ÌâÄ¿ÄÚÈÝ

12£®ÓÐA¡¢B¡¢C¡¢DËÄÖÖÇ¿µç½âÖÊ£¬ËüÃÇÔÚË®ÖеçÀëʱ¿É²úÉúÏÂÁÐÀë×Ó£¨Ã¿ÖÖÎïÖÊÖ»º¬Ò»ÖÖÑôÀë×ÓºÍÒ»ÖÖÒõÀë×Ó£¬ÇÒ»¥²»Öظ´£©£®
ÑôÀë×ÓK+¡¢Na+¡¢Ba2+¡¢NH4+
ÒõÀëÁËCH3COO-¡¢Cl-¡¢OH-¡¢SO42-
¼ºÖª£º¢ÙA¡¢CÈÜÒºµÄpH¾ù´óÓÚ7£¬BÈÜÒºµÄpHСÓÚ7£¬A¡¢BÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÏàͬ£¬DÈÜÒºµÄÑæÉ«·´Ó¦£¨Í¸¹ýÀ¶É«îܲ£Á§£©ÏÔ×ÏÉ«£®
¢ÚCÈÜÒººÍDÈÜÒºÏàÓöʱֻÉú³É°×É«³Áµí£¬BÈÜÒººÍCÈÜÒºÏàÓöʱֻÉú³ÉÓд̼¤ÐÔÆøÎ¶µÄÆøÌ壬AÈÜÒººÍDÈÜÒº»ìºÏʱÎÞÃ÷ÏÔÏÖÏó£®
£¨1£©AµÄ»¯Ñ§Ê½ÎªCH3COONa£®
£¨2£©ÓÃÀë×Ó·½³Ìʽ±í²»BÈÜÒºµÄpHСÓÚ7µÄÔ­Òò£ºNH4++H2O?NH3•H2O+H+£®
£¨3£©Ð´³öCÈÜÒººÍDÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£ºBa2++SO42-=BaSO4¡ý£®
£¨4£©½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄBÈÜÒººÍCÈÜÒº»ìºÏ£¬·´Ó¦ºóÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨OH-£©£¾c£¨Ba2+£©=c£¨Cl-£©£¾c£¨NH4+£©£¾c£¨H+£©£®

·ÖÎö A¡¢CÈÜÒºµÄpH¾ù´óÓÚ7£¬Ó¦Îª´×ËáÑκͼîÈÜÒº£¬BÈÜÒºµÄpHСÓÚ7£¬Ó¦Îªï§ÑÎÈÜÒº£¬A¡¢BµÄÈÜÒºÖÐË®µÄµçÀë³Ì¶ÈÏàͬ£¬ÔòÈÜҺˮ½â³Ì¶ÈÏàͬ£»DÈÜÒºÑæÉ«·´Ó¦£¨Í¸¹ýÀ¶É«îܲ£Á§£©ÏÔ×ÏÉ«£¬ÈÜÒºÖк¬ÓмØÀë×Ó£»
CÈÜÒººÍDÈÜÒºÏàÓöʱֻÉú³É°×É«³Áµí£¬BÈÜÒººÍCÈÜÒºÏàÓöʱֻÉú³É´Ì¼¤ÐÔÆøÎ¶µÄÆøÌ壬AÈÜÒººÍDÈÜÒº»ìºÏʱÎÞÃ÷ÏÔÏÖÏó£¬Ôò˵Ã÷CΪBa£¨OH£©2£¬DΪK2SO4£¬ÔòBΪNH4Cl£¬×ÛÉÏËùÊö£¬AΪCH3COONa£¬BΪNH4Cl£¬CΪBa£¨OH£©2£¬DΪK2SO4£¬ÒÔ´ËÀ´½â´ð£®

½â´ð ½â£º£¨1£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬AµÄ»¯Ñ§Ê½ÎªCH3COONa£¬
¹Ê´ð°¸Îª£ºCH3COONa£» 
£¨2£©BΪNH4Cl£¬ï§¸ùÀë×ÓË®½âʹÈÜÒº³ÊËáÐÔ£¬ÓÃÀë×Ó·½³Ìʽ±íʾΪNH4++H2O?NH3•H2O+H+£¬
¹Ê´ð°¸Îª£ºNH4++H2O?NH3•H2O+H+£»
£¨3£©CÈÜÒººÍDÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪBa2++SO42-=BaSO4¡ý£¬
¹Ê´ð°¸Îª£ºBa2++SO42-=BaSO4¡ý£»
£¨4£©ÂÈ»¯ï§ºÍÇâÑõ»¯±µµÄÎïÖʵÄÁ¿ÏàµÈ£¬»ìºÏºó£¬ÈÜÒºÖеÄÈÜÖÊÊÇһˮºÏ°±¡¢ÂÈ»¯±µºÍÇâÑõ»¯±µ£¬ÂÈ»¯±µºÍÇâÑõ»¯±µµÄŨ¶ÈÏàµÈ£¬Ò»Ë®ºÏ°±µÄŨ¶ÈÊÇÂÈ»¯±µºÍÇâÑõ»¯±µÅ¨¶ÈµÄ2±¶£¬ÈÜÒºÖÐÇâÑõ¸ùÀë×ÓŨ¶È×î´ó£¬±µÀë×ÓºÍÂÈÀë×ÓŨ¶ÈÏàµÈ£¬Ò»Ë®ºÏ°±ÊÇÈõ¼î²¿·ÖµçÀëµ¼ÖÂÂÈÀë×ÓŨ¶È´óÓÚ笠ùÀë×ÓŨ¶È£¬ÈÜÒº³Ê¼îÐÔ£¬ÇâÀë×ÓŨ¶È×îС£¬ËùÒÔÈÜÒºÖÐÀë×ÓŨ¶È´óС˳ÐòÊÇc£¨OH-£©£¾c£¨Ba2+£©=c£¨Cl-£©£¾c£¨NH4+£©£¾c£¨H+£©£¬
¹Ê´ð°¸Îª£ºc£¨OH-£©£¾c£¨Ba2+£©=c£¨Cl-£©£¾c£¨NH4+£©£¾c£¨H+£©£®

µãÆÀ ±¾Ì⿼²éÎÞ»úÎïµÄÍÆ¶Ï£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕÎïÖʵÄÐÔÖÊ¡¢·¢ÉúµÄ·´Ó¦¡¢ÑÎÀàË®½âΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓëÍÆ¶ÏÄÜÁ¦µÄ¿¼²é£¬×¢Òâ°×É«³Áµí¼°ÆøÌåµÄÅжϣ¬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
13£®ÁòËáï§ÊÇ»¯¹¤¡¢È¾Ö¯¡¢Ò½Ò©¡¢Æ¤¸ïµÈ¹¤ÒµÔ­ÁÏ£®Ä³ÁòËṤ³§ÀûÓø±²úÆ·Y´¦ÀíÎ²ÆøSO2µÃµ½CaSO4£¬ÔÙÓëÏàÁڵĺϳɰ±¹¤³§ÁªºÏÖÆ±¸£¨NH4£©2SO4£¬¹¤ÒÕÁ÷³ÌÈçͼ£º

Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©ÏÂÁÐÓйأ¨NH4£©2SO4ÈÜÒºµÄ˵·¨ÕýÈ·µÄÊÇ
A£®µçÀë·½³Ìʽ£º£¨NH4£©2SO4?2NH4++SO42-
B£®Ë®½âÀë×Ó·½³Ìʽ£ºNH4++H2O?NH3•H2O+H+
C£®Àë×ÓŨ¶È¹ØÏµ£ºc£¨NH4+£©+c£¨H+£©¨Tc£¨SO42-£©+c£¨OH-£©
D£®Î¢Á£Å¨¶È´óС£ºc£¨NH4+£©£¾c£¨SO42-£©£¾c£¨H+£©£¾c£¨NH3•H2O£©£¾c£¨OH-£©
£¨2£©ÁòËṤҵÖУ¬V2O5×÷´ß»¯¼Áʱ·¢Éú·´Ó¦2SO2+O2?2SO3£¬SO2µÄת»¯ÂÊÓëζȡ¢Ñ¹Ç¿Óйأ¬Çë¸ù¾ÝϱíÐÅÏ¢£¬½áºÏ¹¤ÒµÉú²úʵ¼Ê£¬Ñ¡Ôñ±íÖÐ×îºÏÊʵÄζȺÍѹǿ·Ö±ðÊÇ420¡æ¡¢1.01¡Á105Pa£®¸Ã·´Ó¦420¡æÊ±µÄƽºâ³£Êý£¾520¡æÊ±µÄƽºâ³£Êý£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
1.01¡Á105Pa5.05¡Á105Pa1.01¡Á106Pa
420¡æ0.99610.99720.9984
520¡æ0.96750.97670.9852
620¡æ0.85200.88970.9276
£¨3£©ÔÚ2LÃܱÕÈÝÆ÷ÖÐÄ£Äâ½Ó´¥·¨ÖƱ¸ÈýÑõ»¯Áòʱ£¬ÈôµÚ12·ÖÖÓÇ¡ºÃ´ïµ½Æ½ºâ£¬²âµÃÉú³ÉSO3µÄÎïÖʵÄÁ¿Îª1.2mol£¬¼ÆËãǰ12·ÖÖÓÓÃÑõÆø±íʾ·´Ó¦ËÙÂÊv£¨O2£©Îª0.025mol/£¨L£®min£©£®
£¨4£©¸±²úÆ·YÊÇÑõ»¯¸Æ£®³Áµí³ØÖз¢ÉúµÄÖ÷Òª·´Ó¦·½³ÌʽÊÇCaSO4+CO2+2NH3+H2O¡úCaCO3¡ý+£¨NH4£©2SO4£®
£¨5£©´ÓÂÌÉ«»¯Ñ§ºÍ×ÊÔ´×ÛºÏÀûÓõĽǶÈ˵Ã÷ÉÏÊöÁ÷³ÌµÄÖ÷ÒªÓŵãÊDzúÉúµÄCO2Ñ­»·Ê¹Óã¬ÎïÖʳä·ÖÀûÓ㬸±²úÆ·ÓÐÓã¬ÎÞÎÛȾÐÔÎïÖʲúÉú£®
4£®Ä³Í¬Ñ§½øÐÐÁËÁò´úÁòËáÄÆÓëÁòËá·´Ó¦µÄÓйØÊµÑ飬ʵÑé¹ý³ÌµÄÊý¾Ý¼Ç¼Èç±í£¨¼û±í¸ñ£©£¬Çë½áºÏ±íÖÐÐÅÏ¢£¬»Ø´ðÓйØÎÊÌ⣺£¨ Na2S2O3+H2SO4¨TNa2SO4+S¡ý+SO2¡ü+H2O£©
ʵÑé
ÐòºÅ
·´Ó¦ÎÂ
¶È/¡æ
²Î¼Ó·´Ó¦µÄÎïÖÊ
Na2S2O3H2SO4H2O
V/mLc/mol•L-1V/mLc/mol•L-1V/mL
A20100.1100.10
B2050.1100.15
C20100.150.15
D4050.1100.15
£¨1£©¸ù¾ÝÄãËùÕÆÎÕµÄ֪ʶÅжϣ¬ÔÚÉÏÊöʵÑéÖУ¬·´Ó¦ËÙÂÊ×î¿ìµÄ¿ÉÄÜÊÇD£¨ÌîʵÑéÐòºÅ£©£®
£¨2£©ÔڱȽÏijһÒòËØ¶ÔʵÑé²úÉúµÄÓ°Ïìʱ£¬±ØÐëÅųýÆäËûÒòËØµÄ±ä¶¯ºÍ¸ÉÈÅ£¬¼´ÐèÒª¿ØÖƺÃÓëʵÑéÓйصĸ÷Ïî·´Ó¦Ìõ¼þ£¬ÆäÖУº
¢ÙÄÜ˵Ã÷ζȶԸ÷´Ó¦ËÙÂÊÓ°ÏìµÄ×éºÏÊÇBD£¨ÌîʵÑéÐòºÅ£¬ÏÂͬ£©£»
¢ÚAºÍB¡¢AºÍCµÄ×éºÏ±È½Ï£¬ËùÑо¿µÄÎÊÌâÊÇÏàͬζÈÌõ¼þÏÂŨ¶È¶Ô¸Ã·´Ó¦ËÙÂʵÄÓ°Ï죻
¢ÛBºÍCµÄ×éºÏ±È½Ï£¬ËùÑо¿µÄÎÊÌâÊÇÏàͬζÈÌõ¼þÏ£¬¸Ã·´Ó¦ËÙÂʸü´ó³Ì¶ÈÉÏÈ¡¾öÓÚÄÄÖÖ·´Ó¦ÎïµÄŨ¶È
£¨3£©½Ì²ÄÖÐÀûÓÃÁ˳öÏÖ»ÆÉ«³ÁµíµÄ¿ìÂýÀ´±È½Ï·´Ó¦ËÙÂʵĿìÂý£¬ÇëÄã·ÖÎöΪºÎ²»²ÉÓÃÅÅË®·¨²âÁ¿µ¥Î»Ê±¼äÄÚÆøÌåÌå»ýµÄ´óС½øÐбȽϣºSO2¿ÉÈÜÓÚË®£¬²â¶¨²»¾«È·»òʵÑé×°Öýϸ´ÔÓ£¬²»Ò׿ØÖÆ£®

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

¾«Ó¢¼Ò½ÌÍø