ÌâÄ¿ÄÚÈÝ

Ϊ²â¶¨Ä³H2C2O4ÈÜÒºµÄŨ¶È£¬È¡25£®00mL¸ÃÈÜÒºÓÚ׶ÐÎÆ¿ÖУ¬¼ÓÈëÊÊÁ¿Ï¡H2SO4ºó£¬ÓÃŨ¶ÈΪc mol/L KMnO4±ê×¼ÈÜÒºµÎ¶¨¡£µÎ¶¨Ô­ÀíΪ£º
2KMnO4+5H2C2O4+3H2SO4=K2SO4+10CO2¡ü+2MnSO4+8H2O
£¨1£©µÎ¶¨Ê±£¬KMnO4ÈÜҺӦװÔÚ                £¨Ìî¡°ËáʽµÎ¶¨¹Ü¡±»ò¡°¼îʽµÎ¶¨¹Ü¡± £©ÖУ¬´ïµ½µÎ¶¨ÖÕµãµÄÏÖÏóΪ                                                             ¡£
£¨2£©ÈôµÎ¶¨Ê±£¬Ã»Óñê׼ҺϴµÓµÎ¶¨¹Ü£¬»áʹµÃ²ÝËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È_              _        (Ìî¡°Æ«¸ß¡±¡°Æ«µÍ¡±¡°ÎÞÓ°Ï족)
£¨3£©ÈôµÎ¶¨Ê±£¬·´Ó¦Ç°ºóµÄÁ½´Î¶ÁÊý·Ö±ðΪaºÍb£¬ÔòʵÑé²âµÃËùÅä²ÝËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ           mol/L¡£
£¨4£©ÔÚ¸ÃÈÜÒºÓëKOHÈÜÒº·´Ó¦ËùµÃµÄ0£®1 mol/L KHC2O4ÈÜÒºÖУ¬c(C2O42-)£¾c£¨H2C2O4£©£¬ÏÂÁйØϵÕýÈ·µÄÊÇ               ¡£

A£®c£¨K+£©+c£¨H+£©=c(HC2O4-)+c(OH-)+c(C2O42-)
B£®c(HC2O4-)+ c (C2O42-)+ c£¨H2C2O4£©=0£®1mol/L
C£®c(H+)£¼c£¨OH-£©
D£®c£¨K+£©=c£¨H2C2O4£©+c(HC2O4-)+c(C2O42-)

£¨1£©ËáʽµÎ¶¨¹Ü  (2·Ö)      µÎ¼Ó×îºóÒ»µÎKMnO4ºó£¬×¶ÐÎÆ¿ÖÐÈÜÒº´ÓÎÞÉ«Í»±äΪ×ϺìÉ«£¬ÇÒ°ë·ÖÖÓÄÚÑÕÉ«²»»Ö¸´¡£  (2·Ö)
£¨2£©Æ«¸ß  (2·Ö)
£¨3£©0£®1c(b-a)  (3·Ö)
£¨4£©BD  (3·Ö)

