ÌâÄ¿ÄÚÈÝ

ÏÂÁÐÒ»¾ä»°ÖÐÐðÊöÁËÁ½¸öÖµ£¬Ç°Õß¼ÇΪM£¬ºóÕß¼ÇΪN£¬MºÍNµÄ¹Øϵ´ÓA¡¢B¡¢C¡¢D ÖÐÑ¡Ôñ£º
A£®M£¾N         B£®M£¼N       C£®M=N     D£®ÎÞ·¨±È½Ï
£¨1£©ÏàͬζÈÏ£¬1L 1mol/L µÄNH4ClÈÜÒºÖеÄNH4+¸öÊýºÍ2L 0.5mol?L-1NH4ClÈÜÒºÖÐNH4+µÄ¸öÊý£º
A
A
£»
£¨2£©ÏàͬζÈÏ£¬pHֵΪ12µÄÉÕ¼îÈÜÒºÖÐË®µÄµçÀë¶ÈºÍpHֵΪ12µÄCH3COONaÈÜÒºÖÐË®µÄµçÀë¶È£º
B
B
£»
£¨3£©Á½·ÝÊÒÎÂʱµÄ±¥ºÍʯ»ÒË®£¬Ò»·ÝÉýε½50¡æ£»ÁíÒ»·Ý¼ÓÈëÉÙÁ¿CaO£¬»Ö¸´ÖÁÊÒΣ¬Á½ÈÜÒºÖеÄc£¨Ca2+£©£º
B
B
£»
£¨4£©³£ÎÂÏÂÁ½·ÝµÈŨ¶ÈµÄ´¿¼îÈÜÒº£¬½«µÚ¶þ·ÝÉý¸ßζȣ¬Á½ÈÜÒºÖÐc£¨HCO3-£©£º
B
B
£»
£¨5£©½«pHֵΪ2µÄ´×ËáºÍÑÎËᶼϡÊÍÏàͬ±¶ÊýËùµÃÏ¡ÈÜÒºµÄpHÖµ£º
B
B
£»
£¨6£©³£ÎÂÏÂ0.1mol/LµÄCH3COOHÓë0.1mol/LCH3COONaµÈÌå»ý»ìºÏºóÈÜÒºÖÐc£¨Na+£©ºÍc£¨CH3COO-£©£º
B
B
£»
£¨7£©Í¬Î¶ÈÏ£¬0.1mol/LFeCl3ÈÜÒºÖÐFe3+Ë®½â°Ù·ÖÂÊÓë0.01mol?L-1FeCl3ÈÜÒºÖÐFe3+ µÄË®½â°Ù·ÖÂÊ£º
B
B
£»
£¨8£©ÊÒÎÂÏÂijǿËáºÍijǿ¼îÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpHֵΪ7£¬Ô­ËáÈÜÒººÍÔ­¼îÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£º
D
D
£»
£¨9£©pHÖµÏàͬµÄ´×ËáºÍÑÎËᣬ·Ö±ðÓÃÕôÁóˮϡÊÍÖÁÔ­À´µÄM±¶ºÍN±¶£¬Ï¡ÊͺóÁ½ÈÜÒºµÄPHÖµÈÔÈ»Ïàͬ£¬ÔòMºÍNµÄ¹ØϵÊÇ£º
A
A
£®
·ÖÎö£º£¨1£©ÂÈ»¯ï§ÖУ¬ï§¸ùÀë×ÓË®½â£¬Å¨¶ÈÔ½´óË®½â³Ì¶ÈԽС£»
£¨2£©¼ÆËãpHֵΪ12µÄÉÕ¼îÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶È´óСºÍpHֵΪ12µÄCH3COONaÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶È´óСÀ´Åжϣ»
£¨3£©ÇâÑõ»¯¸ÆµÄÈܽâ¶ÈËæ×ÅζȵÄÉý¸ß¶ø¼õС£»
£¨4£©Ì¼ËáÄÆÈÜÒºÖУ¬Ì¼Ëá¸ùµÄË®½â³Ì¶ÈËæ×ÅζȵÄÉý¸ß¶øÔö´ó£»
£¨5£©´×ËáÊÇÈõËᣬϡÊÍ´Ù½øµçÀ룬ÑÎËáÊÇÇ¿ËᣬϡÊ͹ý³ÌŨ¶È¼õС£»
£¨6£©´×ËáºÍ´×ËáÄƵĻìºÏÒºÖУ¬´×ËáµÄµçÀë³Ì¶È´óÓÚ´×Ëá¸ùµÄË®½â³Ì¶È£»
£¨7£©ÈÜҺԽϡÀë×ÓµÄË®½â³Ì¶ÈÔ½´ó£»
£¨8£©Ç¿ËáºÍÇ¿¼îÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpHֵΪ7£¬ÈÜÒºÏÔʾÖÐÐÔ£¬¸ù¾ÝÖкÍÔ­ÀíÀ´»Ø´ð£»
£¨9£©´×ËáÊÇÈõËᣬϡÊÍ´Ù½øµçÀ룬ÑÎËáÊÇÇ¿ËᣬϡÊ͹ý³ÌŨ¶È¼õС£®
½â´ð£º½â£º£¨1£©ÂÈ»¯ï§ÖУ¬ï§¸ùÀë×ÓË®½â£¬ï§¸ùÀë×ÓŨ¶ÈÔ½´óË®½â³Ì¶ÈԽС£¬¼´ÏàͬζÈÏ£¬1L 1mol/L µÄNH4ClÈÜÒºÖеÄNH4+¸öÊý´óÓÚ2L 0.5mol?L-1NH4ClÈÜÒºÖÐNH4+µÄ¸öÊý£¬¹Ê´ð°¸Îª£ºA£»
£¨2£©pHֵΪ12µÄÉÕ¼îÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ10-12mol/L£¬pHֵΪ12µÄCH3COONaÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ10-2mool/L£¬ËùÒÔÏàͬζÈÏ£¬pHֵΪ12µÄÉÕ¼îÈÜÒºÖÐË®µÄµçÀë¶ÈСÓÚpHֵΪ12µÄCH3COONaÈÜÒºÖÐË®µÄµçÀë¶È£¬¹Ê´ð°¸Îª£ºB£»
£¨3£©Á½·ÝÊÒÎÂʱµÄ±¥ºÍʯ»ÒË®£¬Ò»·ÝÉýε½50¡æ£¬ÇâÑõ»¯¸ÆµÄÈܽâ¶È¼õС£¬¸ÆÀë×ÓÊý¼õС£¬ÁíÒ»·Ý¼ÓÈëÉÙÁ¿CaO£¬»á·¢Éú·´Ó¦Éú³ÉÇâÑõ»¯¸Æ£¬»Ö¸´ÖÁÊÒΣ¬ÈÜÒºÈÔÈ»ÊDZ¥ºÍµÄ£¬¸ÆÀë×ÓÊý²»±ä£¬¹Ê´ð°¸Îª£ºB£»
£¨4£©Ì¼ËáÄÆÈÜÒºÖУ¬Ì¼Ëá¸ùµÄË®½â³Ì¶ÈËæ×ÅζȵÄÉý¸ß¶øÔö´ó£¬¼°Éý¸ßζÈÒÔ»ÆÆÒ£¬Ì¼ËáÇâ¸ùµÄŨ¶ÈÔö´ó£¬¹Ê´ð°¸Îª£ºB£»
£¨5£©´×ËáÊÇÈõËᣬϡÊÍ´Ù½øµçÀ룬ÑÎËáÊÇÇ¿ËᣬϡÊ͹ý³ÌŨ¶È¼õС£¬ËùÒÔ½«pHֵΪ2µÄ´×ËáºÍÑÎËᶼϡÊÍÏàͬ±¶ÊýËùµÃÏ¡ÈÜÒºµÄpHÖµÊÇ´×ËáµÄСÓÚÑÎËáµÄ£¬¹Ê´ð°¸Îª£ºB£»
£¨6£©´×ËáºÍ´×ËáÄƵĻìºÏÒºÖУ¬´×ËáµÄµçÀë³Ì¶È´óÓÚ´×Ëá¸ùµÄË®½â³Ì¶È£¬µ¼ÖÂÈÜÒºÏÔʾËáÐÔ£¬c£¨H+£©£¾c£¨OH-£©£¬¸ù¾ÝµçºÉÊغ㣬ËùÒÔc£¨Na+£©£¼c£¨CH3COO-£©£¬¹Ê´ð°¸Îª£ºB£»
£¨7£©ÈÜҺԽϡ£¬¼´Å¨¶ÈԽС£®ÌúÀë×ÓµÄË®½â³Ì¶ÈÔ½´ó£¬ËùÒÔ0.1mol/LFeCl3ÈÜÒºÖÐFe3+Ë®½â°Ù·ÖÂÊСÓÚ0.01mol?L-1FeCl3ÈÜÒºÖÐFe3+ µÄË®½â°Ù·ÖÂÊ£¬¹Ê´ð°¸Îª£ºB£»
£¨8£©Ç¿ËáºÍÇ¿¼îÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpHֵΪ7£¬ÔòÈÜÒºÏÔʾÖÐÐÔ£¬ËùÒÔÔ­À´ÈÜÒºÖÐÇâÀë×ÓºÍÇâÑõ¸ùÀë×ÓµÄÎïÖʵÄÁ¿ÊÇÏàµÈµÄ£¬µ«ÊÇËáºÍ¼îµÄŨ¶È´óС²»ÄÜÈ·¶¨£¬¹Ê´ð°¸Îª£ºD£»
£¨9£©´×ËáÊÇÈõËᣬϡÊÍ´Ù½øµçÀ룬ÔÚÌå»ýÔö´óµ¼ÖÂŨ¶È¼õСµÄͬʱ»¹»áµçÀë³öÒ»²¿·ÖÇâÀë×Ó£¬ÑÎËáÊÇÇ¿ËᣬϡÊ͹ý³ÌÖÐÌå»ýÔö´ó¶øµ¼ÖÂŨ¶È¼õС£¬pHÖµÏàͬµÄ´×ËáºÍÑÎËᣬ·Ö±ðÓÃÕôÁóˮϡÊÍÖÁÔ­À´µÄM±¶ºÍN±¶£¬Ï¡ÊͺóÁ½ÈÜÒºµÄPHÖµÈÔÈ»Ïàͬ£¬Ôò´×Ëá¼ÓË®¶à£¬¹Ê´ð°¸Îª£ºA£®
µãÆÀ£º±¾ÌâÊÇÒ»µÀ¹ØÓÚÑεÄË®½â¹æÂÉ¡¢Èõµç½âÖʵĵçÀëƽºâµÄÓ°ÏìµÈ·½Ãæ֪ʶµÄ×ۺϿ¼²éÌ⣬ҪÇóѧÉú¾ßÓзÖÎöºÍ½â¾öÎÊÌâµÄÄÜÁ¦£¬ÄѶȽϴó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

