ÌâÄ¿ÄÚÈÝ

15£®ÈÜÒºµÄËá¼îÐÔ¿ÉÓÃËá¶È£¨AG£©±íʾ[AG=lg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$]£¬ÊÒÎÂÏ£¬ÏòŨ¶È¾ùΪ0.1mol/LÌå»ý¾ùΪ100mLµÄÁ½ÖÖÒ»ÔªËáHX¡¢HYµÄÈÜÒºÖУ¬·Ö±ð¼ÓÈëNaOH¹ÌÌ壬AGËæ¼ÓÈëNaOHµÄÎïÖʵÄÁ¿µÄ±ä»¯ÈçͼËùʾ£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®HX¡¢HY¾ùΪÈõËá
B£®aµãÓÉË®µçÀë³öµÄc£¨H+£©=1.0¡Ál0-13mol•L-l
C£®cµãÈÜÒºÖУºc£¨Y-£©£¼c£¨Na+£©£¼c£¨HY£©
D£®bµãʱ£¬ÈÜÒºµÄpH=7£¬Ëá¼îÇ¡ºÃÍêÈ«ÖкÍ

·ÖÎö A£®lg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$Ô½´ó£¬ÈÜÒºÖÐÇâÀë×ÓŨ¶ÈÔ½´ó£¬Î´¼ÓNaOHʱ£¬HXÈÜÒºÖÐlg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$µÄÖµ´ó£¬´ÓͼÏóÖмÓÈëÇâÑõ»¯ÄÆÎïÖʵÄÁ¿ºÍÈÜÒºÖÐlg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$µÄÖµ·ÖÎöÅжϣ»
B£®aµãlg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$=12£¬ÔòÈÜÒºÖÐc£¨H+£©=0.1mol/L£»
C£®cµãlg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$=6£¬ÔòÈÜÒºÖÐc£¨H+£©=10-4mol/L£¬´ËʱÏûºÄµÄNaOHΪ0.005mol£¬ÔòÈÜÒºÖеÄÈÜÖÊΪNaYºÍHY£»
D£®Å¨¶È¾ùΪ0.1mol•L-1¡¢Ìå»ý¾ù100mLµÄHYÓëNaOHÇ¡ºÃÖкÍÏûºÄNaOHΪ0.01mol£®

½â´ð ½â£ºA£®lg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$Ô½´ó£¬ÈÜÒºÖÐÇâÀë×ÓŨ¶ÈÔ½´ó£¬Î´¼ÓNaOHʱ£¬HXÈÜÒºÖÐlg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$µÄÖµ´ó£¬ËùÒÔHXµÄËáÐÔ´óÓÚHY£¬¼ÓÈëÇâÑõ»¯ÄÆÎïÖʵÄÁ¿10¡Á10-3mol£¬lg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$=1£¬c£¨H+£©=c£¨OH-£©ËáºÍÇâÑõ»¯ÄÆÈÜҺǡºÃ·´ÈÜÒº³ÊÖÐÐÔ£¬HXΪǿËᣬ¹ÊA´íÎó£»
B£®aµãlg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$=12£¬ÔòÈÜÒºÖÐc£¨H+£©=0.1mol/L£¬ÈÜÒºÖÐË®µçÀëµÄc£¨H+£©=$\frac{1{0}^{-14}}{0.1}$=10-13mol•L-1£¬¹ÊBÕýÈ·£»
C£®cµãlg$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$=6£¬ÔòÈÜÒºÖÐc£¨H+£©=10-4mol/L£¬´ËʱÏûºÄµÄNaOHΪ0.005mol£¬ÔòÈÜÒºÖеÄÈÜÖÊΪNaYºÍHY£¬ÓÉÓÚÈÜÒºÏÔËáÐÔ£¬ËùÒÔHYµÄµçÀë³Ì¶È´óÓÚNaYµÄË®½â³Ì¶È£¬ËùÒÔc£¨Y-£©£¾c£¨HY£©£¬¹ÊC´íÎó£»
D£®Å¨¶È¾ùΪ0.1mol•L-1¡¢Ìå»ý¾ù100mLµÄHYÓëNaOHÇ¡ºÃÖкÍÏûºÄNaOHΪ0.01mol£¬¶øbµãʱÏûºÄµÄNaOHΪ0.008mol£¬ËùÒÔËá¹ýÁ¿£¬¹ÊD´íÎó£®
¹ÊÑ¡B£®