½âÎöÊÔÌâ·ÖÎö£º1£©ÒòΪKMnO4¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬»á¸¯Ê´Ï𽺹ܣ¬¹ÊÓ¦ÓÃËáʽµÎ¶¨¹ÜÊ¢×°£¬¹Ê´ð°¸ÎªËáʽµÎ¶¨¹Ü¡£ÒòKMnO4ÈÜÒº×ÔÉíµÄÑÕÉ«×÷Ϊָʾ¼ÁÅжϵζ¨ÖÕµãʱ£¬ÔٵμÓKnO4ÈÜҺʱ£¬ÈÜÒº½«ÓÉÎÞÉ«±äΪ×ÏÉ«£¬¹Ê´ð°¸Îª£ºµ±µÎÈë×îºóÒ»µÎKMnO4ÈÜҺʱ£¬ÈÜÒºÓÉÎÞÉ«±äΪ×ÏÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊÉ«£¬¼´´ïµÎ¶¨Öյ㣻£¨2£©ÈôµÎ¶¨Ê±£¬Ã»Óñê׼ҺϴµÓµÎ¶¨¹Ü£¬Ï൱Óڰѱê×¼ÈÜÒº½øÐÐÁËÏ¡ÊÍ£¬ÕâÑùÏûºÄµÄµÄ±ê×¼ÒºÌå»ýÆ«¸ß£¬»áʹµÃ²ÝËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈÆ«¸ß¡££¨3£©ÈôµÎ¶¨Ê±£¬·´Ó¦Ç°ºóµÄÁ½´Î¶ÁÊý·Ö±ðΪaºÍb£¬Ôò±ê×¼ÒºµÄÌå»ýΪ(b-a)ml£¬ÇóµÃ±ê×¼ÒºKMnO4µÄÎïÖʵÄÁ¿Îª£ºc(b-a)/1000mol¡£¸ù¾Ý2KMnO4-----5H2C2O4¹Øϵʽ¿ÉÇóµÃËùÅä²ÝËáÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0£®1c(b-a)mol/L¡££¨4£©ÓÉc(C2O42-)£¾c£¨H2C2O4£©£¬ËµÃ÷HC2O4-µçÀë³Ì¶È´óÓÚË®½â³Ì¶È¡£  A£® c£¨K+£©+c£¨H+£©=c(HC2O4-)+c(OH-)+c(C2O42-)µçºÉÊغã¹ØϵʽÊéд´íÎó£»  B£®c (HC2O4-)+ c (C2O42-)+ c£¨H2C2O4£©=0£®1mol/LÊÇÎïÁÏÊغ㣬ÕýÈ·£» HC2O4-µçÀë³Ì¶È´óÓÚË®½â³Ì¶È£¬¹Ê C£® c(H+)£¼c£¨OH-£©´íÎó£»  D£®c£¨K+£©=c£¨H2C2O4£©+c(HC2O4-)+c(C2O42-)ÊÇÊغã±í´ïʽ£¬ÕýÈ·£»
¿¼µã£º±¾Ì⿼²éÁËÖк͵ζ¨ÊµÑ飬ÄѶÈÊÊÖУ¬×¢ÒâÕÆÎÕ²ÝËẬÁ¿µÄ¼ÆËã·½·¨¼°Öк͵ζ¨ÖеÄÎó²î·ÖÎö¼´¿É½â´ð¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ʵÑéÊÒÄ£Äâ»ØÊÕij·Ï¾Éº¬Äø´ß»¯¼Á£¨Ö÷Òª³É·ÖΪNiO£¬Áíº¬Fe2O3¡¢CaO¡¢CuO¡¢BaOµÈ£©Éú²úNi2O3¡£Æ乤ÒÕÁ÷³ÌΪ£º

  
ͼ¢ñ                                        Í¼¢ò
£¨1£©¸ù¾Ýͼ¢ñËùʾµÄXÉäÏßÑÜÉäͼÆ×£¬¿ÉÖª½þ³öÔüº¬ÓÐÈýÖÖÖ÷Òª³É·Ö£¬ÆäÖС°ÎïÖÊX¡±Îª        ¡£Í¼¢ò±íʾÄøµÄ½þ³öÂÊÓëζȵĹØϵ£¬µ±½þ³öζȸßÓÚ70¡æʱ£¬ÄøµÄ½þ³öÂʽµµÍ£¬½þ³öÔüÖÐNi(OH)2º¬Á¿Ôö´ó£¬ÆäÔ­ÒòÊÇ                              ¡£
£¨2£©¹¤ÒÕÁ÷³ÌÖС°¸±²úÆ·¡±µÄ»¯Ñ§Ê½Îª             ¡£
£¨3£©ÒÑÖªÓйØÇâÑõ»¯Î↑ʼ³ÁµíºÍ³ÁµíÍêÈ«µÄpHÈçÏÂ±í£º

ÇâÑõ»¯Îï
Fe(OH)3
Fe(OH)2
Ni(OH)2
¿ªÊ¼³ÁµíµÄpH
1.5
6.5
7.7
³ÁµíÍêÈ«µÄpH
3.7
9.7
9.2
 
²Ù×÷BÊÇΪÁ˳ýÈ¥ÂËÒºÖеÄÌúÔªËØ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈçÏÂʵÑé·½°¸£ºÏò²Ù×÷AËùµÃµÄÂËÒºÖмÓÈëNaOHÈÜÒº£¬µ÷½ÚÈÜÒºpHΪ3.7¡«7.7£¬¾²Ö㬹ýÂË¡£Çë¶Ô¸ÃʵÑé·½°¸½øÐÐÆÀ¼Û£º                £¨ÈôÔ­·½°¸ÕýÈ·£¬Çë˵Ã÷ÀíÓÉ£»ÈôÔ­·½°¸´íÎó£¬Çë¼ÓÒÔ¸ÄÕý£©¡£
£¨4£©²Ù×÷CÊÇΪÁ˳ýÈ¥ÈÜÒºÖеÄCa2+£¬Èô¿ØÖÆÈÜÒºÖÐF£­Å¨¶ÈΪ3¡Á10£­3 mol¡¤L£­1£¬ÔòCa2+µÄŨ¶ÈΪ             ______mol¡¤L£­1¡££¨³£ÎÂʱCaF2µÄÈܶȻý³£ÊýΪ2.7¡Á10£­11£©
£¨5£©µç½â²úÉú2NiOOH¡¤H2OµÄÔ­Àí·ÖÁ½²½£º¢Ù¼îÐÔÌõ¼þÏÂCl£­ÔÚÑô¼«±»Ñõ»¯ÎªClO£­£»¢ÚNi2+±»ClO£­Ñõ»¯²úÉú2NiOOH¡¤H2O³Áµí¡£µÚ¢Ú²½·´Ó¦µÄÀë×Ó·½³ÌʽΪ                   ¡£

