Question

2013Äê1ÔÂ13ÈÕ£®±±¾©Êз¢²¼Á˱±¾©ÆøÏóÊ·ÉÏÊ׸öö²³ÈÉ«Ô¤¾¯£¬²¿·ÖµØÇøPM2.5ʵʱŨ¶ÈÒ»¶È³¬¹ý900΢¿Ë/Á¢·½Ã×£®PM2.5ÊÇÖ¸´óÆøÖÐÖ±¾¶Ð¡ÓÚ»òµÈÓÚ2.5΢Ã׵ĿÅÁ£ÎÓֳơ°¿ÉÈë·Î¿ÅÁ£¡±£¬ÕâÖÖÓж¾¿ÅÁ£ÎïµÄÖ÷ÒªÀ´Ô´ÊÇÆû³µÎ²Æø¡¢Éú»îȼú²úÉúµÄ·ÏÆøÒÔ¼°¹¤Òµ·ÏÆøµÈ£®
£¨1 £©ÔÚÆû³µÎ²ÆøϵͳÖа²×°´ß»¯×ª»»Æ÷£¬¿ÉÓÐЧ¼õÉÙCO¡¢µªÑõ»¯ÎïºÍ̼Ç⻯ºÏÎï µÈÓк¦ÆøÌåµÄÅÅ·Å£®ÔÚת»»Æ÷µÄÇ°°ë²¿£¬COºÍµªÑõ»¯ÎïÔÚ´ß»¯¼ÁµÄ×÷ÓÃÏ£¬Éú³É¿É²Î Óë´óÆøÑ­»·µÄÁ½ÖÖÎÞ¶¾ÆøÌ壬Æ仯ѧ·½³ÌʽΪ

 

£¨µªÑõ»¯ºÏÎïÓÃNO_x±íʾ£©£®ÔÚת»»Æ÷µÄºó°ë²¿£¬²ÎÓëµÄÌþ¼°COÔÚ´ß»¯¼ÁµÄ×÷ÓÃϱ»Ñõ»¯Éú³É

 

Á½ÖÖÎÞ¶¾ÎïÖÊ£®
£¨2£©°±ÆøÒ²¿ÉÓÃÀ´´¦ÀíµªÑõ»¯ºÏÎÈçÔÚÒ»¶¨Ìõ¼þÏ£¬µªÆøÓëNO₂·´Ó¦Éú³ÉµªÆøºÍË®£®Èôij³§ÅųöµÄÆøÌåÖÐNO₂µÄº¬Á¿Îª0.6%£¨Ìå»ý·ÖÊý£©£¬Ôò´¦Àí1m³ÕâÖÖ·ÏÆøÐèÒªÏàͬÌõ¼þÏ°±ÆøµÄÌå»ýÖÁÉÙΪ

 

L£®
£¨3 £©ÀûÓÃÄƼîÑ­»··¨¿ÉÍѳý¹¤Òµ·ÏÆøÖеÄSO₂£®
¢ÙÔÚÄƼîÑ­»··¨ÖУ¬ÀûÓÃÑÇÁòËáÄÆÈÜÒº×÷ΪÎüÊÕÒº£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ

 

£»

n£¨SO32-£©£ºn£¨HSO3-£©	91£º9	1£º1	9£º91
pH	8.2	7.2	6.2

µ±ÎüÊÕÒº³ÊËáÐÔʱ£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈµÄÕýÈ·¹ØϵÊÇ

 

£»
A£®c£¨Na⁺£©+c£¨H⁺£©=c£¨SO₃^2-£©+c£¨HSO₃^-£©+c£¨OH^-£©
B£®c£¨Na⁺£©£¼2c£¨SO₃^2-£©+£¨HSO₃^-£©
C£®c£¨Na⁺£©£¾c£¨HSO₃^-£©£¾c£¨SO₃^2-£©£¾c£¨H⁺£©£¾c£¨OH^-£©
£¨4£©ÖÎÀíº¬SO₂µÄÆøÌåÒ²¿ÉÀûÓÃÔ­µç³ØÔ­Àí£¬ÓÃSO₂ºÍO₂À´ÖƱ¸ÁòËᣬװÖÃÈçͼËùʾ£®µç¼«Îª¶à¿×µÄ²ÄÁÏ£¬ÄÜÎü¸½ÆøÌ壬ͬʱҲÄÜʹÆøÌåÓëµç½âÖÊÈÜÒº³ä·Ö½Ó´¥£®
¢ÙB¼«µÄµç¼«·´Ó¦Ê½Îª