µãÆÀ ±¾Ì⿼²éËá¼î»ìºÏÈÜÒºËá¼îÐÔÅжϼ°ÈÜÒºÖÐÀë×ÓŨ¶È´óС±È½Ï£¬Éæ¼°ÑÎÀàµÄË®½âºÍÈõËáµÄµçÀëµÈ£¬ÌâÄ¿ÄѶÈÖеȣ¬²àÖØÓÚ¿¼²éѧÉúµÄ·ÖÎöÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
4£®Èý²ÝËáºÏÌúËá¼Ø¾§Ì壨K3[Fe£¨C2O4£©3]•xH2O£©ÊÇÒ»ÖÖ¹âÃô²ÄÁÏ£¬ÔÚ110¡æ¿ÉÍêȫʧȥ½á¾§Ë®£®Îª²â¶¨¸Ã¾§ÌåÖÐÌúµÄº¬Á¿ºÍ½á¾§Ë®µÄº¬Á¿£¬Ä³ÊµÑéС×é×öÁËÈçÏÂʵÑ飺
£¨1£©Ìúº¬Á¿µÄ²â¶¨
²½ÖèÒ»£º³ÆÁ¿5.00gÈý²ÝËáºÏÌúËá¼Ø¾§Ì壬ÅäÖÆ³É250mLÈÜÒº£®
²½Öè¶þ£ºÈ¡ËùÅäÈÜÒº25.00mLÓÚ×¶ÐÎÆ¿ÖУ¬¼ÓÏ¡H2SO4Ëữ£¬µÎ¼ÓKMnO4ÈÜÒºÖÁ²ÝËá¸ùÇ¡ºÃÈ«²¿Ñõ»¯³É¶þÑõ»¯Ì¼£¬Í¬Ê±£¬MnO4-±»»¹Ô­³ÉMn2+£®Ïò·´Ó¦ºóµÄÈÜÒºÖмÓÈëһС³×п·Û£¬¼ÓÈÈÖÁ»ÆÉ«¸ÕºÃÏûʧ£¬¹ýÂË£¬Ï´µÓ£¬½«¹ýÂ˼°Ï´µÓËùµÃÈÜÒºÊÕ¼¯µ½×¶ÐÎÆ¿ÖУ¬´Ëʱ£¬ÈÜÒºÈÔ³ÊËáÐÔ£®
²½ÖèÈý£ºÓÃ0.010mol/L KMnO4ÈÜÒºµÎ¶¨²½Öè¶þËùµÃÈÜÒºÖÁÖյ㣬ÏûºÄKMnO4ÈÜÒº20.02mL£¬µÎ¶¨ÖÐMnO4-±»»¹Ô­³ÉMn2+£®
ÖØ¸´²½Öè¶þ¡¢²½ÖèÈý²Ù×÷£¬µÎ¶¨ÏûºÄ0.010mol/LKMnO4ÈÜÒº19.98ml£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÅäÖÆÈý²ÝËáºÏÌúËá¼ØÈÜÒºµÄ²Ù×÷²½ÖèÒÀ´ÎÊÇ£º³ÆÁ¿¡¢ÈÜ½â¡¢×ªÒÆ¡¢Ï´µÓ²¢×ªÒÆ¡¢¶¨ÈÝÒ¡ÔÈ£®
¢Ú¼ÓÈëп·ÛµÄÄ¿µÄÊǽ«Fe3+Ç¡ºÃ»¹Ô­³ÉFe2+£®
¢Ûд³ö²½ÖèÈýÖз¢Éú·´Ó¦µÄÀë×Ó·½³Ìʽ5Fe2++MnO4-+8H+=5Fe3++Mn2++4H2O£®
¢ÜʵÑé²âµÃ¸Ã¾§ÌåÖÐÌúµÄÖÊÁ¿·ÖÊýΪ11.20%£®ÔÚ²½Öè¶þÖУ¬Èô¼ÓÈëµÄKMnO4ÈÜÒºµÄÁ¿²»¹»£¬Ôò²âµÃµÄÌúº¬Á¿Æ«¸ß£®£¨Ñ¡Ìî¡°Æ«µÍ¡±¡°Æ«¸ß¡±¡°²»±ä¡±£©
