ÌâÄ¿ÄÚÈÝ

12£®·ÏǦÐîµç³ØÁ¿¼±ËÙÔö¼ÓËùÒýÆðµÄǦÎÛȾÈÕÒæÑÏÖØ£®¹¤ÒµÉÏ´Ó·ÏǦÐîµç³ØµÄǦ¸à»ØÊÕǦµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

ÒÑÖª£ºKsp£¨PbSO4£©=1.6¡Á10-5£¬Ksp£¨PbCO3£©=3.3¡Á10-14£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³ö²½Öè¢ÙÖÐPbSO4ת»¯ÎªPbCO3¹ý³ÌµÄƽºâ³£Êý±í´ïʽK=$\frac{c£¨S{{O}_{4}}^{2-}£©}{c£¨C{{O}_{3}}^{2-}£©}$£¬ÎªÌá¸ß²½Öè¢ÙµÄ·´Ó¦ËÙÂʺÍǦ½þ³öÂÊ£¬ÄãÈÏΪ¿É²ÉÈ¡µÄÁ½Ìõ´ëÊ©Êdzä·Ö½Á°è¡¢Êʵ±Éý¸ßζȣ®
£¨2£©²½Öè¢ÙÖз¢ÉúµÄÑõ»¯»¹Ô­·´Ó¦µÄÀë×Ó·½³ÌʽΪPbO2+SO32-+H2O=PbSO4+2OH-£®
£¨3£©Ð´³ö²½Öè¢ÜÓöèÐԵ缫µç½âµÄÒõ¼«·´Ó¦Ê½Pb2++2e-=Pb£®
£¨4£©PbO2ÔÚ¼ÓÈȹý³Ì·¢Éú·Ö½âµÄÊ§ÖØÇúÏßÈçͼ1Ëùʾ£¬ÒÑÖªÊ§ÖØÇúÏßÉϵÄaµãΪÑùÆ·Ê§ÖØ4.0%£¨¼´$\frac{ÑùÆ·ÆðʼÖÊÁ¿-aµã¹ÌÌåÖÊÁ¿}{ÑùÆ·ÆðʼÖÊÁ¿}$¡Á100%£©µÄ²ÐÁô¹ÌÌ壬Èôaµã¹ÌÌå×é³É±íʾΪPbOx£¬¼ÆËãx=1.4£®
£¨5£©Ç¦µÄ¼Ó¹¤Í¬Ñù»áʹˮÌåÖÐÖØ½ðÊôǦµÄº¬Á¿Ôö´óÔì³ÉÑÏÖØÎÛȾ£®Ë®ÈÜÒºÖÐǦµÄ´æÔÚÐÎ̬Ö÷ÒªÓÐPb2+¡¢Pb£¨OH£©+¡¢Pb£¨OH£©2¡¢Pb£¨OH£©3-¡¢Pb£¨OH£©42-£®¸÷ÐÎ̬µÄŨ¶È·ÖÊý¦ÁËæÈÜÒºpH±ä»¯µÄ¹ØÏµÈçͼ2Ëùʾ£¬Ä³¿ÎÌâ×éÖÆ±¸ÁËÒ»ÖÖÐÂÐÍÍÑǦ¼Á£¬ÄÜÓÐЧȥ³ýË®ÖеĺÛÁ¿Ç¦£¬ÊµÑé½á¹ûÈç±í

 Àë×Ó/£¨mol£®L-1£© Pb2+ Ca2+ Fe3+ Mn2+ Cl-
 ´¦ÀíǰŨ¶È 0.100 29.8 0.120 0.087 51.9
 ´¦ÀíºóŨ¶È 0.004 22.6 0.040 0.053 49.9
ÔòÉϱíÖгýPb2+Í⣬¸ÃÍÑǦ¼Á¶ÔÆäËûÀë×ÓµÄÈ¥³ýЧ¹û×îºÃµÄÊÇFe3+£¬Èç¹û¸ÃÍÑǦ¼Á£¨ÓÃEH±íʾ£©ÍÑǦ¹ý³ÌÖÐÖ÷Òª·¢ÉúµÄ·´Ó¦Îª£º2EH£¨s£©+Pb2+?E2Pb£¨s£©+2H+£®ÔòÍÑǦµÄ×îºÏÊÊpH·¶Î§ÎªB£¨Ìî×Öĸ£©£®
A£®4¡«5     B£®6¡«7    C£®9¡«10    D£®11¡«12£®

