ÌâÄ¿ÄÚÈÝ

18£®¸õ»¯Ñ§·á¸»¶à²Ê£¬ÎÞÂÛÊǸõµ¥ÖÊ»¹ÊǺ¬¸õ»¯ºÏÎÔÚÉú²úºÍ¿ÆѧÑо¿Öж¼¾ßÓÐÖØÒªµÄ¼ÛÖµ£¬Çë¸ù¾ÝÏà¹ØÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣮
£¨1£©ÔÚÏÂͼװÖÃÖУ¬¹Û²ìµ½Í¼1×°ÖÃÍ­µç¼«ÉϲúÉú´óÁ¿µÄÎÞÉ«ÆøÅÝ£¬¶øͼ2×°ÖÃÖÐÍ­µç¼«ÉÏÎÞÆøÌå²úÉú£¬¸õµç¼«¸½½ü²úÉú´óÁ¿ÓÐÉ«ÆøÌ壮¸ù¾ÝÉÏÊöÏÖÏóÊÔÍƲâ½ðÊô¸õµÄÁ½¸öÖØÒª»¯Ñ§ÐÔÖʽðÊô¸õµÄ»î¶¯ÐÔ±ÈÍ­Ç¿£¬ÄܺÍÏ¡ÁòËá·´Ó¦Éú³É H2¡¢½ðÊô¸õÒ×±»Ï¡ÏõËá¶Û»¯£»

£¨2£©Cr£¨OH£©3ºÍAl£¨OH£©3ÀàËÆ£¬Ò²ÊÇÁ½ÐÔÇâÑõ»¯ÎÔÚË®ÖдæÔÚËáʽºÍ¼îʽµçÀëƽºâ£¬ÆäËáʽµçÀë·½³ÌʽÊÇCr£¨OH£©3?H++CrO2-+H2O£»
£¨3£©¾Æºó¼Ý³µÊÇÒý·¢½»Í¨Ê¹ʵÄÖØÒªÔ­Òò£®½»¾¯¶Ô¼ÝʻԱ½øÐкôÆø¾Æ¾«¼ì²âµÄÔ­ÀíÊÇ£º³ÈÉ«µÄK2Cr2O7ËáÐÔË®ÈÜÒºÓöÒÒ´¼Ñ¸ËÙÉú³ÉÀ¶ÂÌÉ«Cr3+£®·´Ó¦Ô­ÀíΪ£º
K2Cr2O7+C2H5OH+H2SO4Ò»K2SO4+Cr2£¨SO4£©3+CH3COOH+H2O
Çëд³öÆäÅäƽºóµÄÀë×Ó·½³Ìʽ2Cr2O72-+3C2H5OH+16H+=4Cr3++3CH3COOH+11H2O£¬
µ±·´Ó¦ÖÐÓÐ0.03molµç×ÓתÒÆʱ£¬±»Ñõ»¯µÄÒÒ´¼ÓÐ0.0075mol£»
£¨4£©¹¤ÒµÉϾ»»¯´¦Àí¸õÎÛȾ·½·¨Ö®Ò»ÊÇ£º½«º¬K2Cr2O7ËáÐÔ·ÏË®·ÅÈëµç½â²ÛÄÚ£¬¼ÓÈëÊÊÁ¿µÄNaCl£¬ÒÔFeºÍʯīΪµç¼«½øÐеç½â£®¾­¹ýÒ»¶Îʱ¼äºó£¬Éú³ÉCr£¨OH£©3ºÍFe£¨OH£©3³Áµí³ýÈ¥£¨ÒÑÖªKsp[Fe£¨OH£©3]=4.0¡Á10-38£¬Ksp[Cr£¨OH£©3]=6.0¡Á10-31£©£®
¢ÙÔÚ×°ÖÃÖÐÓ¦¸ÃÓÃÌú£¨Ìî²ÄÁÏÃû³Æ£©×öµç½â³ØµÄÑô¼«£¬Òõ¼«Éϵ缫·´Ó¦Îª2H++2e-=H2¡ü£¬
¢ÚÒÑÖªµç½âºóµÄÈÜÒºÖÐc£¨Fe3+£©Îª2.0¡Á10-13mol/L£¬ÔòÈÜÒºÖÐc£¨Cr3+£©Îª3¡Á10-6mol/L£®

