ÌâÄ¿ÄÚÈÝ
CuSO4ÈÜÒºÓëK2C2O4ÈÜÒº»ìºÏ·´Ó¦£¬²úÎïÖ®Ò»ÊÇÖ»º¬Ò»ÖÖÒõÀë×ÓµÄÀ¶É«¼ØÑÎË®ºÏÎͨ¹ýÏÂÊöʵÑéÈ·¶¨¸Ã¾§ÌåµÄ×é³É£®
²½Öèa£º³ÆÈ¡0.6720gÑùÆ·£¬·ÅÈë׶ÐÎÆ¿£¬¼ÓÈëÊÊÁ¿2mol?L-1Ï¡ÁòËᣬ΢ÈÈʹÑùÆ·Èܽ⣮ÔÙ¼ÓÈë30mlË®¼ÓÈÈ£¬ÓÃ0.2000mol?L-1 KMnO4ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ8.00ml£®
²½Öèb£º½Ó׎«ÈÜÒº³ä·Ö¼ÓÈÈ£¬Ê¹µ×ϺìÉ«Ïûʧ£¬ÈÜÒº×îÖÕ³ÊÏÖÀ¶É«£®ÀäÈ´ºó£¬µ÷½ÚpH²¢¼ÓÈë¹ýÁ¿µÄKI¹ÌÌ壬ÈÜÒº±äΪ×ØÉ«²¢²úÉú°×É«³ÁµíCuI£®ÓÃ0.2500mol?L-1 Na2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ8.00ml£®
ÒÑÖªÉæ¼°µÄ²¿·ÖÀë×Ó·½³ÌʽΪ£º
²½Öèa£º2MnO4-+5C2O42-+16H+=2Mn2++8H2O+10CO2¡ü
²½Öèb£º2Cu2++4I-=2CuI¡ý+I2 I2+2S2O32-=2I-+S4O62-
£¨1£©ÒÑÖªÊÒÎÂÏÂCuIµÄKsp=1.27¡Á10-12£¬ÓûʹÈÜÒºÖÐc£¨Cu+£©¡Ü1.0¡Á10-6mol?L-1£¬Ó¦±£³ÖÈÜÒºÖÐc£¨I-£©¡Ý
£¨2£©MnO4-ÔÚËáÐÔÌõ¼þÏ£¬¼ÓÈÈÄÜ·Ö½âΪO2£¬Í¬Ê±Éú³ÉMn2+£®
¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ
ÈôÎ޸òÙ×÷£¬Ôò²â¶¨µÄCu2+µÄº¬Á¿½«»á
£¨3£©²½ÖèbÓõí·ÛÈÜÒº×öָʾ¼Á£¬ÔòµÎ¶¨ÖÕµã¹Û²ìµ½µÄÏÖÏóΪ
£¨4£©Í¨¹ý¼ÆËãÈ·¶¨ÑùÆ·¾§ÌåµÄ×é³É£®
²½Öèa£º³ÆÈ¡0.6720gÑùÆ·£¬·ÅÈë׶ÐÎÆ¿£¬¼ÓÈëÊÊÁ¿2mol?L-1Ï¡ÁòËᣬ΢ÈÈʹÑùÆ·Èܽ⣮ÔÙ¼ÓÈë30mlË®¼ÓÈÈ£¬ÓÃ0.2000mol?L-1 KMnO4ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ8.00ml£®
²½Öèb£º½Ó׎«ÈÜÒº³ä·Ö¼ÓÈÈ£¬Ê¹µ×ϺìÉ«Ïûʧ£¬ÈÜÒº×îÖÕ³ÊÏÖÀ¶É«£®ÀäÈ´ºó£¬µ÷½ÚpH²¢¼ÓÈë¹ýÁ¿µÄKI¹ÌÌ壬ÈÜÒº±äΪ×ØÉ«²¢²úÉú°×É«³ÁµíCuI£®ÓÃ0.2500mol?L-1 Na2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㣬ÏûºÄ8.00ml£®
ÒÑÖªÉæ¼°µÄ²¿·ÖÀë×Ó·½³ÌʽΪ£º
