ÌâÄ¿ÄÚÈÝ

³£ÎÂÏ£¬½«0.1 mol/L¡¡HAÈÜÒºÓë0.1 mol/L¡¡NaOHÈÜÒºµÈÌå»ý»ìºÏºó(ºöÂÔ»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯)£¬²âµÃ»ìºÏÈÜÒºµÄpHµÈÓÚ8£®ÊԻشðÏÂÁÐÎÊÌ⣺

(1)»ìºÏÈÜÒºµÄpHµÈÓÚ8µÄÔ­ÒòÊÇ(ÓÃÀë×Ó·½³Ìʽ±íʾ)________£®

(2)»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)________0.1 mol/L¡¡NaOHÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»òÕß¡°µÈÓÚ¡±)£®

(3)Çó³ö»ìºÏÈÜÒºÖÐÏÂÁÐËãʽµÄ¾«È·¼ÆËã½á¹û(Ìî¾ßÌåÊý×Ö)£º

c(Na+)£­c(A£­)£½________mol/L£»

c(OH£­)£­c(HA)£½________mol/L£®

´ð°¸£º
½âÎö£º

¡¡¡¡´ð°¸

¡¡¡¡

¡¡¡¡½²Îö£º

¡¡¡¡


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

´×ËáºÍÑÎËáÊÇÈÕ³£Éú»îÖм«Îª³£¼ûµÄËᣬÔÚÒ»¶¨Ìõ¼þÏÂCH3COOHÈÜÒºÖдæÔÚµçÀëÆ½ºâ£ºCH3COOHCH3COO£­+H+  ¦¤H£¾0

£¨1£©³£ÎÂÏ£¬pH£½5´×ËáÈÜÒºÖУ¬c(CH3COO£­)£½______mol/L(¾«È·Öµ£¬ÒªÇóÁÐʽ²»±Ø»¯¼ò)£»

£¨2£©ÏÂÁз½·¨ÖпÉÒÔʹ0.10 mol¡¤L-1 CH3COOHµÄµçÀë³Ì¶ÈÔö´óµÄÊÇ          

a£®¼ÓÈëÉÙÁ¿0.10 mol¡¤L£­1Ï¡ÑÎËá     b£®¼ÓÈÈCH3COOHÈÜÒº   c£®¼ÓˮϡÊÍÖÁ0.010 mol¡¤L£­1  

d£®¼ÓÈëÉÙÁ¿±ù´×Ëá     e£®¼ÓÈëÉÙÁ¿ÂÈ»¯ÄƹÌÌå           f£®¼ÓÈëÉÙÁ¿0.10 mol¡¤L£­1 NaOHÈÜÒº

£¨3£©½«µÈÖÊÁ¿µÄпͶÈëµÈÌå»ýÇÒpH¾ùµÈÓÚ3µÄ´×ËáºÍÑÎËáÈÜÒºÖУ¬¾­¹ý³ä·Ö·´Ó¦ºó£¬·¢ÏÖÖ»ÔÚÒ»ÖÖÈÜÒºÖÐÓÐп·ÛÊ£Ó࣬ÔòÉú³ÉÇâÆøµÄÌå»ý£ºV(ÑÎËá)_________V(´×Ëá)£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©

£¨4£©ÓÃNaOHÈÜÒº·Ö±ðµÎ¶¨20.00mL0.1mol/LÑÎËáºÍ20.00mL0.1mol/L´×ËáÈÜÒº£¬µÃµ½ÈçͼËùʾÁ½ÌõµÎ¶¨ÇúÏߣ¬ÓÃNaOHÈÜÒºµÎ¶¨´×ËáÈÜÒºµÄÇúÏßÊÇ                 £¨Ìͼ1¡±»ò¡°Í¼2¡±£©

£¨5£©³£ÎÂÏ£¬½«0.1 mol/LÑÎËáºÍ0.1 mol/L´×ËáÄÆÈÜÒº»ìºÏ£¬ËùµÃÈÜҺΪÖÐÐÔ£¬Ôò»ìºÏÈÜÒºÖи÷Àë×ÓµÄŨ¶È°´ÓÉ´óµ½Ð¡ÅÅÐòΪ_______________________________¡£

£¨6£©ÏÂͼ±íʾÈÜÒºÖÐc(H£«)ºÍc(OH£­)µÄ¹ØÏµ

¢ÙMÇøÓòÄÚ£¨ÒõÓ°²¿·Ö£©ÈÎÒâµãc(H£«)______c(OH£­)£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©

¢ÚÔÚT2ζÈÏ£¬½«pH£½9 NaOHÈÜÒºÓëpH£½4 HClÈÜÒº»ìºÏ£¬ÈôËùµÃ»ìºÏÈÜÒºµÄpH£½7£¬ÔòNaOHÈÜÒºÓëHClÈÜÒºµÄÌå»ý±ÈΪ        ¡££¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©

 

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

¾«Ó¢¼Ò½ÌÍø