ÌâÄ¿ÄÚÈÝ

¼×È©ÊÇÊÀ½çÎÀÉú×éÖ¯£¨WHO£©È·ÈϵÄÖ°©ÎïºÍÖ»ûÎïÖÊÖ®Ò»£®ÎÒ¹ú¹æ¶¨£ºÊÒÄÚ¼×È©º¬Á¿²»µÃ³¬¹ý0.08mg?m-3£®Ä³ÖÐѧһÑо¿ÐÔѧϰС×éÓûÀûÓÃËáÐÔKMnO4ÈÜÒº²â¶¨Ä³ÐÂ×°ÐÞ·¿¼ä¿ÕÆøÖм×È©µÄº¬Á¿£®

£¨1£©ÊµÑé²½ÖèÈçÏ£¨ÇëÌîд¿Õ°×£©£º
¢ÙÓÃ
ËáʽµÎ¶¨¹Ü
ËáʽµÎ¶¨¹Ü
Á¿È¡4.00mL 0.01mol?L-1KMnO4ÈÜÒºÓÚÏ´ÆøÆ¿ÖУ¬²¢µÎÈ뼸µÎÏ¡H2SO4£¬¼ÓˮϡÊÍÖÁ20mL±¸Óã®
¢Ú½«1.00¡Á10-3mol?L-1²ÝËᣨH2C2O4£©±ê×¼ÈÜÒºÖÃÓÚËáʽµÎµí¹ÜÖб¸Óã®
¢Û×°ºÃʵÑé×°Öã¨ÈçͼËùʾ£¬Í¼ÖÐXΪһÈÝ»ýΪ5LµÄÏðƤ´¢Æø×°Öã©£¬´ò¿ª
a
a
£¬¹Ø±Õ
b
b
£¨Ìî¡°a¡±»ò¡°b¡±£©£¬ÓÃÏðƤÄÒÎüÈ¡ÐÂ×°Ð޵ķ¿ÎÝÄÚ¿ÕÆø£¬È»ºó¹Ø±Õ
a
a
£¬´ò¿ª
b
b
£¨Ìî¡°a¡±»ò¡°b¡±£©£¬ÔÙÂýÂýѹËõƤÄÒ£¬½«ÆäÖÐÆøÌåÈ«²¿Ñ¹ÈëËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖУ¬Ê¹Æä³ä·Ö·´Ó¦£®ÆäÖظ´50´Î£®
¢ÜÈ¡Ï´ÆøÆ¿ÖÐÈÜÒº5mLµ½×¶ÐÎÆ¿ÖУ¬Óñê×¼²ÝËáÈÜÒº½øÐе樣¬²¢¼Ç¼µÎ¶¨ËùÏûºÄµÄ²ÝËáÈÜÒºµÄÌå»ý£®ÔÙÖظ´ÊµÑé2´Î£¨Ã¿´ÎËùÓõĸßÃÌËá¼ØÈÜÒº¾ùΪ5.00mL£©£®
¢Ý¼ÆËã3´ÎʵÑéÏûºÄ²ÝËáÈÜÒºµÄƽ¾ùÌå»ý£¬ÌîÈëϱíÖУ®
µÎ¶¨
´ÎÊý
Á¿È¡KMnO4ÈÜÒºµÄÌå»ý£¨mL£© ±ê×¼²ÝËáÈÜÒºµÄÌå»ý ƽ¾ùÖµ
£¨mL£©
µÎ¶¨Ç°¿Ì¶È µÎ¶¨ºó¿Ì¶È ʵ¼ÊÌå»ý£¨mL£©
µÚÒ»´Î 5.00 0.00 19.60
µÚ¶þ´Î 5.00 0.20 19.60
µÚÈý´Î 5.00 0.06 18.06
£¨2£©¸ÃµÎ¶¨ÊÇ·ñÐèÁí¼Óָʾ¼Á
²»ÐèÒª
²»ÐèÒª
£¬Ô­ÒòÊÇ
KMnO4ÈÜҺΪ×ϺìÉ«£¬µÎ¶¨µ½ÖÕµãʱÑÕÉ«ÓÉ×ϺìÉ«±äΪÎÞÉ«¿ÉָʾµÎ¶¨
KMnO4ÈÜҺΪ×ϺìÉ«£¬µÎ¶¨µ½ÖÕµãʱÑÕÉ«ÓÉ×ϺìÉ«±äΪÎÞÉ«¿ÉָʾµÎ¶¨
£»
£¨3£©Í¨¹ý¼ÆËãÅжϸþÓÊÒÄÚ¿ÕÆøÖм×È©µÄŨ¶ÈÊÇ·ñ´ï±ê£¬¼òҪд³ö¼ÆËã¹ý³Ì£®Óõ½µÄ»¯Ñ§·½³ÌʽÓУº4MnO4-+5HCHO+12H+=4Mn2++5CO2¡ü+6H2O£»2MnO4-+5H2C2O4+6H+=2Mn2++10CO2¡ü+8H2O
£¨4£©Ä³Í¬Ñ§Óø÷½·¨²âÁ¿µÄ¿ÕÆøÖм×È©º¬Á¿µÄÊýÖµ±ÈС×éµÄƽ¾ùÖµµÍÐí¶à£¬ÇëÄã¶ÔÆä¿ÉÄܵÄÔ­Òò£¨¼ÙÉèÈÜÒºÅäÖÆ¡¢³ÆÁ¿»òÁ¿»òÁ¿È¡¼°µÎ¶¨ÊµÑé¾ùÎÞ´íÎó£©Ìá³öºÏÀí¼ÙÉ裨ÖÁÉÙ´ð³ö2ÖÖ¿ÉÄÜÐÔ£©
δ°ÑÏðƤÄÒÖÐÆøÌåÍêȫѹ³ö
δ°ÑÏðƤÄÒÖÐÆøÌåÍêȫѹ³ö
£¬
ѹËÍÆøÌåËٶȹý¿ì¡¢×°ÖÃÆøÃÜÐԽϲµ¼Æø¹ÜÉìÈëKMnO4ÈÜÒºÖÐ̫dz¡¢Í¬Ò»µØµãÈ¡Ñù´ÎÊýÌ«¶àµÈ
ѹËÍÆøÌåËٶȹý¿ì¡¢×°ÖÃÆøÃÜÐԽϲµ¼Æø¹ÜÉìÈëKMnO4ÈÜÒºÖÐ̫dz¡¢Í¬Ò»µØµãÈ¡Ñù´ÎÊýÌ«¶àµÈ
£»