´×ËáºÍÑÎËáÊÇÖÐѧ»¯Ñ§Öг£¼ûµÄËᣬÔÚÒ»¶¨Ìõ¼þÏ£¬CH3COOHÈÜÒºÖдæÔÚµçÀëƽºâ£ºCH3COOHCH3COO-+H+  ¦¤H£¾0¡£
£¨1£©³£ÎÂÏ£¬ÔÚ pH =5µÄÏ¡´×ËáÈÜÒºÖУ¬c(CH3COO-)£½____________(ÁÐʽ£¬²»±Ø»¯¼ò)£»ÏÂÁз½·¨ÖУ¬¿ÉÒÔʹ0£®10 mol¡¤L-1 CH3COOHµÄµçÀë³Ì¶ÈÔö´óµÄÊÇ______?
a£®¼ÓÈëÉÙÁ¿0£®10 mol¡¤L-1µÄÏ¡ÑÎËá      b£®¼ÓÈÈCH3COOHÈÜÒº
c£®¼ÓˮϡÊÍÖÁ0£®010 mol¡¤L-1           d£®¼ÓÈëÉÙÁ¿±ù´×Ëá
e£®¼ÓÈëÉÙÁ¿ÂÈ»¯ÄƹÌÌå                 f£®¼ÓÈëÉÙÁ¿0£®10 mol¡¤L-1µÄNaOHÈÜÒº
£¨2£©½«µÈÖÊÁ¿µÄпͶÈëµÈÌå»ýÇÒpH¾ùµÈÓÚ3µÄ´×ËáºÍÑÎËáÈÜÒºÖУ¬¾­¹ý³ä·Ö·´Ó¦ºó£¬·¢ÏÖÖ»ÔÚÒ»ÖÖÈÜÒºÖÐÓÐп·ÛÊ£Ó࣬ÔòÉú³ÉÇâÆøµÄÌå»ý£ºV(ÑÎËá)_________V(´×Ëá)£¬·´Ó¦µÄ×î³õËÙÂÊΪ£ºv(ÑÎËá)_________v(´×Ëá)¡£(Ìîд¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±)
£¨3£©Ä³Í¬Ñ§ÓÃ0£®1000mol/LNaOHÈÜÒº·Ö±ðµÎ¶¨20£®00mL 0£®1000mol/LHClºÍ20£®00mL0£®1000mol/L CH3COOH£¬µÃµ½ÈçͼËùʾÁ½ÌõµÎ¶¨ÇúÏߣ¬ÇëÍê³ÉÓйØÎÊÌ⣺

¢ÙNaOHÈÜÒºµÎ¶¨CH3COOHÈÜÒºµÄÇúÏßÊÇ               £¨Ìͼ1¡±»ò¡°Í¼2¡±£©£»
¢Úa£½                mL¡£
£¨4£©³£ÎÂÏ£¬½«0£®1 mol/LÑÎËáºÍ0£®1 mol/L´×ËáÄÆÈÜÒº»ìºÏ£¬ËùµÃÈÜҺΪÖÐÐÔ£¬Ôò»ìºÏÈÜÒºÖи÷Àë×ÓµÄŨ¶È°´ÓÉ´óµ½Ð¡ÅÅÐòΪ_______________________________¡£
£¨5£©ÒÑÖª£º90¡æʱ£¬Ë®µÄÀë×Ó»ý³£ÊýΪKw = 3£®80¡Á10-13£¬ÔÚ´ËζÈÏ£¬½«pH=3µÄÑÎËáºÍ
pH = 11µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏ£¬Ôò»ìºÏÈÜÒºÖеÄc(H+)=____________(±£ÁôÈýλÓÐЧÊý×Ö)mol/L¡£

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

¾«Ó¢¼Ò½ÌÍø