£¨2£©½á¾§Ë®µÄ²â¶¨
½«ÛáÛöÏ´¾»£¬ºæ¸ÉÖÁºãÖØ£¬¼Ç¼ÖÊÁ¿£»ÔÚÛá¹øÖмÓÈëÑÐϸµÄÈý²ÝËáºÏÌúËá¼Ø¾§Ì壬³ÆÁ¿²¢¼Ç¼ÖÊÁ¿£»¼ÓÈÈÖÁ110¡æ£¬ºãÎÂÒ»¶Îʱ¼ä£¬ÖÃÓÚ¿ÕÆøÖÐÀäÈ´£¬³ÆÁ¿²¢¼Ç¼ÖÊÁ¿£»¼ÆËã½á¾§Ë®º¬Á¿£®Çë¾ÀÕýʵÑé¹ý³ÌÖеÄÁ½´¦´íÎ󣺼ÓÈȺóµÄ¾§ÌåÒªÔÚ¸ÉÔïÆ÷ÖÐÀäÈ´£»Á½´Î³ÆÁ¿ÖÊÁ¿²î²»³¬¹ý0.1 g£®
1£®µª£¨N£©¡¢Á×£¨P£©¡¢É飨As£©µÈ¶¼ÊÇ¢õA×åµÄÔªËØ£¬¸Ã×åÔªËØµÄ»¯ºÏÎïÔÚÑо¿ºÍÉú²úÖÐÓÐÐí¶àÖØÒªÓÃ;£®»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©»¯ºÏÎïN2H4µÄµç×ÓʽΪ£®
£¨2£©AsÔ­×ӵĺËÍâµç×ÓÅŲ¼Ê½Îª1s22s22p63s23p63d104s24p3£®
£¨3£©PºÍSÊÇͬһÖÜÆÚµÄÁ½ÖÖÔªËØ£¬PµÄµÚÒ»µçÀëÄܱÈS´ó£¬Ô­ÒòÊÇPµÄpÑDzãÊǰë³äÂú״̬£¬±È½ÏÎȶ¨£¬ËùÒÔµÚÒ»µçÀëÄܱÈÁòµÄ´ó£®
£¨4£©NH4+ÖÐH-N-HµÄ½¡½Ç±ÈNH3ÖÐH-N-HµÄ¼ü½Ç´ó£¬Ô­ÒòÊÇNH4+ÖеĵªÔ­×ÓÉϾùΪ³É¼üµç×Ó£¬¶øNH3·Ö×ÓÖеĵªÔ­×ÓÉÏÓÐÒ»¶Ô¹Â¶Ôµç×Ó£¬¹Â¶Ôµç×Ӻͳɼüµç×ÓÖ®¼äµÄÅųâÁ¦Ç¿Óڳɼüµç×Ӻͳɼüµç×ÓÖ®¼äµÄÅųâÁ¦£®
£¨5£©Na3AsO4Öк¬ÓеĻ¯Ñ§¼üÀàÐͰüÀ¨Àë×Ó¼ü¡¢¹²¼Û¼ü£»AsO43-¿Õ¼ä¹¹ÐÍΪÕýËÄÃæÌ壬As4O6µÄ·Ö×ӽṹÈçͼ1Ëùʾ£¬ÔòÔڸû¯ºÏÎïÖÐAsµÄÔÓ»¯·½Ê½ÊÇsp3£®
£¨6£©°×Á×£¨P4£©µÄ¾§ÌåÊôÓÚ·Ö×Ó¾§Ì壬Æä¾§°û½á¹¹Èçͼ2£¨Ð¡Ô²È¦±íʾ°×Á×·Ö×Ó£©£®¼ºÖª¾§°ûµÄ±ß³¤Îªacm£¬°¢·ü¼ÓµÂÂÞ³£ÊýΪNAmol-1£¬Ôò¸Ã¾§°ûÖк¬ÓеÄPÔ­×ӵĸöÊýΪ16£¬¸Ã¾§ÌåµÄÃܶÈΪ$\frac{496}{{a}^{3}{N}_{A}}$  g•cm-3£¨Óú¬NA¡¢aµÄʽ×Ó±íʾ£©£®

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

¾«Ó¢¼Ò½ÌÍø