£¨5£©ÊµÑé½áÊøºó£¬¸ÃС×é³ÉÔ±ÔÚÏ໥½»Á÷µÄ¹ý³ÌÖÐÒ»ÖÂÈÏΪʵÑé×°ÖÃÓ¦¼ÓÒԸĽø£®ÀýÈç¿É½«²åÈëKMnO4ÈÜÒºµÄ¹Ü×Ó϶˸ijɾßÓжà¿×µÄÇòÅÝ£¨Èçͼ£©£¬ÓÐÀûÓÚÌá¸ßʵÑéµÄ׼ȷ¶È£¬ÆäÀíÓÉÊÇ
¿ÉÔö´óÆøÌåÓëÈÜÒºµÄ½Ó´¥Ãæ»ý£¬Ê¹¿ÕÆøÖеÄHCHO±»³ä·ÖÎüÊÕ
¿ÉÔö´óÆøÌåÓëÈÜÒºµÄ½Ó´¥Ãæ»ý£¬Ê¹¿ÕÆøÖеÄHCHO±»³ä·ÖÎüÊÕ
£®
·ÖÎö£º£¨1£©¸ßÃÌËá¼ØÈÜÒº³ÊËáÐÔ£¬ÒªÓÃËáʽµÎ¶¨¹ÜÁ¿È¡£»
´ò¿ªa¹Ø±ÕbÊÕ¼¯ÊÒÄÚ¿ÕÆø£¬È»ºó¹Ø±Õa´ò¿ªb½«ÏðƤÄÒÖеÄÆøÌåѹËõµ½ËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖУ»
£¨2£©¸ù¾Ý¸Ã·´Ó¦ÊÇ·ñÓÐÃ÷ÏÔÏÖÏóÅжϣ»
£¨4£©´ÓʵÑé¸÷¸ö»·½Ú¿ÉÄܳöÏÖµÄÎÊÌâ·ÖÎö£»
£¨5£©·´Ó¦ÎïµÄ½Ó´¥Ãæ»ýÔ½´ó£¬·´Ó¦Ô½³ä·Ö£®
½â´ð£º½â£º£¨1£©¸ßÃÌËá¼ØÈÜÒº³ÊËáÐÔÇÒÌå»ýÊÇСÊýµãºóÁ½Î»£¬ËùÒÔÒªÓÃËáʽµÎ¶¨¹ÜÁ¿È¡£¬´ò¿ªa¹Ø±ÕbÊÕ¼¯ÊÒÄÚ¿ÕÆø£¬È»ºó¹Ø±Õa´ò¿ªb½«ÏðƤÄÒÖеÄÆøÌåѹËõµ½ËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖУ¬
¹Ê´ð°¸Îª£ºËáʽµÎ¶¨¹Ü£¨»òÒÆÒº¹Ü£©  a¡¢b£¬a¡¢b£»
£¨2£©KMnO4ÈÜҺΪ×ϺìÉ«£¬µÎ¶¨µ½ÖÕµãʱÑÕÉ«ÓÉ×ϺìÉ«±äΪÎÞÉ«¿ÉָʾµÎ¶¨£¬ËùÒÔ²»ÐèҪָʾ¼Á£¬
¹Ê´ð°¸Îª£º²»ÐèÒª£¬KMnO4ÈÜҺΪ×ϺìÉ«£¬µÎ¶¨µ½ÖÕµãʱÑÕÉ«ÓÉ×ϺìÉ«±äΪÎÞÉ«¿ÉָʾµÎ¶¨£»
£¨4£©´ÓʵÑé¸÷¸ö»·½Ú¿ÉÄܳöÏÖµÄÎÊÌâ·ÖÎö£¬È磺δ°ÑÏðƤÄÒÖÐÆøÌåÍêȫѹ³ö¡¢Ñ¹ËÍÆøÌåËٶȹý¿ì¡¢×°ÖÃÆøÃÜÐԽϲµ¼Æø¹ÜÉìÈëKMnO4ÈÜÒºÖÐ̫dz¡¢Í¬Ò»µØµãÈ¡Ñù´ÎÊýÌ«¶àµÈ£¬
¹Ê´ð°¸Îª£ºÎ´°ÑÏðƤÄÒÖÐÆøÌåÍêȫѹ³ö¡¢Ñ¹ËÍÆøÌåËٶȹý¿ì¡¢×°ÖÃÆøÃÜÐԽϲµ¼Æø¹ÜÉìÈëKMnO4ÈÜÒºÖÐ̫dz¡¢Í¬Ò»µØµãÈ¡Ñù´ÎÊýÌ«¶àµÈ£»
£¨5£©·´Ó¦ÎïµÄ½Ó´¥Ãæ»ýÔ½´ó£¬·´Ó¦Ô½³ä·Ö£¬Ôò²âÁ¿Ô½×¼È·£¬
¹Ê´ð°¸Îª£º¿ÉÔö´óÆøÌåÓëÈÜÒºµÄ½Ó´¥Ãæ»ý£¬Ê¹¿ÕÆøÖеÄHCHO±»³ä·ÖÎüÊÕ£®
µãÆÀ£º±¾Ì⿼²éÁËÎïÖʺ¬Á¿µÄ²â¶¨£¬¸ù¾ÝÎïÖʵÄÐÔÖÊÀ´Éè¼ÆʵÑ飬ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¼×È©ÊÇÊÀ½çÎÀÉú×éÖ¯£¨WHO£©È·ÈϵÄÖ°©ÎïºÍÖ»ûÎïÖÊÖ®Ò»£®ÎÒ¹ú¹æ¶¨£ºÊÒÄÚ¼×È©º¬Á¿²»µÃ³¬¹ý0.08mg?m-3£®Ä³Ñо¿ÐÔѧϰС×éÓûÀûÓÃËáÐÔKMnO4ÈÜÒº²â¶¨¿ÕÆøÖм×È©µÄº¬Á¿£¬ÇëÄã²ÎÓ벢ЭÖúËûÃÇÍê³ÉÏà¹ØѧϰÈÎÎñ£®
¡¾²â¶¨Ô­Àí¡¿KMnO4£¨H+£©ÈÜҺΪǿÑõ»¯¼Á£¬¿ÉÑõ»¯¼×È©ºÍ²ÝËᣮ
4MnO4-+5HCHO+H+=4Mn2++5CO2¡ü+11H2O
2MnO4-+5H2C2O4+6H+=2Mn2++10CO2¡ü+8H2O
¡¾²â¶¨×°Öῲ¿·Ö×°ÖÃÈçͼ1Ëùʾ