 

£®
¢ÚÈÜÒºÖÐH⁺µÄÒƶ¯·½ÏòΪÓÉ

 

¼«Çøµ½

 

¼«Çø£®£¨Ìî¡°A¡±»ò¡°B¡±£©
£¨5£©ÎªÁ˲ⶨijÁòË᳧ÖÜΧµÄ¿ÕÆøÖжþÑõ»¯ÁòµÄº¬Á¿£¬Ä³Ñ§Ï°Ð¡×éÈ¡1m³£¨ÒÑÕÛË㠳ɱê×¼×´¿ö£©¸ÃµØÇø¿ÕÆø£¬Í¨Èë×°ÓйýÁ¿äåË®ÈÜÒºµÄÎüÊÕÆ¿ÖУ¬³ä·Ö·´Ó¦£®ÔÚÎüÊÕºóµÄË® ÈÜÒºÖмÓÈË×ãÁ¿BaCl₂ÈÜÒº£¬Éú³É°×É«³Áµí0.12g£®Ôò¸Ã¿ÕÆøÖÐSO₂µÄÌå»ý·ÖÊýΪ

 

£®

Answer 1

¿¼µã£º³£¼ûµÄÉú»î»·¾³µÄÎÛȾ¼°ÖÎÀí,»¯Ñ§·½³ÌʽµÄÓйؼÆËã,³£¼û»¯Ñ§µçÔ´µÄÖÖÀ༰Æ乤×÷Ô­Àí,Àë×ÓŨ¶È´óСµÄ±È½Ï

רÌ⣺»ù±¾¸ÅÄîÓë»ù±¾ÀíÂÛ,µç»¯Ñ§×¨Ìâ,ÔªËؼ°Æ仯ºÏÎï

·ÖÎö£º£¨1£©¸ù¾Ý·´Ó¦ÎïºÍÉú³ÉÎïµÄÐÔÖÊÌصãºÍÔªËØÊغã½øÐзÖÎö£»
£¨2£©½áºÏ·´Ó¦µÄ·½³Ìʽ¼ÆË㣻
£¨3£©¢Ù¸ù¾ÝÔ­ÁÏÅäƽ¼´¿É£»
 ¢Ú¸ù¾ÝÊغã¹Øϵ¼°±íÖÐÊý¾Ý½øÐзÖÎö£»
£¨4£©±¾ÌâΪSO₂ÓëO₂·´Ó¦Éú³ÉSO₃£¬SO₃ÔÙÓëË®»¯ºÏÉú³ÉÁòËᣬ¸ù¾ÝÁòËáµÄ³ö¿ÚÅжÏÕý¸º¼«£¬¸º¼«·¢ÉúÑõ»¯·´Ó¦£¬Õý¼«·¢Éú»¹Ô­·´Ó¦£¬Ô­µç³Ø·Åµçʱ£¬µç½âÖÊÈÜÒºÖÐÑôÀë×ÓÏòÕý¼«Òƶ¯£»
£¨5£©ÀûÓùØϵʽ£ºSO₂¡«SO₄^2-¡«BaSO₄£¬¸ù¾ÝBaSO₄³ÁµíµÄÖÊÁ¿Çó³öSO₂µÄÌå»ý£¬×îºóÇó³öÌå»ý·ÖÊý£®

½â´ð£º ½â£º£¨1£©¸Ã·´Ó¦Öз´Ó¦ÎïÊǵªÑõ»¯ÎïºÍÒ»Ñõ»¯Ì¼£¬Éú³ÉÎïÊǵªÆøºÍ¶þÑõ»¯Ì¼£¬·´Ó¦Ìõ¼þÊÇ´ß»¯¼ÁºÍ¼ÓÈÈ£¬ËùÒԸ÷´Ó¦·½³ÌʽΪ2NO_x+2xCO

´ß»¯¼Á  .

N₂+2xCO₂£¬
¹Ê´ð°¸Îª£º2NO_x+2xCO

´ß»¯¼Á  .

N₂+2xCO₂  CO₂ºÍ H₂O
£¨2£©·¢Éú·´Ó¦Îª8NH₃+6NO₂¨T7N₂+12H₂0£¬
¹¤³§ÅųöµÄÆøÌåÖжþÑõ»¯µªÎª1¡Ám³¡Á0.6%=0.006m³=6L£¬
ÓÉ8NH₃+6NO₂¨T7N₂+12H₂0£¬¿ÉÖªÐèÒª°±Æø6L¡Á

8

6

=8L£¬
¹Ê´ð°¸Îª£º8£»
£¨3£©¢ÙÀûÓÃÑÇÁòËáÄÆÈÜÒº×÷ΪÎüÊÕÒºÎüÊÕSO₂£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ SO₃^2-+SO₂+H₂O¨T2HSO₃^-£¬¹Ê´ð°¸Îª£ºSO₃^2-+SO₂+H₂O¨T2HSO₃^-£»
¢ÚA¡¢ÀûÓÃÎïÁÏÊغãÁÐʽ£ºc£¨Na⁺£©+c£¨H⁺£©=2c£¨SO₃^2-£©+c£¨HSO₃^-£©+c£¨OH^-£©£¬¹ÊA´íÎó£»
B¡¢ÒòΪÈÜÒºÏÔËáÐÔ£¬¹Êc£¨H⁺£©£¾c£¨OH^-£©£¬½áºÏµçºÉÊغãʽc£¨Na⁺£©+c£¨H⁺£©=2c£¨SO₃^2-£©+c£¨HSO₃^-£©+c£¨OH^-£©£¬¿ÉµÃc£¨Na⁺£©£¼2c£¨SO₃^2-£©+£¨HSO₃^-£©£¬¹ÊBÕýÈ·
C¡¢µ±n£¨SO₃²£©£ºn£¨HSO₃^-£©=1£º1ʱ£¬pHΪ7.2£¬¸ù¾Ý±íÖб仯¹æÂÉ¿ÉÖªËáÐÔʱ£¬n£¨SO₃²£©£¼n£¨HSO₃^-£©£¬ËùÒÔc£¨Na⁺£©£¾c£¨HSO₃^-£©£¾c£¨SO₃^2-£©£¾c£¨H⁺£©£¾c£¨OH^-£©£¬¹ÊCÕýÈ·£»
¹Ê´ð°¸Îª£ºBC£»
£¨4£©¸ÃÔ­µç³ØÖУ¬¸º¼«ÉÏʧµç×Ó±»Ñõ»¯£¬ËùÒÔ¸º¼«Í¨ÈëµÄÆøÌåÊǶþÑõ»¯Áò£¬¶þÑõ»¯Áòʧµç×ÓºÍË®·´Ó¦Éú³ÉÁòËá¸ùÀë×ÓºÍÇâÀë×Ó£»Õý¼«ÉÏͨÈëµÄÊÇÑõÆø£¬Õý¼«ÉÏÑõÆøµÃµç×ÓºÍÇâÀë×Ó·´Ó¦Éú³ÉË®£¬¸ù¾ÝÁòËáºÍË®µÄ³ö¿Ú·½ÏòÖª£¬B¼«ÊǸº¼«£¬A¼«ÊÇÕý¼«£¬ËùÒÔB¼«Éϵĵ缫·´Ó¦Ê½Îª£ºSO₂-2e^-+2H₂O¨TSO₄^2-+4H⁺£¬Ô­µç³Ø·Åµçʱ£¬ÇâÀë×ÓÓɸº¼«BÒÆÏòÕý¼«A£¬¹Ê´ð°¸Îª£ºSO₂+2H₂O-2e^-¨TSO₄^2-+4H⁺£»B£»A£»
£¨5£©ÀûÓùØϵʽ£ºSO₂ ¡«SO₄^2-¡«BaSO₄£¬
             n£¨SO₂£©=n£¨BaSO₄£©=

0.12

233

mol
1L¿ÕÆøÖÐSO₂µÄÌå»ýΪ22.4L/mol¡Á

0.12

233

mol=0.0115L£¬SO₂µÄÌå»ý·ÖÊýΪ

0.0115L

1L

¡Á100%=1.15¡Á10^-3%£¬
¹Ê´ð°¸Îª£º1.15¡Á10^-3%£®

µãÆÀ£º±¾Ì⿼²éÁË»¯Ñ§·½³Ìʽ¡¢Àë×Ó·½³Ìʽ¡¢µç¼«·´Ó¦Ê½µÄÊéд¡¢Àë×ÓŨ¶È´óСµÈÎÊÌ⣬×ÛºÏÐÔ½ÏÇ¿£®