ÌâÄ¿ÄÚÈÝ

ÔÚ25¡æʱ£¬½«ËáHAÓë¼îMOHµÈÌå»ý»ìºÏ.
£¨1£©Èô0.01mol/LµÄÇ¿ËáHAÓë0.01mol/LÇ¿¼îMOH»ìºÏ£¬ÔòËùµÃÈÜÒºÏÔ      £¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£¬ÏÂͬ£©¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ                                         
£¨2£©ÈôPH=3µÄÇ¿ËáHAÓëPH=11µÄÈõ¼îMOH»ìºÏ£¬ÔòËùµÃÈÜÒºÏÔ         £¬ÀíÓÉÊÇ£º          
£¨3£©Èô0.01mol/LµÄÇ¿ËáHAÓë0.01mol/LÈõ¼îMOH»ìºÏ£¬ÔòËùµÃÈÜÒºÏÔ              £¬½âÊÍÕâÒ»ÏÖÏóµÄÀë×Ó·½³ÌʽÊÇ                                
£¨1£©ÖÐÐÔ   H++OH-=H2O
£¨2£©¼îÐÔ  ËáHAÓë¼îMOHÖкͺ󣬼î¹ýÁ¿£¬»¹»áµçÀë³öOH-(2·Ö£¬ÆäËûºÏÀí´ð°¸Ò²¿É¸ø·Ö)
£¨3£© ËáÐÔ    M++H2OMOH+H+(2·Ö)

ÊÔÌâ·ÖÎö£º£¨1£©Èô0.01mol/LµÄÇ¿ËáHAÓë0.01mol/LÇ¿¼îMOH»ìºÏ£¬¶þÕßÇ¡ºÃÍêÈ«·´Ó¦Éú³ÉÇ¿ËáÇ¿¼îÑÎMA£¬ÔòËùµÃÈÜÒºÏÔÖÐÐÔ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪH++OH-=H2O£»£¨2£©ÈôPH=3µÄÇ¿ËáHAÓëPH=11µÄÈõ¼îMOH»ìºÏ£¬ÔòËùµÃÈÜÒºÏÔ¼îÐÔ£¬ÀíÓÉÊÇ£ºPH=3µÄÇ¿ËáHAÓëPH=11µÄÈõ¼îMOHÖкͺ󣬼î¹ýÁ¿£¬»¹»áµçÀë³öOH-£»£¨3£©Èô0.01mol/LµÄÇ¿ËáHAÓë0.01mol/LÈõ¼îMOH»ìºÏ£¬¶þÕßÇ¡ºÃÍêÈ«·´Ó¦Éú³ÉÇ¿ËáÈõ¼îÑÎMA£¬Èõ¼îÑôÀë×ÓË®½âÏÔËáÐÔ£¬½âÊÍÕâÒ»ÏÖÏóµÄÀë×Ó·½³ÌʽÊÇ
M++H2OMOH+H+¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÖÓÐÎåÖÖ¿ÉÈÜÐÔÎïÖÊA¡¢B¡¢C¡¢D¡¢E£¬ËüÃÇËùº¬Òõ¡¢ÑôÀë×Ó»¥²»Ïàͬ£¬·Ö±ðº¬ÓÐÎåÖÖÑôÀë×ÓNa£«¡¢Al3£«¡¢Mg2£«¡¢Ba2£«¡¢Fe3£«ºÍÎåÖÖÒõÀë×ÓCl£­¡¢OH£­¡¢NO3¡ª¡¢CO32¡ªµÄÒ»ÖÖ¡£
(1)ijͬѧͨ¹ý±È½Ï·ÖÎö£¬ÈÏΪÎÞÐè¼ìÑé¾Í¿ÉÅжÏÆäÖбØÓеÄÁ½ÖÖÎïÖÊÊÇ________ºÍ________(Ìѧʽ)¡£
(2)ΪÁËÈ·¶¨X£¬ÏÖ½«(1)ÖеÄÁ½ÖÖÎïÖʼÇΪAºÍB£¬º¬XµÄÎïÖʼÇΪC£¬µ±CÓëBµÄÈÜÒº»ìºÏʱ£¬²úÉúºìºÖÉ«³ÁµíºÍÎÞÉ«ÎÞζÆøÌ壻µ±CÓëAµÄÈÜÒº»ìºÏʱ²úÉú×Ø»ÆÉ«³Áµí£¬Ïò¸Ã³ÁµíÖеÎÈëÏ¡ÏõËá³Áµí²¿·ÖÈܽ⣬×îºóÁôÓа×É«³Áµí²»ÔÙÈܽ⡣ÔòXΪ________¡£
A£®SO32¡ª     B£®SO4¡ª C£®CH3COO£­    D£®SiO32¡ª
(3)½«CuͶÈëµ½×°ÓÐDÈÜÒºµÄÊÔ¹ÜÖУ¬Cu²»Èܽ⣻ÔٵμÓÏ¡H2SO4£¬CuÖð½¥Èܽ⣬¹Ü¿Ú¸½½üÓкì×ØÉ«ÆøÌå³öÏÖ¡£ÔòÎïÖÊDÒ»¶¨º¬ÓÐÉÏÊöÀë×ÓÖеÄ________ (ÌîÏàÓ¦µÄÀë×Ó·ûºÅ)¡£Óйط´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________________________¡£
(4)ÀûÓÃÉÏÊöÒѾ­È·¶¨µÄÎïÖÊ£¬¿ÉÒÔ¼ìÑé³öD¡¢EÖеÄÑôÀë×Ó¡£Çë¼òÊöʵÑé²Ù×÷²½Öè¡¢ÏÖÏó¼°½áÂÛ£º______________________________________________

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

¾«Ó¢¼Ò½ÌÍø