[²â¶¨²½Öè]
¢ÙÓÃ
ËáʽµÎ¶¨¹Ü
ËáʽµÎ¶¨¹Ü
Á¿È¡5.00mL1.00¡Á10-3mol?L-1KMnO4ÈÜÒºÓÚÏ´ÆøÆ¿ÖУ¬²¢µÎÈ뼸µÎÏ¡H2SO4Ëữ£®
¢Ú½«1.00¡Á10-3 mol?L-1µÄ²ÝËá±ê×¼ÈÜÒºÖÃÓÚËáʽµÎ¶¨¹ÜÖб¸Óã®
¢Û´ò¿ª
a
a
£¬¹Ø±Õ
b
b
£¨Ìî¡°a¡±»ò¡°b¡±£©£¬ÓÃ×¢ÉäÆ÷³éÈ¡100mLÐÂ×°Ð޵ķ¿ÎÝÊÒÄÚ¿ÕÆø£®¹Ø±Õ
a
a
£¬´ò¿ª
b
b
£¨Ìî¡°a¡±»ò¡°b¡±£©£¬ÔÙÍƶ¯×¢ÉäÆ÷£¬½«ÆøÌåÈ«²¿ÍÆÈëËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖУ¬Ê¹Æä³ä·Ö·´Ó¦£®ÔÙÖظ´¢Û²Ù×÷4´Î£®
¢Ü½«Ï´ÆøÆ¿ÖÐÈÜҺתÒƵ½×¶ÐÎÆ¿ÖУ¨°üÀ¨Ï´µÓÒº£©£¬ÔÙÓñê×¼²ÝËáÈÜÒº½øÐе樣¬¼Ç¼µÎ¶¨ËùÏûºÄµÄ²ÝËáÈÜÒºµÄÌå»ý£®
¢ÝÔÙÖظ´ÊµÑé2´Î£¬ÊµÑéËùÏûºÄ²ÝËáÈÜÒºµÄÌå»ýƽ¾ùֵΪ12.38mL£®
¡¾½»Á÷ÌÖÂÛ¡¿
£¨1£©¼ÆËã¸Ã¾ÓÊÒÄÚ¿ÕÆøÖм×È©µÄŨ¶È
18
18
mg?m-3£¬¸Ã¾ÓÊҵļ×È©
ÊÇ
ÊÇ
£¨ÌîÊÇ»ò·ñ£©³¬±ê£®
£¨2£©Ä³Í¬Ñ§Óø÷½·¨²âÁ¿¿ÕÆøÖм×È©µÄº¬Á¿Ê±£¬Ëù²âµÃµÄÊýÖµ±Èʵ¼Êº¬Á¿µÍ£¬ÇëÄã¶ÔÆä¿ÉÄܵÄÔ­Òò£¨¼ÙÉèÈÜÒºÅäÖÆ¡¢³ÆÁ¿»òÁ¿È¡¼°µÎ¶¨ÊµÑé¾ùÎÞ´íÎó£©Ìá³öºÏÀí¼ÙÉ裺
×¢ÉäÆ÷ѹËÍÆøÌåËٶȹý¿ì¡¢×°ÖÃÆøÃÜÐԽϲµ¼Æø¹ÜÉìÈëKMnO4ÈÜÒºÖÐ̫dzµÈ
×¢ÉäÆ÷ѹËÍÆøÌåËٶȹý¿ì¡¢×°ÖÃÆøÃÜÐԽϲµ¼Æø¹ÜÉìÈëKMnO4ÈÜÒºÖÐ̫dzµÈ
£¨ÖÁÉÙ´ð³ö2ÖÖ¿ÉÄÜÐÔ£©£®
£¨3£©ÊµÑé½áÊøºó£¬¸ÃС×é³ÉÔ±ÔÚÏ໥½»Á÷µÄ¹ý³ÌÖÐÒ»ÖÂÈÏΪ£ºÊµÑé×°ÖÃÓ¦¼ÓÒԸĽø£®ÓÐͬѧÌáÒ飺¿É½«²åÈëKMnO4ÈÜÒºµÄ¹Ü×Ó϶˸ijɾßÓжà¿×µÄÇòÅÝ£¨Í¼2£©£¬ÓÐÀûÓÚÌá¸ßʵÑéµÄ׼ȷ¶È£¬ÆäÀíÓÉÊÇ
Ôö´ó¿ÕÆøÓë¸ßÃÌËá¼ØÈÜÒºµÄ½Ó´¥Ãæ»ý£¬Ê¹¿ÕÆøÖеļ×È©·´Ó¦³ä·Ö
Ôö´ó¿ÕÆøÓë¸ßÃÌËá¼ØÈÜÒºµÄ½Ó´¥Ãæ»ý£¬Ê¹¿ÕÆøÖеļ×È©·´Ó¦³ä·Ö
£®
¼×È©Òà³Æ¡°ÒÏÈ©¡±£®º¬¼×È©37¡«40%¡¢¼×´¼8%µÄË®ÈÜÒºË׳ơ°¸£¶ûÂíÁÖ¡±£®¼×È©ÊÇÖØÒªµÄÓлúºÏ³ÉÔ­ÁÏ£¬´óÁ¿ÓÃÓÚÉú²úÊ÷Ö¬¡¢ºÏ³ÉÏËά¡¢Ò©ÎͿÁϵȣ®¼×È©ÊÇÊÀ½çÎÀÉú×éÖ¯£¨WHO£©È·ÈϵÄÖ°©ÎïºÍÖ»ûÎïÖÊÖ®Ò»£®ÎÒ¹ú¹æ¶¨£ºÊÒÄÚ¼×È©º¬Á¿²»µÃ³¬¹ý0.08mg?m-3£®
£¨1£©ÏÂÁÐ˵·¨»ò×ö·¨²»ºÏÀíµÄÊÇ
 