 

µ±Î¶ȸßÓÚ40¡æʱ£¬Í­µÄƽ¾ùÈܽâËÙÂÊËæ×Å·´Ó¦Î¶ÈÉý¸ß¶øϽµ£¬ÆäÖ÷ÒªÔ­ÒòÊÇ                   ¡£

¢ÈÔÚÌá´¿ºóµÄCuSO4ÈÜÒºÖмÓÈëÒ»¶¨Á¿µÄNa2SO3ºÍNaClÈÜÒº£¬¼ÓÈÈ£¬Éú³ÉCuCl³Áµí¡£ÖƱ¸CuClµÄÀë×Ó·½³ÌʽÊÇ                                              ¡£

28. (9·Ö)ÏÂÁÐÒ»¾ä»°ÖÐÐðÊöÁËÁ½¸öÖµ£¬Ç°Õß¼ÇΪM£¬ºóÕß¼ÇΪN£¬MºÍNµÄ¹Øϵ´ÓA¡¢B¡¢C¡¢DÖÐÑ¡Ôñ

A. M>N                    B. M<N                C. M=N         D. ÎÞ·¨±È½Ï

¢ÅÏàͬζÈÏ£¬1L 1mol/L µÄNH4ClÈÜÒºÖеÄNH4£«¸öÊýºÍ2 L 0.5mol¡¤L£­1NH4ClÈÜÒºÖÐNH4£«µÄ¸öÊý£º      £»

¢ÆÏàͬζÈÏ£¬pHֵΪ12µÄÉÕ¼îÈÜÒºÖÐË®µÄµçÀë¶ÈºÍpHֵΪ12µÄCH3COONaÈÜÒºÖÐË®µÄµçÀë¶È£º      £»

¢ÇÁ½·ÝÊÒÎÂʱµÄ±¥ºÍʯ»ÒË®£¬Ò»·ÝÉýε½50¡æ£»ÁíÒ»·Ý¼ÓÈëÉÙÁ¿CaO£¬»Ö¸´ÖÁÊÒΣ¬Á½ÈÜÒºÖеÄc(Ca2+)£º    £»

¢È³£ÎÂÏÂÁ½·ÝµÈŨ¶ÈµÄ´¿¼îÈÜÒº£¬½«µÚ¶þ·ÝÉý¸ßζȣ¬Á½ÈÜÒºÖÐc(HCO3£­)£º       £»

¢É½«pHֵΪ2µÄ´×ËáºÍÑÎËᶼϡÊÍÏàͬ±¶ÊýËùµÃÏ¡ÈÜÒºµÄpHÖµ£º         £»

¢Ê³£ÎÂÏÂ0.1mol/LµÄCH3COOHÓë0.1mol/LCH3COONaµÈÌå»ý»ìºÏºóÈÜÒºÖÐc(Na+)ºÍc(CH3COO£­)£º      £»

¢ËͬζÈÏ£¬0.1mol/LFeCl3ÈÜÒºÖÐFe3+Ë®½â°Ù·ÖÂÊÓë0.01mol¡¤L£­1FeCl3ÈÜÒºÖÐFe3+ µÄË®½â°Ù·ÖÂÊ£º        £»

¢ÌÊÒÎÂÏÂijǿËáºÍijǿ¼îÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpHֵΪ7£¬Ô­ËáÈÜÒººÍÔ­¼îÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£º   £»

¢ÍPHÖµÏàͬµÄ´×ËáºÍÑÎËᣬ·Ö±ðÓÃÕôÁóˮϡÊÍÖÁÔ­À´µÄM±¶ºÍN±¶£¬Ï¡ÊͺóÁ½ÈÜÒºµÄPHÖµÈÔÈ»Ïàͬ£¬ ÔòMºÍNµÄ¹ØϵÊÇ£º        ¡£

 

ÏÂÁÐÒ»¾ä»°ÖÐÐðÊöÁËÁ½¸öÖµ£¬Ç°Õß¼ÇΪM£¬ºóÕß¼ÇΪN£¬MºÍNµÄ¹Øϵ´ÓA¡¢B¡¢C¡¢D ÖÐÑ¡Ôñ£º
A£®M£¾N         B£®M£¼N       C£®M=N     D£®ÎÞ·¨±È½Ï
£¨1£©ÏàͬζÈÏ£¬1L 1mol/L µÄNH4ClÈÜÒºÖеÄNH4+¸öÊýºÍ2L 0.5mol?L-1NH4ClÈÜÒºÖÐNH4+µÄ¸öÊý£º    £»
£¨2£©ÏàͬζÈÏ£¬pHֵΪ12µÄÉÕ¼îÈÜÒºÖÐË®µÄµçÀë¶ÈºÍpHֵΪ12µÄCH3COONaÈÜÒºÖÐË®µÄµçÀë¶È£º    £»
£¨3£©Á½·ÝÊÒÎÂʱµÄ±¥ºÍʯ»ÒË®£¬Ò»·ÝÉýε½50¡æ£»ÁíÒ»·Ý¼ÓÈëÉÙÁ¿CaO£¬»Ö¸´ÖÁÊÒΣ¬Á½ÈÜÒºÖеÄc£¨Ca2+£©£º    £»
£¨4£©³£ÎÂÏÂÁ½·ÝµÈŨ¶ÈµÄ´¿¼îÈÜÒº£¬½«µÚ¶þ·ÝÉý¸ßζȣ¬Á½ÈÜÒºÖÐc£¨HCO3-£©£º    £»
£¨5£©½«pHֵΪ2µÄ´×ËáºÍÑÎËᶼϡÊÍÏàͬ±¶ÊýËùµÃÏ¡ÈÜÒºµÄpHÖµ£º    £»
£¨6£©³£ÎÂÏÂ0.1mol/LµÄCH3COOHÓë0.1mol/LCH3COONaµÈÌå»ý»ìºÏºóÈÜÒºÖÐc£¨Na+£©ºÍc£¨CH3COO-£©£º    £»
£¨7£©Í¬Î¶ÈÏ£¬0.1mol/LFeCl3ÈÜÒºÖÐFe3+Ë®½â°Ù·ÖÂÊÓë0.01mol?L-1FeCl3ÈÜÒºÖÐFe3+ µÄË®½â°Ù·ÖÂÊ£º    £»
£¨8£©ÊÒÎÂÏÂijǿËáºÍijǿ¼îÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpHֵΪ7£¬Ô­ËáÈÜÒººÍÔ­¼îÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£º    £»
£¨9£©pHÖµÏàͬµÄ´×ËáºÍÑÎËᣬ·Ö±ðÓÃÕôÁóˮϡÊÍÖÁÔ­À´µÄM±¶ºÍN±¶£¬Ï¡ÊͺóÁ½ÈÜÒºµÄPHÖµÈÔÈ»Ïàͬ£¬ÔòMºÍNµÄ¹ØϵÊÇ£º    £®

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

¾«Ó¢¼Ò½ÌÍø