ÌâÄ¿ÄÚÈÝ

ÈçͼΪÏò25mL 0.1mol?L-1NaOHÈÜÒºÖÐÖðµÎµÎ¼Ó0.2mol?L-1CH3COOHÈÜÒº¹ý³ÌÖÐÈÜÒºpHµÄ±ä»¯ÇúÏߣ®Çë»Ø´ð£º
£¨1£©BµãÈÜÒº³ÊÖÐÐÔ£¬ÓÐÈ˾ݴËÈÏΪ£¬ÔÚBµãʱNaOHÓëCH3COOHÇ¡ºÃÍêÈ«·´Ó¦£¬ÕâÖÖ¿´·¨ÊÇ·ñÕýÈ·£¿£¨Ñ¡Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©
 
£®Èô²»ÕýÈ·£¬Ôò¶þÕßÇ¡ºÃÍêÈ«·´Ó¦µÄµãÊÇÔÚABÇø¼ä»¹ÊÇBDÇø¼äÄÚ£¿
 
£¨ÈôÕýÈ·£¬´ËÎʲ»´ð£©
£¨2£©ABÇø¼ä£¬c£¨OH-£©£¾c£¨H+£©£¬Ôòc£¨OH-£©Óëc£¨CH3COO-£©´óС¹ØÏµÊÇ
 

A£®c£¨OH-£©Ò»¶¨´óÓÚc£¨CH3COO-£©
B£®c£¨OH-£©Ò»¶¨Ð¡ÓÚc£¨CH3COO-£©
C£®c£¨OH-£©Ò»¶¨µÈÓÚc£¨CH3COO-£©
D£®c£¨OH-£©´óÓÚ¡¢Ð¡ÓÚ»òµÈÓÚc£¨CH3COO-£©
£¨3£©ÔÚDµãʱ£¬ÈÜÒºÖÐc£¨CH3COO-£©+c£¨CH3COOH£©
 
2c£¨Na+£©£¨¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©
£¨4£©³£ÎÂÏ£¬½«VmL¡¢0.1000mol?L-1ÇâÑõ»¯ÄÆÈÜÒºÖðµÎ¼ÓÈëµ½20.00mL¡¢0.1000mol?L-1´×ËáÈÜÒºÖУ¬³ä·Ö·´Ó¦£®»Ø´ðÏÂÁÐÎÊÌ⣮£¨ºöÂÔÈÜÒºÌå»ýµÄ±ä»¯£©
¢ÙÈç¹ûÈÜÒºpH=7£¬´ËʱVµÄȡֵ
 
20.00£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£¬¶øÈÜÒºÖÐc£¨Na+£©¡¢c£¨CH3COO-£©¡¢c£¨H+£©¡¢c£¨OH-£©µÄ´óС¹ØÏµÎª
 