·ÖÎö £¨1£©PbSO4ת»¯ÎªPbCO3·½³ÌʽΪ£ºPbSO4£¨s£©+CO32-£¨aq£©?PbCO3£¨s£©+SO42-£¨aq£©£¬¹ÌÌåºÍ´¿ÒºÌåµÄŨ¶ÈÊǶ¨Öµ²»³öÏÖÔÚ±í´ïʽÖУ»
³ä·Ö½Á°è£¬Êʵ±Éý¸ßζȣ¬Ôö´ó̼ËáÄÆµÄŨ¶ÈµÈ¿ÉÒÔ¼Ó¿ì·´Ó¦ËÙÂÊ£»
£¨2£©²½Öè¢ÙPbO2ÓëÑÇÁòËáÄÆ·¢ÉúÑõ»¯·´Ó¦Éú³ÉPbSO4£¬¸ù¾ÝµçºÉÊØºã£¬¿ÉÖª»¹Éú³ÉNaOH£»
£¨3£©Òõ¼«·¢Éú»¹Ô­·´Ó¦£¬Pb2+»ñµÃµç×ÓÉú³ÉPb£»
£¨4£©¼ÓÈÈ·Ö½â¹ý³Ì¹ÌÌåÖÐPbÔªËØÖÊÁ¿²»±ä£»
£¨5£©Àë×ÓÈ¥³ýÂÊ=$\frac{Àë×ÓŨ¶È±ä»¯Á¿}{Àë×ÓÆðʼŨ¶È}$¡Á100%£¬È¥³ýÂÊÔ½´ó£¬È¥³ýЧ¹ûÔ½ºÃ£»
²Î¼Ó·´Ó¦µÄÊÇPb2+£¬ÓÉͼÏó¿ÉÖª£¬Ñ¡ÔñPHҪʹǦÒÔPb2+ÐÎʽ´æÔÚ£¬ËáÐÔ̫ǿ£¬Õý·´Ó¦»á±»ÒÖÖÆ£®

½â´ð ½â£º£¨1£©PbSO4ת»¯ÎªPbCO3·½³ÌʽΪ£ºPbSO4£¨s£©+CO32-£¨aq£©?PbCO3£¨s£©+SO42-£¨aq£©£¬Æ½ºâ³£ÊýK=$\frac{c£¨S{{O}_{4}}^{2-}£©}{c£¨C{{O}_{3}}^{2-}£©}$£¬ÎªÁ˼ӿ췴ӦËÙÂʺÍǦ½þ³öÂÊ£¬¿ÉÒÔ²ÉÈ¡´ëÊ©ÓУº³ä·Ö½Á°è£»Êʵ±Éý¸ßζȣ»Ôö´ó̼ËáÄÆºÍÑÇÁòËáÄÆµÄŨ¶ÈµÈ£¬
¹Ê´ð°¸Îª£º$\frac{c£¨S{{O}_{4}}^{2-}£©}{c£¨C{{O}_{3}}^{2-}£©}$£»³ä·Ö½Á°è£»Êʵ±Éý¸ßζȣ»Ôö´ó̼ËáÄÆºÍÑÇÁòËáÄÆµÄŨ¶È£¨ÈÎдÁ½µã£©£»
£¨2£©²½Öè¢ÙPbO2ÓëÑÇÁòËáÄÆ·¢ÉúÑõ»¯·´Ó¦Éú³ÉPbSO4£¬¸ù¾ÝµçºÉÊØºã£¬¿ÉÖª»¹Éú³ÉNaOH£¬·´Ó¦Àë×Ó·½³ÌΪ£ºPbO2+SO32-+H2O=PbSO4+2OH-£¬
¹Ê´ð°¸Îª£ºPbO2+SO32-+H2O=PbSO4+2OH-£»
£¨3£©Òõ¼«·¢Éú»¹Ô­·´Ó¦£¬Pb2+»ñµÃµç×ÓÉú³ÉPb£¬Òõ¼«µç¼«·´Ó¦Ê½£ºPb2++2e-=Pb£¬
¹Ê´ð°¸Îª£ºPb2++2e-=Pb£»
£¨4£©¼ÓÈÈ·Ö½â¹ý³Ì¹ÌÌåÖÐPbÔªËØÖÊÁ¿²»±ä£¬Ôò100¡Á$\frac{207}{207+32}$=96¡Á$\frac{207}{207+16x}$£¬½âµÃx=1.4£¬
¹Ê´ð°¸Îª£º1.4£»
£¨5£©Fe3+È¥³ýÂÊΪ$\frac{0.12-0.04}{0.12}$¡Á100%=67%£¬Ca2+È¥³ýÂÊΪ$\frac{29.8-22.6}{29.8}$¡Á100%=24%£¬Mn2+È¥³ýÂÊΪ$\frac{0.087-0.053}{0.087}$¡Á100%=39%£¬Cl-È¥³ýÂÊΪ$\frac{51.9-49.9}{51.9}$¡Á100%=3.9%£¬Ôò³ýPb2+Í⣬¸ÃÍÑǦ¼Á¶ÔÆäËüÀë×ÓµÄÈ¥³ýЧ¹û×îºÃµÄÊÇFe3+£¬
·´Ó¦Îª2EH£¨s£©+Pb2+?E2Pb£¨s£©+2H+£¬²Î¼Ó·´Ó¦µÄÊÇPb2+£¬ÓÉͼÏó¿ÉÖª£¬Ñ¡ÔñPHҪʹǦÒÔPb2+ÐÎʽ´æÔÚ£¬ËáÐÔ̫ǿ£¬Õý·´Ó¦±»ÒÖÖÆ£¬È¥³ýЧ¹û½µµÍ£¬ÔòÍÑǦʱ×îºÏÊʵÄpHÔ¼6¡«7£¬
¹Ê´ð°¸Îª£ºB£®

