúÏÔÉ«·´Ó¦£®
д³öÆäÖÐÈÎÒâÒ»¸öͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ
д³öËÄÕßÖ®Ò»¼´¿É
д³öËÄÕßÖ®Ò»¼´¿É

£¨5£©³£ÎÂÏ£¬½«CÈÜÒººÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºpHÈçÏÂ±í£º
ʵÑé±àºÅ CÎïÖʵÄÁ¿Å¨¶È£¨mol?L-1£© NaOHÎïÖʵÄÁ¿Å¨¶È£¨mol?L-1£© »ìºÏÈÜÒºµÄpH
m 0.1 0.1 pH=9
n 0.2 0.1 pH£¼7
´Óm×éÇé¿ö·ÖÎö£¬ËùµÃ»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨OH-£©=
10-5
10-5
mol?L-1£®
n×é»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©
c£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÏÂͼΪijЩ³£¼ûÎïÖÊÖ®¼äµÄת»¯¹Øϵ£®ÒÑÖª£ºA¡¢B¡¢IÖк¬ÓÐÏàͬµÄÑôÀë×ÓÇÒ¶¼ÊÇXY2ÐÍ»¯ºÏÎÇÒIÊÇʵÑéÊÒ³£ÓõĸÉÔï¼Á£»CΪֱÏßÐÍ·Ö×Ó£»E¡¢FΪ·Ç½ðÊôÆøÌåµ¥ÖÊ£®
Çë°´ÒªÇóÌî¿Õ£º

£¨1£©¢ÙBµÄµç×ÓʽÊÇ
£¬¢ÚKµÄ½á¹¹Ê½ÊÇ
H-O-Cl
H-O-Cl
£»
£¨2£©DÓëG·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2NH3¡ü+2H2O
2NH4Cl+Ca£¨OH£©2
  ¡÷  
.
 
CaCl2+2NH3¡ü+2H2O
£»
£¨3£©ÒÑÖªCµÄȼÉÕÈÈÊÇ1300kJ/mol£¬±íʾCµÄȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽÊÇ
C2H2£¨g£©+
5
2
O2£¨g£©=2CO2£¨g£©+H2O£¨l£©¡÷H=-1300kJ/mol
C2H2£¨g£©+
5
2
O2£¨g£©=2CO2£¨g£©+H2O£¨l£©¡÷H=-1300kJ/mol
£»
£¨4£©½«GÈÜÓÚË®Åä³ÉÈÜÒº£¬¼òÊö¼ìÑé¸ÃÈÜÒºGÖÐËùº¬ÑôÀë×ӵIJÙ×÷·½·¨£º
È¡ÉÙÁ¿µÄGÈÜÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬¼ÓÈÈÊԹܣ¬½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿڣ¬ÈôÊÔÖ½±ä³ÉÀ¶É«£¬Ôòº¬ÓÐNH4+£¬£¨ÆäËûºÏÀí´ð°¸¾ù¸ø·Ö£©
È¡ÉÙÁ¿µÄGÈÜÒºÓÚÊÔ¹ÜÖУ¬¼ÓÈëÊÊÁ¿µÄÇâÑõ»¯ÄÆÈÜÒº£¬¼ÓÈÈÊԹܣ¬½«ÊªÈóµÄºìɫʯÈïÊÔÖ½·ÅÔÚÊԹܿڣ¬ÈôÊÔÖ½±ä³ÉÀ¶É«£¬Ôòº¬ÓÐNH4+£¬£¨ÆäËûºÏÀí´ð°¸¾ù¸ø·Ö£©
£»
£¨5£©³£ÎÂÏÂ0.1mol/LµÄJÈÜÒºÖÐc£¨H+£©/c£¨OH-£©=1¡Á10-8£¬ÏÂÁÐÐðÊö´íÎóµÄÊÇ
BCE
BCE
£»
A£®¸ÃÈÜÒºµÄpH=11£»
B£®¸ÃÈÜÒºÖеÄÈÜÖʵçÀë³öµÄÑôÀë×ÓŨ¶È0.1mol/L
C£®¸ÃÈÜÒºÖÐË®µçÀë³öµÄc£¨H+£©Óëc£¨OH-£©³Ë»ýΪ1¡Á10-22
D£®pH=3µÄÑÎËáÈÜÒºV1 LÓë¸Ã0.1mol/LµÄJÈÜÒºV2 L»ìºÏ£¬Èô»ìºÏÈÜÒºpH=7£¬Ôò£ºV1£¾V2
E£®½«ÒÔÉÏÈÜÒº¼ÓˮϡÊÍ100±¶ºó£¬pHֵΪ9£»
£¨6£©µ¥ÖÊFÔÚ¹¤ÒµÉÏÓÐÖØÒªµÄÓÃ;ÊÇ
ÖÆƯ°×·Û£¬ÖÆÑÎËᣨÖÆƯ°×ÒºµÈ£©
ÖÆƯ°×·Û£¬ÖÆÑÎËᣨÖÆƯ°×ÒºµÈ£©
£®£¨ÖÁÉÙÁ½ÖÖ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(5·Ö) ³£ÎÂÏ£¬ÓÐŨ¶È¾ùΪ0.5mol/LµÄËÄÖÖÈÜÒº£º¢ÙNa2CO3¢ÚNaHCO3¢ÛHCl ¢ÜNH3・H2O

£¨1£©ÉÏÊöÈÜÒºÖУ¬¿É·¢ÉúË®½âµÄÊÇ          £¨ÌîÐòºÅ£©¡£

£¨2£©ÏòÈÜÒº¢ÜÖмÓÈëÉÙÁ¿ÂÈ»¯ï§¹ÌÌ壬´Ëʱc(NH4+)/c(OH£­)µÄÖµ        £¨ÌîÔö´ó¡¢¼õС»ò²»±ä£©¡£

£¨3£©Èô½«¢ÛºÍ¢ÜµÄÈÜÒº»ìºÏºóÈÜҺǡºÃ³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°¢ÛµÄÌå»ý      ¢ÜµÄÌå»ý£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©£¬´ËʱÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ                   ¡£

£¨4£©È¡10mLÈÜÒº¢Û£¬¼ÓˮϡÊ͵½500mL,Ôò¸ÃÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)=        ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê¹ã¶«Ê¡¶«Ý¸Êи߶þµÚ¶þѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§A¾í ÌâÐÍ£ºÌî¿ÕÌâ

(5·Ö) ³£ÎÂÏ£¬ÓÐŨ¶È¾ùΪ0.5mol/LµÄËÄÖÖÈÜÒº£º¢ÙNa2CO3 ¢ÚNaHCO3 ¢ÛHCl ¢ÜNH3・H2O

£¨1£©ÉÏÊöÈÜÒºÖУ¬¿É·¢ÉúË®½âµÄÊÇ          £¨ÌîÐòºÅ£©¡£

£¨2£©ÏòÈÜÒº¢ÜÖмÓÈëÉÙÁ¿ÂÈ»¯ï§¹ÌÌ壬´Ëʱc(NH4+)/c(OH£­)µÄÖµ         £¨ÌîÔö´ó¡¢¼õС»ò²»±ä£©¡£

£¨3£©Èô½«¢ÛºÍ¢ÜµÄÈÜÒº»ìºÏºóÈÜҺǡºÃ³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°¢ÛµÄÌå»ý      ¢ÜµÄÌå»ý£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©£¬´ËʱÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ                    ¡£

£¨4£©È¡10mLÈÜÒº¢Û£¬¼ÓˮϡÊ͵½500mL,Ôò¸ÃÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)=         ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