·ÖÎö £¨1£©Í¼1×°ÖÃÍ­µç¼«ÉϲúÉú´óÁ¿µÄÎÞÉ«ÆøÅÝ£¬ËµÃ÷CrµÄ»îÆÃÐÔ´óÓÚCu£¬ÄܺÍÏ¡ÁòËá·´Ó¦Éú³É H2£»¶øͼ 2×°ÖÃÖÐÍ­µç¼«ÉÏÎÞÆøÌå²úÉú£¬¸õµç¼«ÉϲúÉú´óÁ¿ÓÐÉ«ÆøÌ壬˵Ã÷CrºÍÏõËáÄܲúÉú¶Û»¯ÏÖÏó£»
£¨2£©Àà±ÈAl£¨OH£©3µÄËáʽµçÀë½øÐÐÊéд£»
£¨3£©¸ù¾Ýµç×ÓµÃʧÊغ㣬ÅäƽÏà¹ØϵÊý£¬·½³ÌʽΪ£º2K2Cr2O7+3CH3CH2OH+8H2SO4=2Cr2£¨SO4£©3+3CH3COOH+2K2SO4+11H2O£¬È»ºóÊéдÀë×Ó·½³Ìʽ£»·´Ó¦ÖÐCrÔªËصĻ¯ºÏ¼Û½µµÍ£¬ÒÒ´¼ÖÐCÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬½áºÏÔªËØ»¯ºÏ¼ÛµÄ±ä»¯·ÖÎöµç×ÓתÒÆÊý£¬´Ó¶øÈ·¶¨±»Ñõ»¯µÄÒÒ´¼µÄÎïÖʵÄÁ¿ÒÔ´ËÀ´½â´ð£»
£¨4£©¢ÙÑô¼«·¢ÉúÑõ»¯·´Ó¦£¬²úÉúÑÇÌúÀë×Ó£¬ËùÒԵ缫²ÄÁÏΪÌú£»Òõ¼«Éϵ缫·´Ó¦ÎªÊÇÇâÀë×Ó·¢Éú»¹Ô­·´Ó¦Éú³ÉÇâÆø£»
¢Ú¸ù¾Ýµç½âºóÈÜÒºÖÐc£¨Fe3+£©£¬ÓÉKsP[Fe£¨OH£©3]¼ÆËãÈÜÒºÖÐc£¨OH-£©£¬ÔÙ¸ù¾ÝKsP[Cr£¨OH£©3]¼ÆËãÈÜÒºÖÐc£¨Cr3+£©£®

