ÌâÄ¿ÄÚÈÝ

11£®ÊÒÎÂÏ£¬½«0.4mol/L HAÈÜÒººÍ0.2mol/LNaOHÈÜÒºµÈÌå»ý»ìºÏ£¨ºöÂÔ»ìºÏʱÈÜÒºÌå»ýµÄ±ä»¯£©²âµÃ»ìºÏÈÜÒºµÄpH=5£¬ÔòÏÂÁÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©=1¡Á10-5mol/L
B£®c£¨A-£©+c£¨HA£©=2c £¨Na+£©=0.4 mol/L
C£®HAÈÜÒºÖÐ$\frac{c£¨{A}^{-}£©}{c£¨HA£©•c£¨O{H}^{-}£©}$ÓëÉÏÊö»ìºÏÈÜÒºÖÐ$\frac{c£¨{A}^{-}£©}{c£¨HA£©•c£¨O{H}^{-}£©}$ÏàµÈ
D£®c£¨A-£©-c£¨HA£©=2 c £¨H+£©-c £¨OH-£©

·ÖÎö ½«0.4mol/L HAÈÜÒººÍ0.2mol/LNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ÈÜÒºÖÐÈÜÖÊΪµÈÁ¿µÄNaA¡¢HA£¬²âµÃ»ìºÏÈÜÒºµÄpH=5£¬HAµçÀë´óÓÚNaAË®½â£¬
A£®µçÀëÏÔËáÐÔ£¬ÒÖÖÆË®µÄµçÀ룻
B£®»ìºÏºóÌå»ý±äΪԭÀ´µÄ2±¶£¬½áºÏÎïÁÏÊØºã·ÖÎö£»
C.$\frac{c£¨{A}^{-}£©}{c£¨HA£©•c£¨O{H}^{-}£©}$=$\frac{c£¨{A}^{-}£©}{c£¨HA£©•\frac{Kw}{c£¨{H}^{+}£©}}$=$\frac{Ka}{Kw}$£»
D£®ÓÉÎïÁÏÊØºã¿ÉÖª£¬c£¨A-£©+c£¨HA£©=2c £¨Na+£©£¬ÓɵçºÉÊØºã¿ÉÖª£¬c£¨A-£©+c£¨OH-£©=c £¨H+£©+c £¨Na+£©£¬ÒÔ´ËÀ´½â´ð£®