£®£¨Ìî×Öĸ£©
a£®Óü×È©ÈÜÒº½þÅÝË®²úÆ·ÒÔ³¤Ê±¼ä±£³ÖË®²úÆ·µÄ¡°ÐÂÏÊ¡±
b£®¸Õ×°ÐÞµÄз¿Èëסǰ·¿¼äÄÚ±£³ÖÒ»¶¨Î¶Ȳ¢×¢Òâͨ·ç
c£®¶ÔÈËÌ彡¿µÓÐΣº¦µÄµõ°×¿éµÄºÏ³É·´Ó¦NaHSO3+HCHO¡úNaO-CH2-SO3HµÄ·´Ó¦ÀàÐÍÊǼӳɷ´Ó¦
d£®¸£¶ûÂíÁÖ¿ÉÓÃÓÚÖÆ×÷¶¯Îï±ê±¾£¨»ò±£´æʬÌ壩
£¨2£©Ä³Ñо¿ÐÔѧϰС×éÄâÓü×È©·¨²â¶¨³£¼ûï§Ì¬µª·ÊµÄº¬µªÁ¿[×ÊÁÏ£º4NH4++6HCHO=£¨CH2£©6N4H++3H++6H2O£¬ËùÉú³ÉµÄH+ºÍ£¨CH2£©6N4H+¿ÉÓÃNaOH±ê×¼ÈÜÒºµÎ¶¨£¬²ÉÓ÷Ó̪×÷ָʾ¼Á]£®Óü×È©·¨²â¶¨º¬µªÁ¿£¬²»ÊʺϵÄï§ÑÎÊÇ
 
