ÊÒÎÂÏ£¬ÔÚ25 mL 0.1 mol£¯LNaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol£¯L CH3COOHÈÜÒº£¬ÇúÏßÈçÓÒͼËùʾ£¬ÈôºöÂÔÁ½ÈÜÒº»ìºÏʱµÄÌå»ý±ä»¯£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½Ï´íÎóµÄÊÇ

A£®ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÒ»¶¨¶¼ÓУºc(Na+)+c(H+)= c(CH3COO£­)+c(OH£­)
B£®ÔÚBµã£ºa>12.5£¬ÇÒÓÐc(Na+)=c(CH3COO£­)=c(OH£­)=c(H+)
C£®ÔÚCµã£ºc(Na+)>c(CH3COO£­) >c(H+)>c(OH£­)
D£®ÔÚDµã£ºc(CH3COO£­)+c(CH3COOH)£½0.1mol/L

BC

½âÎö¿¼µã£ºËá¼î»ìºÏʱµÄ¶¨ÐÔÅжϼ°ÓйØphµÄ¼ÆË㣻Àë×ÓŨ¶È´óСµÄ±È½Ï£®
·ÖÎö£ºÔÚ25mL 0.1mol£®L-1NaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2mol?L-1 CH3COOH ÈÜÒº£¬¶þÕßÖ®¼äÏ໥·´Ó¦£¬µ±Ç¡ºÃÍêÈ«·´Ó¦Ê±£¬ËùÐè´×ËáµÄÌå»ýΪ12.5mL£¬µ±·´Ó¦ÖÁÈÜÒºÏÔÖÐÐÔʱ£¬´×ËáÓ¦ÉÔ¹ýÁ¿£¬ÇÒc£¨OH-£©=c£¨H+£©£¬×¢Òâ¸ù¾ÝµçºÉÊغã˼ÏëÀ´±È½ÏÀë×ÓŨ¶È´óС£®
½â£ºA¡¢ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÖ»´æÔÚËÄÖÖÀë×ÓÓÐNa+¡¢H+¡¢CH3COO-¡¢OH-£¬¸ù¾ÝµçºÉÊغãÔòÓУº
c£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬¹ÊAÕýÈ·£»
B¡¢ÔÚBµãÈÜÒºÏÔÖÐÐÔ£¬Ôò½á¹ûÊÇc£¨OH-£©=c£¨H+£©£¬¸ù¾ÝµçºÉÊغãc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬ÔòÒ»¶¨ÓÐ
c£¨Na+£©=c£¨CH3COO-£©£¬ÈÜÒºµÄ³É·ÖΪ£º·´Ó¦Éú³ÉµÄ´×ËáÄƺÍÊ£ÓàµÄ´×Ëᣬ´×ËáÄƵÄË®½â³Ì¶ÈºÍ´×ËáµÄµçÀë³Ì¶ÈÏàµÈ£¬¹ÊÓУºc£¨Na+£©=c£¨CH3COO-£©£¾c£¨OH-£©=c£¨H+£©£¬¹ÊB´íÎó£»
C¡¢ÔÚCµã£¬ÈÜÒºÏÔËáÐÔ£¬¹ÊÓÐc£¨OH-£©£¼c£¨H+£©£¬¸ù¾ÝµçºÉÊغ㣺c£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬¹Êc£¨Na+£©£¼c£¨CH3COO-£©£¬¹ÊC´íÎó£»
D¡¢ÔÚDµãʱ£¬´×ËáÊ£Ó࣬ʣÓàµÄ´×ËáµÄŨ¶ÈºÍÉú³ÉµÄ´×ËáÄÆŨ¶ÈÏàµÈ¾ùΪ0.05mol/l£¬¸ù¾ÝÎïÁÏÊغ㣬Ôò£º
c£¨CH3COO-£©+c£¨CH3COOH£©=0.1mol?L-1£¬¹ÊDÕýÈ·£»
¹ÊÑ¡BC£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º¹ã¶«Ê¡ÉÇÍ·ÊнðɽÖÐѧ2011£­2012ѧÄê¸ß¶þÉÏѧÆÚ12ÔÂÔ¿¼»¯Ñ§ÊÔÌâ ÌâÐÍ£º021