½â´ð ½â£ºA£®µçÀëÏÔËáÐÔ£¬ÒÖÖÆË®µÄµçÀ룬Ôò»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨H+£©=$\frac{1{0}^{-14}}{1{0}^{-5}}$=1¡Á10-9mol/L£¬¹ÊA´íÎó£»
B£®ÓÉÌå»ý¼°ÎïÁÏÊØºã¿ÉÖªc£¨A-£©+c£¨HA£©=2c £¨Na+£©=$\frac{0.4mol/L¡ÁV}{2V}$=0.2 mol/L£¬¹ÊB´íÎó£»
C.$\frac{c£¨{A}^{-}£©}{c£¨HA£©•c£¨O{H}^{-}£©}$=$\frac{c£¨{A}^{-}£©}{c£¨HA£©•\frac{Kw}{c£¨{H}^{+}£©}}$=$\frac{Ka}{Kw}$£¬Ka¡¢Kw¾ùÓëζÈÓйأ¬ÔòHAÈÜÒºÖÐ$\frac{c£¨{A}^{-}£©}{c£¨HA£©•c£¨O{H}^{-}£©}$ÓëÉÏÊö»ìºÏÈÜÒºÖÐ$\frac{c£¨{A}^{-}£©}{c£¨HA£©•c£¨O{H}^{-}£©}$ÏàµÈ£¬¹ÊCÕýÈ·£»
D£®ÓÉÎïÁÏÊØºã¿ÉÖª£¬c£¨A-£©+c£¨HA£©=2c £¨Na+£©£¬ÓɵçºÉÊØºã¿ÉÖª£¬c£¨A-£©+c£¨OH-£©=c £¨H+£©+c £¨Na+£©£¬Ôò2c£¨OH-£©=2 c £¨H+£©+c£¨HA£©£¬¹ÊD´íÎó£»
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²éËá¼î»ìºÏµÄ¶¨ÐÔÅжϣ¬Îª¸ßƵ¿¼µã£¬°ÑÎÕÈÜÒºÖеÄÈÜÖÊ¡¢µçºÉ¼°ÎïÁÏÊØºãΪ½â´ð±¾Ìâ¹Ø¼ü£¬²àÖØ·ÖÎöÓëÓ¦ÓÃÄÜÁ¦µÄ¿¼²é£¬×¢ÒâÑ¡ÏîCΪ½â´ðµÄÄѵ㣬ÌâÄ¿ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
1£®Èý²ÝËáºÏÌúËá¼Ø¾§Ì壨K3[Fe£¨C2O4£©3]•xH2O£©ÊÇÒ»ÖÖ¹âÃô²ÄÁÏ£¬110¡æÊ±Ê§È¥È«²¿½á¾§Ë®£®Ä³ÊµÑéС×éΪ²â¶¨¸Ã¾§ÌåÖÐÌúµÄº¬Á¿£¬×öÁËÈçÏÂʵÑ飬Íê³ÉÏÂÁÐÌî¿Õ£º
²½ÖèÒ»£º³ÆÈ¡5.0gÈý²ÝËáºÏÌúËá¼Ø¾§Ì壬ÅäÖÆ³É250mLÈÜÒº£®
²½Öè¶þ£ºÈ¡ËùÅäÈÜÒº25.00mLÓÚ×¶ÐÎÆ¿ÖУ¬¼ÓH2SO4Ëữ£¬µÎ¼ÓKMnO4ÈÜÒº£¬·¢Éú·´Ó¦
2MnO4-+16H++5C2O42-=2Mn2++10CO2¡ü+8H2O£¨Mn2+ÊÓΪÎÞÉ«£©£®Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈëп·Û£¬
¼ÓÈÈÖÁ»ÆÉ«¸ÕºÃÏûʧ£¬¹ýÂ˲¢Ï´µÓ£¬½«¹ýÂ˼°Ï´µÓËùµÃÈÜÒºÊÕ¼¯µ½×¶ÐÎÆ¿ÖУ¬´ËʱÈÜÒºÈÔ³ÊËáÐÔ£®
²½ÖèÈý£ºÓÃ0.010mol/L KMnO4ÈÜÒºµÎ¶¨²½Öè¶þËùµÃÈÜÒºÖÁÖյ㣬µÎ¶¨ÖÐMnO4-±»»¹Ô­³ÉMn2+£®ÏûºÄKMnO4ÈÜÒºÌå»ýÈç±íËùʾ£º
µÎ¶¨´ÎÊýµÎ¶¨Æðʼ¶ÁÊý£¨mL£©µÎ¶¨ÖÕµã¶ÁÊý£¨mL£©
µÚÒ»´Î1.08¼ûÓÒͼ
µÚ¶þ´Î2.0224.52
µÚÈý´Î1.0020.98
£¨1£©²½ÖèÒ»ËùÓÃÒÇÆ÷ÒÑÓÐÍÐÅÌÌìÆ½£¨´øíÀÂ룩¡¢²£Á§°ô¡¢ÉÕ±­¡¢Ò©³×£¬»¹È±ÉÙµÄÒÇÆ÷250mLÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü£®
£¨2£©¼ÓÈëп·ÛµÄÄ¿µÄÊǽ«Fe3+Ç¡ºÃ»¹Ô­³ÉFe2+£¬Ê¹Fe2+ÔÚ²½ÖèÈýÖÐÓëKMnO4·¢ÉúÑõ»¯»¹Ô­·´Ó¦£®
£¨3£©Ð´³ö²½ÖèÈýÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ5Fe2++MnO4-+8H+=5Fe3++Mn2++4H2O£®
£¨4£©ÊµÑé²âµÃ¸Ã¾§ÌåÖÐÌúÔªËØµÄÖÊÁ¿·ÖÊýΪ11.2%£®ÔÚ²½Öè¶þÖУ¬ÈôµÎ¼ÓµÄKMnO4ÈÜÒºµÄÁ¿²»¹»£¬Ôò²âµÃÌúµÄº¬Á¿Æ«¸ß£®£¨Ñ¡Ìî¡°Æ«µÍ¡±¡¢¡°Æ«¸ß¡±»ò¡°²»±ä¡±£©
19£®ÎåÑõ»¯¶þµª£¨N2O5£©ÊÇÓлúºÏ³ÉÖг£ÓõÄÂÌÉ«Ïõ»¯¼Á£¬¿ÉÓÃNO2ÓëO3·´Ó¦ÖƵã®Ä³»¯Ñ§ÐËȤС×éÉè¼ÆÈçÏÂʵÑé×°ÖÃÖÆ±¸N2O5£¨²¿·Ö×°ÖÃÂÔÈ¥£©£®

£¨1£©Í­ºÍŨÏõËá·´Ó¦µÄÀë×Ó·½³ÌʽΪCu+4H++2NO3-=Cu2++2NO2¡ü+2H2O£®¼ÓÈëŨÏõËáºó£¬×°ÖâñÖÐÉÕÆ¿Éϲ¿¿É¹Û²ìµ½µÄÏÖÏóÊÇÓкì×ØÉ«ÆøÌåÉú³É£®
£¨2£©ÒÇÆ÷aµÄÃû³ÆÎªÇòÐθÉÔï¹Ü£¬¸ÃÒÇÆ÷ÖÐÊ¢×°µÄÊÔ¼ÁÊÇD£¨Ìî±êºÅ£©£®
A£®Éúʯ»ÒB£®Å¨ÁòËáC£®¼îʯ»ÒD£®ÎåÑõ»¯¶þÁ×
£¨3£©ÒÑÖªÏÂÁÐÎïÖʵÄÓйØÊý¾Ý£®
ÎïÖÊÈÛµã/¡æ·Ðµã/¡æ
N2O54132 £¨Éý»ª£©
N2O4-1124
×°ÖâòÖгÖÐøÍ¨ÈëO3£¬ÎªµÃµ½¾¡¿ÉÄܶàµÄ´¿¾»µÄN2O5£¬Î¶ȿØÖÆÔÚ24¡æ〜32¡æ£®ÊÕ¼¯µ½µÄN2O5¹ÌÌåÖк¬ÓÐҺ̬ÎïÖÊ£¬¿ÉÄÜÔ­ÒòÊǸÉÔï²»ÍêÈ«£¬µ¼ÖÂÓÐÏõËáÉú³É£®
£¨4£©×°ÖâóµÄ×÷ÓÃÊǰ²È«Æ¿£¬·ÀÖ¹µ¹Îü£®
£¨5£©»¯Ñ§ÐËȤС×éÓÃÖÆµÃµÄN2O5ÖÆ±¸ÉÙÁ¿¶ÔÏõ»ù¼×±½£¨£¬Ïà¶Ô·Ö×ÓÖÊÁ¿137£©£®²½ÖèÈçÏ£ºÔÚÈý¿ÚÉÕÆ¿ÖзÅÈë´ß»¯¼ÁºÍ30mL N2O5µÄCH2Cl2ÈÜÒº£¨N2O5µÄŨ¶ÈΪlmol•L-1£©£¬30¡æÊ±£¬µÎ¼Ó15mL¼×±½£¬³ä·Ö·´Ó¦µÃ¶ÔÏõ»ù¼×±½ 1.73g£®»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÖÆ±¸¶ÔÏõ»ù¼×±½µÄ»¯Ñ§·½³ÌʽΪ+N2O5$¡ú_{¡÷}^{´ß»¯¼Á}$+H2O£®
¢ÚN2O5Éú³É¶ÔÏõ»ù¼×±½µÄת»¯ÂÊΪ21.05%£®
16£®ÇâÇèËᣨHCN£©ÊÇÒ»ÖÖ¾ßÓпàÐÓÈÊÆøÎ¶µÄÎÞɫҺÌ壬Ò×ÈÜÓÚË®£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ò»¶¨µÄζȺʹ߻¯¼Á×÷ÓÃÏ£¬ÀûÓü×Íé¡¢°±ÆøÎªÔ­ÁÏÑõ»¯ºÏ³ÉÇâÇèËᣮ
¢Ù°±ÆøµÄµç×ÓʽΪ£®
¢ÚºÏ³ÉÇâÇèËáµÄ»¯Ñ§·½³ÌʽΪ2CH4+2NH3+3O2$\frac{\underline{\;Ò»¶¨Ìõ¼þ\;}}{´ß»¯¼Á}$2HCN+6H2O
£¨2£©ÒÑÖª25¡æÊ±HCNºÍH2CO3µÄµçÀë³£Êý£¨Ka£©Èç±í£º
ÎïÖʵçÀë³£Êý£¨Ka£©
HCNKa=5¡Á10-10
H2CO3Ka1=4.5¡Á10-7    Ka2=4.7¡Á10-11
25¡æÊ±£¬ÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1mol•L-1µÄNaCN¡¢NaHCO3ºÍNa2CO3ÈýÖÖÈÜÒº£¬ÆäpH×î´óµÄÊÇNa2CO3£¨Ìѧʽ£©£®
£¨3£©-¶¨Ìõ¼þÏ£¬HCNÓëH2ºÍH2O·´Ó¦ÈçÏ£º
I£®HCN£¨g£©+3H2£¨g£©?NH3£¨g£©+CH4£¨g£©¡÷H1
¢ò£®HCN£¨g£©+H2O£¨g£©?NH3£¨g£©+CO£¨g£©¡÷H2
¢Ù·´Ó¦¢ó£¬CO£¨g£©+3H2£¨g£©?CH4£¨g£©+H2O£¨g£©µÄ¡÷H=¡÷H1-¡÷H2 £¨Óá÷H1¡¢¡÷H2±íʾ£©£®
¢Ú¶ÔÓÚ·´Ó¦¢ò£¬¼õСѹǿ£¬HCNµÄת»¯Âʲ»±ä£¨Ìî¡°Ìá¸ß¡±¡¢¡°²»±ä¡±»ò¡°½µµÍ¡±£©£®
¢Û·´Ó¦I¡¢¢òµÄƽºâ³£Êý¶ÔÊýÖµ£¨lgK£©ÓëζȵĹØÏµÈçͼ¼×Ëùʾ£¬ÔòT1Kʱ£¬·´Ó¦¢óµÄƽºâ³£Êý¶ÔÊýÖµlgK=10£®

£¨4£©µç½â·¨´¦Àíº¬Çè·ÏË®µÄÔ­ÀíÈçͼÒÒËùʾ£¬Ñô¼«CN-ÏÈ·¢Éúµç¼«·´Ó¦£ºCN-+2OH--2e-¨TCNO-+H2O£¬CNO-ÔÚÑô¼«ÉϽøÒ»²½Ñõ»¯µÄµç¼«·´Ó¦Ê½Îª2CNO-+4OH--6e-=N2¡ü+2CO2¡ü+2H2O£®

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

¾«Ó¢¼Ò½ÌÍø