½â´ð ½â£º£¨1£©Í¼1×°ÖÃÍ­µç¼«ÉϲúÉú´óÁ¿µÄÎÞÉ«ÆøÅÝ£¬ËµÃ÷CrµÄ»îÆÃÐÔ´óÓÚCu£¬ÄܺÍÏ¡ÁòËá·´Ó¦Éú³É H2£»¶øͼ 2×°ÖÃÖÐÍ­µç¼«ÉÏÎÞÆøÌå²úÉú£¬¸õµç¼«ÉϲúÉú´óÁ¿ÓÐÉ«ÆøÌ壬˵Ã÷CrºÍÏõËáÄܲúÉú¶Û»¯ÏÖÏó£¬
¹Ê´ð°¸Îª£º½ðÊô¸õµÄ»î¶¯ÐÔ±ÈÍ­Ç¿£¬ÄܺÍÏ¡ÁòËá·´Ó¦Éú³É H2£»½ðÊô¸õÒ×±»Ï¡ÏõËá¶Û»¯£»
£¨2£©Cr£¨OH£©3ºÍAl£¨OH£©3ÀàËÆ£¬ÓÉAl£¨OH£©3µÄËáʽµçÀë¿ÉÖª£¬Cr£¨OH£©3µÄËáʽµçÀë·½³ÌʽÊÇ£ºCr£¨OH£©3?H++CrO2-+H2O£¬
¹Ê´ð°¸Îª£ºCr£¨OH£©3?H++CrO2-+H2O£»
£¨3£©¸ù¾Ýµç×ÓµÃʧÊغ㣬ÅäƽÏà¹ØϵÊý£¬·½³ÌʽΪ£º2K2Cr2O7+3CH3CH2OH+8H2SO4=2Cr2£¨SO4£©3+3CH3COOH+2K2SO4+11H2O£¬ËùÒÔÀë×Ó·½³ÌΪ£º2Cr2O72-+3C2H5OH+16H+=4Cr3++3CH3COOH+11H2O£¬·´Ó¦ÖÐCrÔªËصĻ¯ºÏ¼Û½µµÍ£¬K2Cr2O7ÊÇÑõ»¯¼Á£¬Ñõ»¯¼Á±»»¹Ô­µÃµ½»¹Ô­²úÎÔòCr2£¨SO4£©3Ϊ»¹Ô­²úÎÒÒ´¼ÖÐCÔªËصĻ¯ºÏ¼ÛÉý¸ß£¬ÒÒ´¼ÊÇ»¹Ô­¼Á£¬»¹Ô­¼Á¾ßÓл¹Ô­ÐÔ£»K2Cr2O7ÖÐCrÔªËØ´Ó+6¼Û½µµÍµ½+3¼Û£¬Ôò1molÑõ»¯¼ÁתÒÆ6molµç×Ó£¬±»Ñõ»¯µÄCH3CH2OHΪ1.5mol£¬ËùÒÔ0.03molµç×ÓתÒÆʱ£¬±»Ñõ»¯µÄÒÒ´¼µÄÁ¿Îª0.0075mol£¬
¹Ê´ð°¸Îª£º2Cr2O72-+3C2H5OH+16H+=4Cr3++3CH3COOH+11H2O£»0.0075£»
£¨4£©¢ÙÑô¼«·¢ÉúÑõ»¯·´Ó¦£¬²úÉúÑÇÌúÀë×Ó£¬ËùÒԵ缫²ÄÁÏΪÌú£»Òõ¼«Éϵ缫·´Ó¦ÎªÊÇÇâÀë×Ó·¢Éú»¹Ô­·´Ó¦Éú³ÉÇâÆø£¬ËùÒÔÒõ¼«µÄµç¼«·´Ó¦Ê½Îª£º2H++2e-=H2¡ü£¬
¹Ê´ð°¸Îª£ºÌú£»2H++2e-=H2¡ü£»
¢Úµç½âºóÈÜÒºÖÐc£¨Fe3+£©=2.0¡Á10-13mol/L£¬ÔòÈÜÒºÖÐc3£¨OH-£©=$\frac{4¡Á1{0}^{-38}}{2¡Á1{0}^{-13}}$mol/L=2¡Á10-25mol/L£¬¹ÊÈÜÒºÖÐc£¨Cr3+£©=$\frac{6¡Á1{0}^{-31}}{2¡Á1{0}^{-25}}$mol/L=3¡Á10-6mol/L£¬
¹Ê´ð°¸Îª£º3¡Á10-6£®

µãÆÀ ±¾Ì⿼²éÁËÑõ»¯»¹Ô­·´Ó¦¡¢KspµÄÓйؼÆË㣬ÒÔ¼°Í¼Ïó·ÖÎö£¬ÌâÄ¿ÄѶÈÖеȣ¬×¢Òâ¶ÔͼÏóµÄ·ÖÎöºÍÊý¾ÝµÄ´¦Àí£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø