(1)Ôò¸ÃÈÜÒºÖÐc(H+)_______c(OH-)(Ìî¡°£¼¡±¡°=¡±»ò¡°£¾¡±)£¬¼òҪ˵Ã÷ÀíÓÉ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©___________________________________________¡£
£¨2£©ÏÖÏòNH4HSO3ÈÜÒºÖУ¬ÖðµÎ¼ÓÈëÉÙÁ¿º¬ÓзÓ̪µÄNaOHÈÜÒº£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ_______£¬Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ_______________________________£»Èô¼ÌÐøµÎ¼ÓNaOHÈÜÒºÖÁ¹ýÁ¿£¬ÔòÓÖ·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________________________________¡£
£¨1£©£¾ ÒòΪc(SO)£¾c(H2SO3),¹ÊHSOµÄµçÀë(HSOH++SO)³Ì¶È±ÈHSOµÄË®½â(HSO+H2OH2SO3+OH-)³Ì¶È´ó£¬¹Êc(H+)£¾c(OH-)
£¨2£©ºìÉ«ÍÊÈ¥ OH-+HSO====H2O+SO NH+OH-====NH3¡¤H2O
½âÎö£º(1)ÒòΪc(SO)£¾c(H2SO3),¹ÊHSOµÄµçÀë³Ì¶È±ÈHSOµÄË®½â³Ì¶È´ó£¬¹Êc(H+)£¾c(OH-)¡£
£¨2£©NH4HSO3ÈÜÒºÖмÓÈëÉÙÁ¿º¬ÓзÓ̪µÄNaOHÈÜÒº£¬NaOHÊ×ÏȺÍHSO·´Ó¦£¬¼ÌÐøµÎ¼ÓNaOHÖÁ¹ýÁ¿£¬ÔòNaOHÓÖºÍNH·´Ó¦¡£
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨1£©¸ÃÈÜÒºÖÐc(H+)____________c(OH-)(Ìî¡°£¾¡±¡°=¡±»ò¡°£¼¡±)£¬¼òÊöÀíÓÉ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£º_______________________________________________________________¡£
£¨2£©ÏÖÏòNH4HSO3ÈÜÒºÖУ¬ÖðµÎ¼ÓÈëÉÙÁ¿º¬ÓзÓ̪µÄNaOHÈÜÒº£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ____________£»Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º_____________________________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÌì½òÊÐÇà¹âÖÐѧ¸ß¶þµÚһѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
£¨10·Ö£©ÒÑÖªÔÚ0.1 mol/LµÄNaHSO3ÈÜÒºÖÐÓйØ΢Á£Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ£º
c(Na£«)>c(HSO)>c(SO)>c(H2SO3)¡£
(1)Ôò¸ÃÈÜÒºÖÐc(H£«)________c(OH£)(Ìî¡°<¡±¡¢¡°>¡±»ò¡°£½¡±)£¬¼òÊöÀíÓÉ(ÓÃÀë×Ó·½³ÌʽºÍ±ØÒªµÄÎÄ×Ö˵Ã÷)________________ ____________________________________
____________ ________¡£
(2)ÏÖÏòNaHSO3ÈÜÒºÖУ¬ÖðµÎ¼ÓÈëÉÙÁ¿º¬ÓзÓ̪µÄNaOHÈÜÒº£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ________________________________________________________________________£»
д³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ_____________ _________________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨1£©¸ÃÈÜÒºÖÐc£¨H+£©________c£¨OH-£©£¨Ìî¡°£¾¡±¡°=¡±»ò¡°£¼¡±£©£¬¼òÊöÀíÓÉ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©£º_______________________________¡£
£¨2£©ÏÖÏòNH4HSO3ÈÜÒºÖУ¬ÖðµÎ¼ÓÈëÉÙÁ¿º¬ÓзÓ̪µÄNaOHÈÜÒº£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ__________________________£»Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º______________________¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(1)¸ÃÈÜÒºÖУÛH+£Ý £ÛOH-£Ý(Ìî¡°£¾¡±¡°=¡±»ò¡°£¼¡±)£¬ÆäÀíÓÉÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)£º ¡£
(2)ÏÖÏòNH4HSO3ÈÜÒºÖУ¬ÖðµÎ¼ÓÈëÉÙÁ¿º¬ÓзÓ̪µÄNaOHÈÜÒº£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇ £»Ð´³ö¸Ã·´Ó¦µÄÀë×Ó·½³Ìʽ£º ¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com