£®
¢ÚÈç¹ûV=40.00£¬Ôò´ËʱÈÜÒºÖÐc£¨OH-£©-c£¨H+£©-c£¨CH3COOH£©=
 
mol?L-1£®
¿¼µã£ºËá¼î»ìºÏʱµÄ¶¨ÐÔÅжϼ°ÓйØphµÄ¼ÆËã
רÌ⣺
·ÖÎö£º£¨1£©NaOHÓëCH3COOHÇ¡ºÃÍêÈ«·´Ó¦£¬Éú³É´×ËáÄÆ£¬´×ËáÄÆÎªÇ¿¼îÈõËáÑΣ¬Ë®½âºóÈÜÒºÏÔ¼îÐÔ£¬pH£¾7£»
£¨2£©ÔÚABÇø¼äÄÚ£¬°üÀ¨CH3COOHºÍNaOHÇ¡ºÃÍêÈ«·´Ó¦ÒÔ¼°CH3COOH²»×㣬ÇâÑõ»¯ÄÆÈÜÒº¹ýÁ¿ÈÜÒºÏÔ¼îÐÔÁ½ÖÖ¿ÉÄÜÐÔ£»
£¨3£©ÔÚDµãʱ£¬NaOHºÍCH3COOH·´Ó¦ºóÊ£ÓàCH3COOH£¬ÈÜÒºµÄ×é³ÉΪµÈŨ¶ÈµÄCH3COOHºÍCH3COONaµÄ»ìºÏÎ
£¨4£©¢ÙÈÜÒºµÄËá¼îÐÔÊǸù¾ÝÈÜÒºÖÐH+Ũ¶ÈÓëOH-Ũ¶ÈµÄÏà¶Ô´óСÅжϵģ¬Ö»ÒªÈÜÒºÖÐc£¨H+£©=c£¨OH-£©£¬ÈÜÒº¾Í³ÊÖÐÐÔ£¬CH3COOHÊÇÈõµç½âÖÊ£¬µçÀë³Ì¶È²»´ó£¬NaOHÊÇÇ¿µç½âÖÊ£¬ÍêÈ«µçÀ룬·´Ó¦Éú³ÉµÄÒÒËáÄÆÊÇÇ¿¼îÈõËáÑÎË®½â³Ê¼îÐÔ£¬ÐèÈÜÒº³ÊÖÐÐÔ£¬ÐèÉټӼ¸ù¾ÝÈÜÒº³ÊÖÐÐÔpH=7c£¨H+£©=c£¨OH-£©½áºÏµçºÉÊØºãc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©½øÐнâ´ð£»
¢Ú¸ù¾ÝµçºÉÊØºãc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬ÎïÁÏÊØºãc£¨Na+£©=2[c£¨CH3COO-£©+c£¨CH3COOH£©]½øÐнâ´ð£®
½â´ð£º ½â£º£¨1£©NaOHÓëCH3COOHÇ¡ºÃÍêÈ«·´Ó¦£ºNaOH+CH3COOH=CH3COONa+H20£¬Éú³ÉµÄ´×ËáÄÆÎªÇ¿¼îÈõËáÑΣ¬ÈÜÒºÏÔ¼îÐÔ£¬pH£¾7£¬½éÓÚABÖ®¼ä£¬
¹Ê´ð°¸Îª£º·ñ£»AB£»
£¨2£©ÔÚABÇø¼äÄÚ£¬c£¨OH-£©£¾c£¨H-£©£¬ËµÃ÷ÈÜÒºÏÔ¼îÐÔ£¬µ±NaOHºÍCH3COOHÇ¡ºÃ·´Ó¦Ê±£¬ÏÔ¼îÐÔ£¬´ËʱÉú³ÉµÄÈÜÒº¿ÉÄÜΪ´×ËáÄÆ£¬c£¨OH-£©Ð¡ÓÚc£¨CH3COO-£©£»µ±NaOHºÍCH3COOH·´Ó¦ºóÊ£ÓàNaOH£¬ÈÜÒºÈÔÈ»ÏÔ¼îÐÔ£¬´ËʱÈôÊ£ÓàµÄNaOHÁ¿ºÜ´ó£¬Ôòc£¨OH-£©´óÓÚc£¨CH3COO-£©£¬Ò²ÓпÉÄÜÊ£ÓàµÄNaOHºÍCH3COONaÖÐCH3COO-Ë®½âÖ®ºóÊ£ÓàµÄCH3COO-µÄŨ¶ÈÏàµÈ£¬
¹ÊÑ¡D£»
£¨3£©ÔÚDµãʱ£¬·´Ó¦ºóCH3COOHÊ£Ó࣬ÈÜÒºµÄ×é³ÉΪµÈŨ¶ÈµÄCH3COOHºÍCH3COONaµÄ»ìºÏÎ¸ù¾ÝÎïÁÏÊØºã£¬´Ëʱ£ºc£¨CH3COO-£©+c£¨CH3COOH£©=2c£¨Na+£©£¬
¹Ê´ð°¸Îª£º=£»
£¨4£©¢ÙCH3COOHÊÇÈõµç½âÖÊ£¬µçÀë³Ì¶È²»´ó£¬NaOHÊÇÇ¿µç½âÖÊ£¬ÍêÈ«µçÀ룬·´Ó¦Éú³ÉµÄÒÒËáÄÆÊÇÇ¿¼îÈõËáÑΣ¬Ë®½â³Ê¼îÐÔ£¬ÐèÈÜÒº³ÊÖÐÐÔpH=7£¬ÐèÉټӼËùÒÔ³£ÎÂÏ£¬½«V mL¡¢0.1000mol?L-1ÇâÑõ»¯ÄÆÈÜÒºÖðµÎ¼ÓÈëµ½20.00mL¡¢0.1000mol?L-1´×ËáÈÜÒºÖУ¬³ä·Ö·´Ó¦£¬V£¼20.00mLÈÜÒº³ÊÖÐÐÔpH=7£¬c£¨H+£©=c£¨OH-£©£»¸ù¾ÝµçºÉÊØºãc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬c£¨H+£©=c£¨OH-£©£¬ÈÜÒºÖеÄÈÜÖÊΪÒÒËáÄÆÈÜÒº£¬Ë®µÄµçÀëÊÇ΢ÈõµÄ£¬ËùÒÔc£¨Na+£©=c£¨CH3COO-£©£¾c£¨H+£©=c£¨OH-£©£¬
¹Ê´ð°¸Îª£º£¼£»c£¨Na+£©=c£¨CH3COO-£©£¾c£¨H+£©=c£¨OH-£©£» 
¢Ú¸ù¾ÝµçºÉÊØºãc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬ÎïÁÏÊØºãc£¨Na+£©=2[c£¨CH3COO-£©+c£¨CH3COOH£©]£¬µÃµ½c£¨H+£©+c£¨CH3COO-£©+2c£¨CH3COOH£©=c£¨OH-£©£¬Ôòc£¨OH-£©-c£¨H+£©-c£¨CH3COOH£©=c£¨CH3COO-£©+c£¨CH3COOH£©£¬·´Ó¦ºóÈÜÒºµÄÌå»ý±äΪ60mL£¬Ôòc£¨CH3COO-£©+c£¨CH3COOH£©¨T
0.1000mol/L¡Á20ml
60ml
=
1
30
mol/L£¬»ò0.033mol/L£¬
¹Ê´ð°¸Îª£º
1
30
»ò0.033£®
µãÆÀ£º±¾ÌâÖ÷Òª¿¼²éÁËÖк͵樲Ù×÷ÒÔ¼°ÈÜÒºÖÐPHµÄ¼ÆË㣬»¯Ñ§Æ½ºâµÄÓйØÖªÊ¶£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕËá¼îÖкͷ´Ó¦ºóÈÜÒºpHµÄ¼ÆËã·½·¨¡¢Öк͵樲Ù×÷·½·¨£¬ÊÔÌâÅàÑøÁËѧÉúÁé»îÓ¦ÓÃËùѧ֪ʶµÄÄÜÁ¦£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
̼¡¢µª¼°Æä»¯ºÏÎïÔÚ¹¤Å©ÒµÉú²úÉú»îÖÐÓÐ×ÅÖØÒª×÷Óã®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÃCH4´ß»¯»¹Ô­NOx¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ£®ÀýÈ磺
CH4£¨g£©+4NO2£¨g£©=4NO£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H1=-574kJ?mol-1
CH4£¨g£©+4NO£¨g£©=2N2£¨g£©+CO2£¨g£©+2H2O£¨g£©¡÷H2
Èô2mol CH4»¹Ô­NO2ÖÁN2£¬Õû¸ö¹ý³ÌÖзųöµÄÈÈÁ¿Îª1734kJ£¬Ôò¡÷H2=
 