²½Öèa£º2MnO4-+5C2O42-+16H+=2Mn2++8H2O+10CO2¡ü
²½Öèb£º2Cu2++4I-=2CuI¡ý+I2 I2+2S2O32-=2I-+S4O62-
£¨1£©ÒÑÖªÊÒÎÂÏÂCuIµÄKsp=1.27¡Á10-12£¬ÓûʹÈÜÒºÖÐc£¨Cu+£©¡Ü1.0¡Á10-6mol?L-1£¬Ó¦±£³ÖÈÜÒºÖÐc£¨I-£©¡Ý
1.27¡Á10-6
1.27¡Á10-6
mol?L-1£®£¨2£©MnO4-ÔÚËáÐÔÌõ¼þÏ£¬¼ÓÈÈÄÜ·Ö½âΪO2£¬Í¬Ê±Éú³ÉMn2+£®
¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ
4MnO4-+12H+
4Mn2++5O2¡ü+6H2O
| ||
4MnO4-+12H+
4Mn2++5O2¡ü+6H2O
£®
| ||
ÈôÎ޸òÙ×÷£¬Ôò²â¶¨µÄCu2+µÄº¬Á¿½«»á
Æ«¸ß
Æ«¸ß
£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°²»±ä¡±£©£®£¨3£©²½ÖèbÓõí·ÛÈÜÒº×öָʾ¼Á£¬ÔòµÎ¶¨ÖÕµã¹Û²ìµ½µÄÏÖÏóΪ
ÈÜÒºÓÉÀ¶É«±äΪÎÞÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«
ÈÜÒºÓÉÀ¶É«±äΪÎÞÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«
£®£¨4£©Í¨¹ý¼ÆËãÈ·¶¨ÑùÆ·¾§ÌåµÄ×é³É£®
·ÖÎö£º£¨1£©¸ù¾Ýµâ»¯ÑÇ͵ÄÈܶȻý¼ÆËã³öµâÀë×ÓµÄŨ¶È£»
£¨2£©¸ù¾Ý»¯ºÏ¼ÛÉý½µÏàµÈÅäƽ·´Ó¦·½³Ìʽ£»¸ßÃÌËá¼ØÄܹ»½«µâÀë×ÓÑõ»¯³Éµâµ¥ÖÊ£»
£¨3£©I2+2S2O32-=2I-+S4O62-·´Ó¦½áÊøʱ£¬ÈÜÒºÖеⵥÖÊÏûʧ£¬ÈÜÒº±ä³ÉÎÞÉ«£»
£¨4£©¸ù¾ÝÌâÖз´Ó¦ÕÒ³ö¹Øϵʽ2MnO4-¡«5C2O42- 2S2O32-¡«I2¡«2Cu2+£¬ÀûÓõζ¨Êý¾ÝÇó³ön£¨C2O42-£©¡¢n£¨Cu2+£©£¬È»ºóÀûÓõçºÉÊغãÇó³ö¼ØÀë×ÓµÄÎïÖʵÄÁ¿£¬ÔÙÀûÓÃÑùÆ·ÖÊÁ¿Ëã³öË®µÄÎïÖʵÄÁ¿£¬×îºó¼ÆËã³öÑùÆ·¾§ÌåµÄ×é³É£®
£¨2£©¸ù¾Ý»¯ºÏ¼ÛÉý½µÏàµÈÅäƽ·´Ó¦·½³Ìʽ£»¸ßÃÌËá¼ØÄܹ»½«µâÀë×ÓÑõ»¯³Éµâµ¥ÖÊ£»
£¨3£©I2+2S2O32-=2I-+S4O62-·´Ó¦½áÊøʱ£¬ÈÜÒºÖеⵥÖÊÏûʧ£¬ÈÜÒº±ä³ÉÎÞÉ«£»
£¨4£©¸ù¾ÝÌâÖз´Ó¦ÕÒ³ö¹Øϵʽ2MnO4-¡«5C2O42- 2S2O32-¡«I2¡«2Cu2+£¬ÀûÓõζ¨Êý¾ÝÇó³ön£¨C2O42-£©¡¢n£¨Cu2+£©£¬È»ºóÀûÓõçºÉÊغãÇó³ö¼ØÀë×ÓµÄÎïÖʵÄÁ¿£¬ÔÙÀûÓÃÑùÆ·ÖÊÁ¿Ëã³öË®µÄÎïÖʵÄÁ¿£¬×îºó¼ÆËã³öÑùÆ·¾§ÌåµÄ×é³É£®
½â´ð£º½â£º£¨1£©CuIµÄKsp=1.27¡Á10-12=c£¨Cu+£©?c£¨I-£©£¬c£¨I-£©=
=
£¬ÓÉÓÚÈÜÒºÖÐc£¨Cu+£©¡Ü1.0¡Á10-6mol?L-1£¬c£¨I-£©¡Ý1.27¡Á10-6£¬
¹Ê´ð°¸Îª£º1.27¡Á10-6£»
£¨2£©ÃÌÔªËØ»¯ºÏ¼Û+7¡ú+2£¬½µµÍÁË5¼Û£¬Éú³ÉÑõÆø£¬ÑõÔªËØ»¯ºÏ¼Û-2¡ú0£¬Éý¸ßÁË2¡Á2=4£¬ËùÒÔÃÌÀë×ÓÇ°Åä4£¬ÑõÆøÇ°Åä5£¬ÔÙÀûÓù۲취ÅäƽÆäËüÎïÖÊ£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£º4MnO4-+12H+
4Mn2++5O2¡ü+6H2O£»Èô²»¼ÓÈȳýÈ¥¸ßÃÌËá¸ùÀë×Ó£¬¸ßÃÌËá¸ùÀë×ÓÄܹ»Ñõ»¯µâÀë×Ó£¬Ê¹µÃµâµ¥ÖʵÄÎïÖʵÄÁ¿Ôö¼Ó£¬²â¶¨ÍÀë×Óº¬Á¿Æ«¸ß£¬
¹Ê´ð°¸Îª£º4MnO4-+12H+
4Mn2++5O2¡ü+6H2O£»Æ«¸ß£»
£¨3£©¸ù¾Ý·´Ó¦I2+2S2O32-=2I-+S4O62-£¬µ±·´Ó¦½áÊøʱ£¬ÈÜÒºµÄÀ¶É«Ïûʧ±ä³ÉÎÞÉ«ÈÜÒº£¬ËùÒԵζ¨ÖÕµãÊÇ£ºÈÜÒºÓÉÀ¶É«±äΪÎÞÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«£¬
¹Ê´ð°¸Îª£ºÈÜÒºÓÉÀ¶É«±äΪÎÞÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«£»
£¨4£©¸ù¾Ý·´Ó¦2MnO4-+5C2O42-+16H+=2Mn2++8H2O+10CO2¡ü¡¢2Cu2++4I-=2CuI¡ý+I2 I2+2S2O32-=2I-+S4O62-
¿ÉµÃ¹Øϵʽ£º2MnO4-¡«5C2O42- 2S2O32-¡«I2¡«2Cu2+
2 5 2 2
0.2¡Á8¡Á10-3mol n£¨C2O42-£© 0.25¡Á8¡Á10-3mol n£¨Cu2+£©
ËùÒÔn£¨C2O42-£©=4¡Á10-3mol£¬n£¨Cu2+£©=2¡Á10-3mol£¬
¸ù¾ÝµçºÉÊغãÔÀí£ºn£¨K+£©=2n£¨C2O42-£©-2n£¨Cu2+£©=4¡Á10-3mol
¸ù¾ÝÖÊÁ¿ÊغãÔÀí£ºm£¨H2O£©=0.6720-39¡Á4¡Á10-3-64¡Á2¡Á10-3-88¡Á4¡Á10-3=0.036g
n£¨H2O£©=2¡Á10-3£¬n £¨K+£©£ºn£¨Cu2+£©£ºn£¨C2O42-£©£ºn £¨H2O£©=2£º1£º2£º1£¬
ËùÒÔÑùÆ·¾§Ìå×é³ÉÊÇ£ºK2Cu£¨C2O4£©2?H2O£¬
´ð£º¸ÃÑùÆ·¾§ÌåµÄ»¯Ñ§Ê½Îª£ºK2Cu£¨C2O4£©2?H2O£®
Ksp |
c(Cu+) |
1.27¡Á10-12 |
c(Cu+) |
¹Ê´ð°¸Îª£º1.27¡Á10-6£»
£¨2£©ÃÌÔªËØ»¯ºÏ¼Û+7¡ú+2£¬½µµÍÁË5¼Û£¬Éú³ÉÑõÆø£¬ÑõÔªËØ»¯ºÏ¼Û-2¡ú0£¬Éý¸ßÁË2¡Á2=4£¬ËùÒÔÃÌÀë×ÓÇ°Åä4£¬ÑõÆøÇ°Åä5£¬ÔÙÀûÓù۲취ÅäƽÆäËüÎïÖÊ£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ£º4MnO4-+12H+
| ||
¹Ê´ð°¸Îª£º4MnO4-+12H+
| ||
£¨3£©¸ù¾Ý·´Ó¦I2+2S2O32-=2I-+S4O62-£¬µ±·´Ó¦½áÊøʱ£¬ÈÜÒºµÄÀ¶É«Ïûʧ±ä³ÉÎÞÉ«ÈÜÒº£¬ËùÒԵζ¨ÖÕµãÊÇ£ºÈÜÒºÓÉÀ¶É«±äΪÎÞÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«£¬
¹Ê´ð°¸Îª£ºÈÜÒºÓÉÀ¶É«±äΪÎÞÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»±äÉ«£»
£¨4£©¸ù¾Ý·´Ó¦2MnO4-+5C2O42-+16H+=2Mn2++8H2O+10CO2¡ü¡¢2Cu2++4I-=2CuI¡ý+I2 I2+2S2O32-=2I-+S4O62-
¿ÉµÃ¹Øϵʽ£º2MnO4-¡«5C2O42- 2S2O32-¡«I2¡«2Cu2+
2 5 2 2
0.2¡Á8¡Á10-3mol n£¨C2O42-£© 0.25¡Á8¡Á10-3mol n£¨Cu2+£©
ËùÒÔn£¨C2O42-£©=4¡Á10-3mol£¬n£¨Cu2+£©=2¡Á10-3mol£¬
¸ù¾ÝµçºÉÊغãÔÀí£ºn£¨K+£©=2n£¨C2O42-£©-2n£¨Cu2+£©=4¡Á10-3mol
¸ù¾ÝÖÊÁ¿ÊغãÔÀí£ºm£¨H2O£©=0.6720-39¡Á4¡Á10-3-64¡Á2¡Á10-3-88¡Á4¡Á10-3=0.036g
n£¨H2O£©=2¡Á10-3£¬n £¨K+£©£ºn£¨Cu2+£©£ºn£¨C2O42-£©£ºn £¨H2O£©=2£º1£º2£º1£¬
ËùÒÔÑùÆ·¾§Ìå×é³ÉÊÇ£ºK2Cu£¨C2O4£©2?H2O£¬
´ð£º¸ÃÑùÆ·¾§ÌåµÄ»¯Ñ§Ê½Îª£ºK2Cu£¨C2O4£©2?H2O£®
µãÆÀ£º±¾Ì⿼²éÁË̽¾¿ÎïÖÊ×é³É·½·¨£¬ÌâÄ¿µÄ×ÛºÏÐÔ½ÏÇ¿£¬¼ÆËãÁ¿ÉԴ󣬳ä·Ö¿¼²éÁËѧÉú·ÖÎö¡¢Àí½âÐÅÏ¢¡¢ÀûÓÃÐÅÏ¢µÄÄÜÁ¦£¬±¾ÌâÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