£®£¨Ìî×Öĸ£©
a£®NH4HCO3 b£®£¨NH4£©2SO4 c£®NH4Cl
£¨3£©¼×´¼ÍÑÇâÊÇÖƼ×È©×î¼òµ¥µÄ¹¤Òµ·½·¨£º¾«Ó¢¼Ò½ÌÍø
·´Ó¦¢ñ£ºCH3OH£¨g£©¡úHCHO£¨g£©+H2£¨g£©£»¡÷H1=92.09kJ?mol-1£¬Æäƽºâ³£ÊýK1=3.92¡Á10-11
¼×´¼Ñõ»¯ÊÇÖƼ×È©µÄÁíÒ»ÖÖ¹¤Òµ·½·¨£¬¼´¼×´¼ÕôÆøºÍÒ»¶¨Á¿µÄ¿ÕÆøͨ¹ýAg´ß»¯¼Á²ã£¬¼×´¼¼´±»Ñõ»¯µÃµ½¼×È©£º
·´Ó¦¢ò£ºCH3OH£¨g£©+
12
O2£¨g£©¡úHCHO£¨g£©+H2O£¨g£©£»¡÷H2=-149.73kJ?mol-1£¬Æäƽºâ³£ÊýK2=4.35¡Á1029£¨ÉÏÊöÊý¾Ý¾ùΪ298.15Kϲⶨ£®£©
¢ÙÂÌÉ«»¯Ñ§Ìᳫ»¯¹¤Éú²úÓ¦Ìá¸ßÔ­×ÓÀûÓÃÂÊ£®Ô­×ÓÀûÓÃÂʱíʾĿ±ê²úÎïµÄÖÊÁ¿ÓëÉú³ÉÎï×ÜÖÊÁ¿Ö®±È£®·´Ó¦
 
£¨Ìî¡°¢ñ¡±»ò¡°¢ò¡±£©ÖƼ×È©Ô­×ÓÀûÓÃÂʸü¸ß£®´Ó·´Ó¦µÄìʱäºÍƽºâ³£ÊýKÖµ¿´£¬·´Ó¦
 
£¨Ìî¡°¢ñ¡±»ò¡°¢ò¡±£©ÖƼ×È©¸ü¼ÓÓÐÀû£®
¢Ú·´Ó¦¢ò×Ô·¢½øÐеÄÌõ¼þÊÇ
 
£®
a£®¸ßÎÂb£®µÍÎÂc£®ÈκÎÌõ¼þ¶¼×Ô·¢
¢ÛÓÒͼÊǼ״¼ÖƼ×È©Óйط´Ó¦µÄlgK£¨Æ½ºâ³£ÊýµÄ¶ÔÊýÖµ£©ËæζÈTµÄ±ä»¯£®Í¼ÖÐÇúÏߣ¨1£©±íʾµÄÊÇ·´Ó¦
 
£¨Ìî¡°¢ñ¡±»ò¡°¢ò¡±£©£®

¼×È©ÊÇÊÀ½çÎÀÉú×éÖ¯£¨WHO£©È·ÈϵÄÖ°©ÎïºÍÖ»ûÎïÖÊÖ®Ò»¡£ÎÒ¹ú¹æ¶¨£ºÊÒÄÚ¼×È©(HCHO)º¬Á¿²»µÃ³¬¹ý0.08mg¡¤m£­3¡£Ä³Ñо¿ÐÔѧϰС×éÓûÀûÓÃËáÐÔKMnO4ÈÜÒº²â¶¨¿ÕÆøÖм×È©µÄº¬Á¿£¬ÇëÄã²ÎÓ벢ЭÖúËûÃÇÍê³ÉÏà¹ØѧϰÈÎÎñ¡£

Ô­Àí£ºKMnO4 ( H£«)ÈÜҺΪǿÑõ»¯¼Á£¬¿ÉÑõ»¯¼×È©¡£

Àë×Ó·½³ÌʽÊÇ£º4MnO4¨D+5HCHO+H£«£½4Mn2£«+5CO2¡ü+11H2O ²¿·Ö×°ÖÃÈçÏÂͼËùʾ

 

²½Ö裺£¨1£©ÅäÖÆ1.000¡Á10£­4mol/LµÄKMnO4ÈÜÒº£ºµÚÒ»²½£ºÓ÷ÖÎöÌìƽ³ÆÈ¡KMnO4¹ÌÌå1.5800g£¬Åä³É0.01mol/LKMnO4ÈÜÒº£¬³ýÁËÌìƽ¡¢Ò©³×Í⣬ÐèÒªµÄ²£Á§ÒÇÆ÷ÊÇ£º        ¡¢       ¡¢

        ¡¢        ¡£µÚ¶þ²½£ºÓÃÒÆÒº¹ÜÁ¿È¡ÉÏÊöÒÑÅäÖƵÄÈÜÒº    mL£¬ÔÙÓÃÉÏÊöÓõ½µÄÒÇÆ÷ÅäÖÆ1.000¡Á10£­4mol/LµÄKMnO4ÈÜÒº¡£