£»
£¨2£©¾Ý±¨µÀ£¬¿ÆÑ§¼ÒÔÚÒ»¶¨Ìõ¼þÏÂÀûÓÃFe2O3Óë¼×Íé·´Ó¦¿ÉÖÆÈ¡¡°ÄÉÃ×¼¶¡±µÄ½ðÊôÌú£®Æä·´Ó¦ÈçÏ£º
Fe2O3£¨s£©+3CH4£¨g£©?2Fe£¨s£©+3CO£¨g£©+6H2£¨g£©¡÷H£¾0
¢ÙÈô·´Ó¦ÔÚ5LµÄÃܱÕÈÝÆ÷ÖнøÐУ¬1minºó´ïµ½Æ½ºâ£¬²âµÃFe2O3ÔÚ·´Ó¦ÖÐÖÊÁ¿¼õÉÙ3.2g£®Ôò
¸Ã¶Îʱ¼äÄÚCOµÄƽ¾ù·´Ó¦ËÙÂÊΪ
 
£®
¢ÚÈô¸Ã·´Ó¦ÔÚºãκãѹÈÝÆ÷ÖнøÐУ¬ÄܱíÃ÷¸Ã·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ
 
£¨Ñ¡ÌîÐòºÅ£©
a£®CH4µÄת»¯ÂʵÈÓÚCOµÄ²úÂÊ
b£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä
c£®v£¨CO£©Óëv£¨H2£©µÄ±ÈÖµ²»±ä
d£®¹ÌÌåµÄ×ÜÖÊÁ¿²»±ä
¢Û¸Ã·´Ó¦´ïµ½Æ½ºâʱijÎïÀíÁ¿ËæÎ¶ȱ仯Èçͼ1Ëùʾ£¬µ±Î¶ÈÓÉT1Éý¸ßµ½T2ʱ£¬Æ½ºâ³£ÊýKA
 
 KB£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®×Ý×ø±ê¿ÉÒÔ±íʾµÄÎïÀíÁ¿ÓÐÄÄЩ
 
£®

a£®H2µÄÄæ·´Ó¦ËÙÂÊ
b£®CH4µÄÌå»ý·ÖÊý
c£®»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿
d£®COµÄÌå»ý·ÖÊý
£¨3£©¹¤ÒµºÏ³É°±ÆøÐèÒªµÄ·´Ó¦Ìõ¼þ·Ç³£¸ßÇÒ²úÁ¿µÍ£¬¶øÒ»Ð©¿ÆÑ§¼Ò²ÉÓøßÖÊ×Óµ¼µçÐÔµÄSCYÌÕ´É
£¨ÄÜ´«µÝH+ £©ÊµÏÖ°±µÄµç»¯Ñ§ºÏ³É£¬´Ó¶ø´ó´óÌá¸ßÁ˵ªÆøºÍÇâÆøµÄת»¯ÂÊ£®µç»¯Ñ§ºÏ³É°±¹ý³Ì
µÄ×Ü·´Ó¦Ê½Îª£ºN2+3H2
Ò»¶¨Ìõ¼þ
2NH3£¬¸Ã¹ý³ÌÖл¹Ô­·´Ó¦µÄ·½³ÌʽΪ
 
£®
£¨4£©ÈôÍù20mL 0.0lmol?L-lµÄÈõËáHNO2ÈÜÒºÖÐÖðµÎ¼ÓÈëÒ»¶¨Å¨¶ÈµÄÉÕ¼îÈÜÒº£¬²âµÃ»ìºÏÈÜÒºµÄζȱ仯Èçͼ2Ëùʾ£¬ÏÂÁÐÓйØËµ·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
¢Ù¸ÃÉÕ¼îÈÜÒºµÄŨ¶ÈΪ0.02mol?L-1
¢Ú¸ÃÉÕ¼îÈÜÒºµÄŨ¶ÈΪ0.01mol?L-1
¢ÛHNO2µÄµçÀëÆ½ºâ³£Êý£ºbµã£¾aµã
¢Ü´Óbµãµ½cµã£¬»ìºÏÈÜÒºÖÐÒ»Ö±´æÔÚ£ºc£¨Na+£©£¾c£¨NO2-£©£¾c£¨OH-£©£¾c£¨H+£©

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø