Question

9£®Ä³ÊжԴóÆø½øÐмà²â£¬·¢ÏÖ¸ÃÊÐÊ×ÒªÎÛȾÎïΪ¿ÉÎüÈë¿ÅÁ£ÎïPM2.5£¨Ö±¾¶Ð¡ÓÚµÈÓÚ2.5¦ÌmµÄÐü¸¡¿ÅÁ£Î£¬ÆäÖ÷ÒªÀ´Ô´ÎªÈ¼Ãº¡¢»ú¶¯³µÎ²ÆøµÈ£®Òò´Ë£¬¶ÔPM2.5¡¢SO₂¡¢NOxµÈ½øÐÐÑо¿¾ßÓÐÖØÒªÒâÒ壮
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©½«PM2.5Ñù±¾ÓÃÕôÁóË®´¦ÀíÖƳɴý²âÊÔÑù£®Èô²âµÃ¸ÃÊÔÑùËùº¬Àë×ӵĻ¯Ñ§×é·Ö¼°ÆäŨ¶ÈÈçÏÂ±í£º

Àë×Ó	H+	K+	Na+	NH4+	SO42-	NO3-	Cl-
Ũ¶È/mol•L-1	δ²â¶¨	4¡Á10-6	6¡Á10-6	2¡Á10-5	4¡Á10-5	3¡Á10-5	2¡Á10-5

¸ù¾Ý±íÖÐÊý¾ÝÅжÏÊÔÑùµÄpHԼΪ4£®
£¨2£©Îª¼õÉÙSO₂µÄÅÅ·Å£¬³£²ÉÈ¡µÄ´ëÊ©ÓУº¢Ù½«Ãº×ª»¯ÎªÇå½àÆøÌåȼÁÏ£®
ÒÑÖª£º2H₂£¨g£©+O₂£¨g£©¨T2H₂O£¨g£©  K₁2C£¨s£©+O₂£¨g£©=2CO £¨g£©       K₂
2C£¨s£©+2H₂O£¨g£©¨T2CO£¨g£©+2H₂£¨g£©ÔòK=$\sqrt{\frac{{K}_{2}}{{K}_{1}}}$£¨Óú¬K₁¡¢K₂µÄʽ×Ó±íʾ£©£®
¢ÚÏ´µÓº¬SO₂µÄÑÌÆø£®ÒÔÏÂÎïÖÊ¿É×÷Ï´µÓ¼ÁµÄÊÇab£®
A£®Ca£¨OH£©₂       B£®Na₂CO₃      C£®CaCl₂D£®NaHSO₃
£¨3£©Æû³µÎ²ÆøÖÐÓÐNOxºÍCOµÄÉú³É¼°×ª»¯£¬ÒÑÖªÆû¸×ÖÐÉú³ÉNOµÄ·´Ó¦Îª£ºN₂£¨g£©+O₂£¨g£©?2NO£¨g£©¡÷H£¾0Ä¿Ç°£¬ÔÚÆû³µÎ²ÆøϵͳÖÐ×°Öô߻¯×ª»¯Æ÷¿É¼õÉÙCOºÍNOxµÄÎÛȾ£¬Æ仯ѧ·´Ó¦·½³ÌʽΪ2CO+2NO$\frac{\underline{\;´ß»¯¼Á\;}}{\;}$2CO₂+N₂£®

Answer 1

·ÖÎö £¨1£©ÒÀ¾ÝµçºÉÊغ㣬Çó³öÇâÀë×ÓµÄŨ¶È£¬´øÈëpH¼ÆË㹫ʽ£»
£¨2£©¸ù¾Ý·´Ó¦2H₂£¨g£©+O₂£¨g£©¨T2H₂O£¨g£©K₁=$\frac{{c}^{2}£¨H{\;}_{2}O£©}{{\;}^{c2}£¨{H}_{2}£©c£¨O{\;}_{2}£©}$£¬2C£¨s£©+O₂£¨g£©¨T2CO£¨g£©K₂=$\frac{{c}^{2}£¨CO£©}{c£¨{O}_{2}£©}$£¬C£¨s£©+H₂O£¨g£©?CO£¨g£©+H₂£¨g£©K=$\frac{c£¨CO£©c£¨H{\;}_{2}£©}{c£¨{H}_{2}O£©}$£¬ÓÉ´ËÈ·¶¨KµÄÖµ£»
¢ÚÒÀ¾Ý¶þÑõ»¯ÁòµÄÐÔÖÊ£ºÊÇÒ»ÖÖËáÐÔÑõ»¯Îï¿ÉÒԺͼӦ£¬¶þÑõ»¯ÁòÈÜÓÚË®µÃÑÇÁòËᣬËáÐÔÇ¿ÓÚ̼ËᣬËùÒÔ¿ÉÒÔºÍ̼ËáÄÆÈÜÒº·´Ó¦£®
£¨3£©COºÍNO´ß»¯Ñõ»¯Éú³ÉµªÆøºÍ¶þÑõ»¯Ì¼£®

½â´ð ½â£º£¨1£©¸ù¾ÝµçºÉÊغãµÃ£º[H⁺]+[K⁺]+[Na⁺]+[NH₄⁺]=2[SO₄^2-]+[NO₃^-]+[Cl^-]£¬½«±íÖÐÊý¾Ý´øÈëµÃ£º[H⁺]=10^-4£¬pH=-lg[H⁺]=4£¬¹Ê´ð°¸Îª£º4£»
£¨2£©¸ù¾Ý·´Ó¦2H₂£¨g£©+O₂£¨g£©¨T2H₂O£¨g£©K₁=$\frac{{c}^{2}£¨H{\;}_{2}O£©}{{\;}^{c2}£¨{H}_{2}£©c£¨O{\;}_{2}£©}$£¬2C£¨s£©+O₂£¨g£©¨T2CO£¨g£©K₂=$\frac{{c}^{2}£¨CO£©}{c£¨{O}_{2}£©}$£¬C£¨s£©+H₂O£¨g£©?CO£¨g£©+H₂£¨g£©K=$\frac{c£¨CO£©c£¨H{\;}_{2}£©}{c£¨{H}_{2}O£©}$=$\sqrt{\frac{{K}_{2}}{{K}_{1}}}$£¬
¹Ê´ð°¸Îª£º$\sqrt{\frac{{K}_{2}}{{K}_{1}}}$£»
¢Ú¶þÑõ»¯ÁòÊÇËáÐÔÑõ»¯ÎïËùÒÔ¾ßÓÐËáÐÔÑõ»¯ÎïµÄͨÐÔÄܹ»Óë¼î·´Ó¦£ºSO₂+Ca£¨OH£©₂=CaSO₃¡ý+H₂O£»
¶þÑõ»¯ÁòË®ÈÜÒºÊÇÑÇÁòËáËùÒÔÒ²Äܹ»Óë̼ËáÄÆ·´Ó¦£ºSO₂+Na₂CO₃=Na₂SO₃+CO₂¡üËùÒÔ¿ÉÒÔÓÃÇâÑõ»¯¸ÆÈÜÒººÍ̼ËáÄÆÈÜÒº×öÏ´µÓ¼Á£»
¶þÑõ»¯ÁòÓëÂÈ»¯¸ÆÈÜÒº²»·´Ó¦£¬ÇÒÔÚÑÇÁòËáÇâÄÆÈÜÒºÖеÄÈܽâ¶ÈºÜС£¬ËùÒÔ²»ÄÜÓÃÂÈ»¯¸ÆºÍÑÇÁòËáÇâÄÆÈÜÒº×öÏ´µÓÒº£¬
¹Ê´ð°¸Îª£ºab£»
£¨3£©COºÍNO´ß»¯Ñõ»¯Éú³ÉµªÆøºÍ¶þÑõ»¯Ì¼£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2CO+2NO$\frac{\underline{\;´ß»¯¼Á\;}}{\;}$2CO₂+N₂£¬¹Ê´ð°¸Îª£º2CO+2NO$\frac{\underline{\;´ß»¯¼Á\;}}{\;}$2CO₂+N₂£®

µãÆÀ ±¾Ìâ×ۺϿ¼²é»¯Ñ§·´Ó¦Ô­ÀíµÄ»ù´¡ÖªÊ¶£¬Éæ¼°Àë×ÓµÄË®½â¡¢PHÖµµÄ¼ÆËã¡¢¸Ç˹¶¨ÂɵÄÓ¦Óᢻ¯Ñ§Æ½ºâ³£ÊýµÄ¼ÆËã¡¢»¯Ñ§·½³ÌʽµÄÊéдµÈ£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÏà¹Ø֪ʶµÄ»ýÀÛ£®