ÊÒÎÂÏ£¬ÔÚ25 mL¡¡0.1 mol/L¡¡NaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol/L¡¡CH3COOHÈÜÒº£¬ÇúÏßÈçÏÂͼËùʾ£¬ÈôºöÂÔÁ½ÈÜÒº»ìºÏʱµÄÌå»ý±ä»¯£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½Ï´íÎóµÄÊÇ

A£®

ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÒ»¶¨¶¼ÓУºc(Na+)£«c(H+)£½c(CH3COO£­)£«c(OH£­)

B£®

ÔÚBµã£ºa£¾12.5£¬ÇÒÓÐc(Na+)£½c(CH3COO£­)£½c(OH£­)£½c(H+)

C£®

ÔÚCµã£ºc(Na+)£¾c(CH3COO£­)£¾c(H+)£¾c(OH£­)

D£®

ÔÚDµã£ºc(CH3COO£­)£«c(CH3COOH)£½0.1 mol/L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ì¹ã¶«Ê¡ÉÇÍ·Êи߶þ12ÔÂÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÑ¡ÔñÌâ

ÊÒÎÂÏ£¬ÔÚ25 mL 0.1 mol£¯LNaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol£¯L CH3COOHÈÜÒº£¬ÇúÏßÈçÓÒͼËùʾ£¬ÈôºöÂÔÁ½ÈÜÒº»ìºÏʱµÄÌå»ý±ä»¯£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½Ï´íÎóµÄÊÇ

A£®ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÒ»¶¨¶¼ÓУºc(Na+)+c(H+)= c(CH3COO£­)+c(OH£­)

B£®ÔÚBµã£ºa>12.5£¬ÇÒÓÐc(Na+)=c(CH3COO£­)=c(OH£­)=c(H+)

C£®ÔÚCµã£ºc(Na+)>c(CH3COO£­) >c(H+)>c(OH£­)

D£®ÔÚDµã£ºc(CH3COO£­)+c(CH3COOH)£½0.1mol/L

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ

ÊÒÎÂÏ£¬ÔÚ25 mL 0.1 mol/LNaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol/L CH3COOHÈÜÒº£¬ÇúÏßÈçÓÒͼËùʾ£¬ÈôºöÂÔÁ½ÈÜÒº»ìºÏʱµÄÌå»ý±ä»¯£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½Ï´íÎóµÄÊÇ


  1. A.
    ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÒ»¶¨¶¼ÓУºc(Na+)+c(H+)= c(CH3COO-)+c(OH-)
  2. B.
    ÔÚBµã£ºa>12.5£¬ÇÒÓÐc(Na+)=c(CH3COO-)=c(OH-)=c(H+)
  3. C.
    ÔÚCµã£ºc(Na+)>c(CH3COO-) >c(H+)>c(OH-)
  4. D.
    ÔÚDµã£ºc(CH3COO-)+c(CH3COOH)£½0.1mol/L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(08»ÝÖݵ÷ÑÐ)ÊÒÎÂÏ£¬ÔÚ25 mL 0.1 mol£¯LNaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol£¯L CH3COOHÈÜÒº£¬

ÇúÏßÈçÏÂͼËùʾ£¬ÈôºöÂÔÁ½ÈÜÒº»ìºÏʱµÄÌå»ý±ä»¯£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½ÏÕýÈ·µÄÊÇ£¨£©

 

A.ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÒ»¶¨¶¼ÓÐ c(Na+)+ c(H+)= c(CH3COO£­)+c(OH£­)

B.ÔÚBµã£ºa=12.5£¬

C.ÔÚCµã£ºc(CH3COO£­) > c(Na+) >c(H+)>c(OH£­)

D.ÔÚDµã£ºc(CH3COO£­) =c(CH3COOH)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