µãÆÀ ±¾ÌâÒÔ´Ó·ÏǦÐîµç³ØµÄǦ¸à»ØÊÕÇ¦ÎªÔØÌ壬¿¼²éÐÅÏ¢»ñÈ¡ÓëÇ¨ÒÆÔËÓá¢Ä°Éú·½³ÌʽµÄÊéд¡¢·´Ó¦ËÙÂÊÓ°ÏìÒòËØ¡¢µç½âÔ­Àí¡¢¶ÔÊý¾ÝÓëͼÏóµÄ·ÖÎö´¦ÀíµÈ£¬ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
3£®Ä³»¯Ñ§ÊµÑéС×éÏëÒªÁ˽âÊг¡ÉÏËùÊÛʳÓð״ף¨Ö÷ÒªÊÇ´×ËáµÄË®ÈÜÒº£©µÄËá¶È£¬ÏÖ´ÓÊг¡ÉÏÂòÀ´Ò»Æ¿Ä³Æ·ÅÆÊ³Óð״ף¬ÓÃʵÑéÊÒ±ê×¼NaOHÈÜÒº¶ÔÆä½øÐе樣®Ï±íÊÇ4ÖÖ³£¼ûָʾ¼ÁµÄ±äÉ«·¶Î§£º
ָʾ¼ÁʯÈï¼×»ù³È¼×»ùºì·Ó̪
±äÉ«·¶Î§£¨pH£©5.0¡«8.03.1¡«4.44.4¡«6.28.2¡«10.0
£¨1£©¸ÃʵÑéӦѡÓ÷Ó̪×÷ָʾ¼Á£¬Ïò×¶ÐÎÆ¿ÖÐÒÆÈ¡Ò»¶¨Ìå»ýµÄ°×´×ËùÓõÄÒÇÆ÷ÊÇËáʽµÎ¶¨¹Ü£®
£¨2£©Èçͼ±íʾ50mL£¬µÎ¶¨¹ÜÖÐÒºÃæµÄλÖã¬ÈôAÓëC¿Ì¶È¼äÏà²î1mL£¬
A´¦µÄ¿Ì¶ÈΪ25£¬µÎ¶¨¹ÜÖÐÒºÃæ¶ÁÊýӦΪ25.40mL£®£¬´ËʱµÎ¶¨¹ÜÖÐÒºÌåµÄÌå»ý´óÓÚ24.60mL£®
£¨3£©ÎªÁ˼õСʵÑéÎó²î£¬¸Ãͬѧһ¹²½øÐÐÁËÈý´ÎʵÑ飬¼ÙÉèÿ´ÎËùÈ¡°×´×Ìå»ý¾ùΪVmL£¬NaOH±ê×¼ÀìŨ¶ÈΪc mol•L-1£¬Èý´ÎʵÑé½á¹û¼Ç¼ÈçÏ£º
ʵÑé´ÎÊýµÚÒ»´ÎµÚ¶þ´ÎµÚÈý´Î
ÏûºÄNaOHÈÜÒºÌå»ý/mL26.0225.3225.28
´ÓÉϱí¿ÉÒÔ¿´³ö£¬µÚÒ»´ÎʵÑéÖмǼÏûºÄNaOHÈÜÒºµÄÌå»ýÃ÷ÏÔ¶àÓÚºóÁ½´Î£¬ÆäÔ­Òò¿ÉÄÜÊÇBC£®
A£®ÊµÑé½áÊøÊ±¸©Êӿ̶ÈÏß¶ÁÈ¡µÎ¶¨ÖÕµãʱNaOHÈÜÒºµÄÌå»ý
B£®µÎ¶¨Ç°µÎ¶¨¹Ü¼â×ìÓÐÆøÅÝ£¬µÎ¶¨½áÊøÎÞÆøÅÝ
C£®Ê¢×°±ê×¼ÒºµÄµÎ¶¨¹ÜװҺǰÓÃÕôÁóË®ÈóÏ´¹ý£¬Î´Óñê×¼ÒºÈóÏ´
D£®µÚÒ»´ÎµÎ¶¨ÓõÄ×¶ÐÎÆ¿Óôý×°ÒºÈóÏ´¹ý£¬ºóÁ½´ÎδÈóÏ´
E£®µÎ¼ÓNaOHÈÜÒº¹ý¿ì£¬Î´³ä·ÖÕñµ´£¬¸Õ¿´µ½ÈÜÒº±äÉ«£¬Á¢¿ÌÍ£Ö¹µÎ¶¨
£¨4£©¸ù¾ÝËù¸øÊý¾Ý£¬Ð´³ö¼ÆËã¸Ã°×´×ÖÐ×ÜËá¶ÈµÄ±í´ïʽ£¨²»±Ø»¯¼ò£©£º$\frac{\frac{25.28+25.32}{2}¡Ác¡Á0.1¡Á60}{2V}$£®

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

¾«Ó¢¼Ò½ÌÍø