£¨2£©²â¶¨Å¨¶È

¢ÙÓÃÒÆÒº¹ÜÁ¿È¡8.00mL 1.000¡Á10£­4mol¡¤L£­1KMnO4ÈÜÒºÓÚÏ´ÆøÆ¿ÖУ¬²¢µÎÈ뼸µÎÏ¡H2SO4£¬¼ÓË®20mLÏ¡Êͱ¸Óá£

¢Ú´ò¿ª       £¬¹Ø±Õ          £¨Ìî¡°a¡±»ò¡°b¡±£©£¬ÓÃ×¢ÉäÆ÷³éÈ¡100mLÐÂ×°Ð޵ķ¿ÎÝÊÒÄÚ¿ÕÆø¡£¹Ø±Õ      £¬´ò¿ª      £¨Ìî¡°a¡±»ò¡°b¡±£©£¬ÔÙÍƶ¯×¢ÉäÆ÷£¬½«ÆøÌåÈ«²¿ÍÆÈëËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖУ¬Ê¹Æä³ä·Ö·´Ó¦¡£Öظ´µ½µÚ5´Î×¢ÉäÆ÷ÍÆÖÁÒ»°ëʱKMnO4ÈÜÒº¸ÕºÃÍÊÉ«£¨MnO4£­¡úMn2+£©¡£

ÌÖÂÛ£º

£¨1£©¼ÆËã¸Ã¾ÓÊÒÄÚ¿ÕÆøÖм×È©µÄŨ¶È       mg¡¤m£­3£¬¸Ã¾ÓÊҵļ×È©          £¨ÌîÊÇ»ò·ñ£©³¬±ê¡£

£¨2£©Ä³Í¬Ñ§Óø÷½·¨²âÁ¿¿ÕÆøÖм×È©µÄº¬Á¿Ê±£¬Ëù²âµÃµÄÊýÖµ±Èʵ¼Êº¬Á¿µÍ£¬Ì½¾¿Æä¿ÉÄܵÄÔ­Òò£¨¼ÙÉèÈÜÒºÅäÖÆÎÞ´íÎ󣩠                                      ¡£

£¨3£©ÊµÑé½áÊøºó£¬¸ÃС×é³ÉÔ±ÔÚÏ໥½»Á÷µÄ¹ý³ÌÖÐÒ»ÖÂÈÏΪ£ºÊµÑé×°ÖÃÓ¦¼ÓÒԸĽø£ºÓÐͬѧÌáÒ飬¿É½«²åÈëKMnO4ÈÜÒºµÄ¹Ü×Ó϶˸ijɾßÓжà¿×µÄÇòÅÝ£¨ÓÒͼ£©£¬ÓÐÀûÓÚÌá¸ßʵÑéµÄ׼ȷ¶È£¬ÆäÀíÓÉÊÇ                                    ¡£

 

 

(8·Ö)(2011¡¤Î÷ÄþÄ£Äâ)¼×È©ÊÇÊÀ½çÎÀÉú×éÖ¯(WHO)È·ÈϵÄÖ°©ÎïÖʺÍÖ»ûÎïÖÊÖ®Ò»¡£ÎÒ¹ú¹æ¶¨£ºÊÒÄÚ¼×È©º¬Á¿²»µÃ³¬¹ý0.08mg¡¤m£­3¡£Ä³Ñо¿ÐÔѧϰС×é´òËãÀûÓÃËáÐÔKMnO4ÈÜÒº²â¶¨ÐÂ×°ÐÞ·¿ÎÝÄڵĿÕÆøÖм×È©µÄº¬Á¿£¬ÇëÄã²ÎÓ벢ЭÖúËûÃÇÍê³ÉÏà¹ØµÄѧϰÈÎÎñ¡£
²â¶¨Ô­Àí£º
ËáÐÔKMnO4ΪǿÑõ»¯¼Á£¬¿ÉÑõ»¯¼×È©ºÍ²ÝËᣬÆä·´Ó¦·½³ÌʽÈçÏÂËùʾ£º
4MnO4£­£«5HCHO£«12H£«===4Mn2£«£«5CO2¡ü£«11H2O
2MnO4£­£«5H2C2O4£«6H£«===2Mn2£«£«10CO2¡ü£«8H2O
²â¶¨×°Öãº
²¿·Ö×°ÖÃÈçÏÂͼËùʾ£º

²â¶¨²½Ö裺
(1)ÓÃ______________Á¿È¡5.00 mL 1.00¡Á10£­3mol¡¤L£­1KMnO4ÈÜÒº£¬×¢ÈëÏ´ÆøÆ¿ÖУ¬²¢µÎÈ뼸µÎÏ¡H2SO4£¬¼ÓË®20 mLÏ¡ÊÍ£¬±¸Óá£
(2)½«1.00¡Á10£­3 mol¡¤L£­1µÄ²ÝËá±ê×¼ÈÜÒºÖÃÓÚËáʽµÎ¶¨¹ÜÖб¸Óá£
(3)´ò¿ªa£¬¹Ø±Õb£¬ÓÃ×¢ÉäÆ÷³éÈ¡100 mLÐÂ×°ÐÞ·¿ÎÝÄڵĿÕÆø¡£¹Ø±Õ________£¬´ò¿ª________(Ìî¡°a¡±»ò¡°b¡±)£¬ÔÙÍƶ¯×¢ÉäÆ÷£¬½«ÆøÌåÈ«²¿ÍÆÈëËáÐÔ¸ßÃÌËá¼ØÈÜÒºÖУ¬Ê¹Æä³ä·Ö·´Ó¦¡£ÔÙÖظ´4´Î¡£
(4)½«Ï´ÆøÆ¿ÖеÄÈÜҺתÒƵ½×¶ÐÎÆ¿ÖÐ(°üÀ¨Ï´µÓÒº)£¬ÔÙÓòÝËá±ê×¼ÈÜÒº½øÐе樣¬¼Ç¼µÎ¶¨ËùÏûºÄµÄ²ÝËá±ê×¼ÈÜÒºµÄÌå»ý¡£
(5)ÔÙÖظ´ÊµÑé2´Î(ÿ´ÎËùÈ¡µÄ¸ßÃÌËá¼ØÈÜÒºµÄÌå»ý¾ùΪ5.00 mL)¡£3´ÎʵÑéËùÏûºÄµÄ²ÝËá±ê×¼ÈÜÒºµÄÌå»ýƽ¾ùֵΪ12.38 mL¡£
½»Á÷ÌÖÂÛ£º
(1)¼ÆËã¸ÃÐÂ×°ÐÞ·¿ÎÝÄڵĿÕÆøÖм×È©µÄŨ¶ÈΪ________mg¡¤m£­3£¬¸ÃÐÂ×°ÐÞ·¿ÎÝÄڵļ×È©________(Ìî¡°ÊÇ¡±»ò¡°·ñ¡±)³¬±ê£»
(2)ijͬѧÓø÷½·¨²âÁ¿¿ÕÆøÖм×È©µÄº¬Á¿Ê±£¬Ëù²âµÃµÄÊýÖµ±Èʵ¼Êº¬Á¿µÍ£¬ÇëÄã¶ÔÆä¿ÉÄܵÄÔ­Òò(¼ÙÉè³ÆÁ¿»òÁ¿È¡¡¢ÈÜÒºÅäÖƼ°µÎ¶¨ÊµÑé¾ùÎÞ´íÎó)Ìá³öºÏÀí¼ÙÉ裺______________¡¢______________(ÖÁÉÙ´ð³öÁ½ÖÖ¿ÉÄÜÐÔ)£»
(3)ʵÑé½áÊøºó£¬¸ÃС×é³ÉÔ±ÔÚÏ໥½»Á÷µÄ¹ý³ÌÖÐÒ»ÖÂÈÏΪ£º
¢ÙʵÑéÔ­Àí¿ÉÒÔ¼ò»¯
ʵÑéÖпɲ»ÓòÝËá±ê×¼ÈÜÒºµÎ¶¨£¬¿É¶à´ÎÖ±½Ó³éÈ¡ÐÂ×°ÐÞ·¿ÎÝÄڵĿÕÆø£¬ÔÙÍÆË͵½Ï´ÆøÆ¿ÖУ¬Ö±ÖÁ_________________________________________________________£»

¢ÚʵÑé×°ÖÃÓ¦¼ÓÒԸĽø
¿É½«²åÈëËáÐÔKMnO4ÈÜÒºÖеĵ¼¹Ü϶˸ijɾßÓжà¿×µÄÇòÅÝ(ÈçͼËùʾ)£¬ÓÐÀûÓÚÌá¸ßʵÑéµÄ׼ȷ¶È£¬ÆäÀíÓÉÊÇ___________________________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø