ÌâÄ¿ÄÚÈÝ

15£®25¡æÊ±£¬ÔÚ´×ËáºÍ´×ËáÄÆ»ìºÏÈÜÒºÖÐÓÐc£¨CH3COOH£©+c£¨CH3COO-£©=0.1mol/LÇÒc£¨CH3COOH£©¡¢c£¨CH3COO-£©ÓëpHµÄ¹ØÏµÈçͼ£®ÓйØÀë×ÓŨ¶ÈÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®pH=3.5ÈÜÒºÖУºc£¨Na+£©+c£¨H+£©-c£¨OH-£©+c£¨CH3COOH£©=0.1mol/L
B£®pH=5.5ÈÜÒºÖУºc£¨CH3COOH£©£¾c£¨CH3COO-£©£¾c£¨H+£©£¾c£¨OH-£©
C£®Wµã±íʾÈÜÒºÖУºc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©
D£®ÏòWµãËù±íʾÈÜÒºÖÐͨÈë0.05mol HClÆøÌ壨ÈÜÒºÌå»ý±ä»¯¿ÉºöÂÔ£©£ºc£¨H+£©=c£¨CH3COOH£©+c£¨OH-£©

·ÖÎö ¸ù¾ÝͼÏóÖª£¬Ëæ×ÅÈÜÒºpHÔö´ó£¬ÔòÈÜÒºÖд×ËáŨ¶È½µµÍ¡¢´×Ëá¸ùÀë×ÓŨ¶ÈÔö´ó£¬ÔòʵÏßÊÇ´×Ëá¸ùÀë×ÓŨ¶È¡¢ÐéÏßÊÇ´×ËáŨ¶È±ä»¯£¬
A£®pH=3.5µÄÈÜÒºÖдæÔÚÎïÁÏÊØºã£¬¸ù¾ÝÎïÁÏÊØºãÅжϣ»
B£®µ±pH=5.5ʱ£¬c£¨CH3COOH£©£¼c£¨CH3COO-£©£»
C£®WµãËù±íʾµÄÈÜÒºÖÐc£¨CH3COOH£©=c£¨CH3COO-£©£¬ÈÜÒºÖдæÔÚµçºÉÊØºã£»
D£®¸ù¾ÝÎïÁÏÊØºãÓÐ c£¨CH3COOH£©+c£¨CH3COO-£©=0.1 mol•L-1£¬ÓɵçºÉÊØºãÓÐc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©+c£¨Cl-£©£®

½â´ð ½â£º¸ù¾ÝͼÏóÖª£¬Ëæ×ÅÈÜÒºpHÔö´ó£¬ÔòÈÜÒºÖд×ËáŨ¶È½µµÍ¡¢´×Ëá¸ùÀë×ÓŨ¶ÈÔö´ó£¬ÔòʵÏßÊÇ´×Ëá¸ùÀë×ÓŨ¶È¡¢ÐéÏßÊÇ´×ËáŨ¶È±ä»¯£¬
A£®B£®pH=3.5µÄÈÜÒºÖдæÔÚÎïÁÏÊØºã£¬¸ù¾ÝÎïÁÏÊØºã¡¢µçºÉÊØºãµÃc£¨CH3COOH£©+c£¨CH3COO-£©=c£¨Na+£©+c£¨H+£©-c£¨OH-£©+c£¨CH3COOH£©=0.1mol•L-1£¬¹ÊAÕýÈ·£»
B£®µ±pH=5.5ʱ£¬¸ù¾ÝͼÏóÖªc£¨CH3COOH£©£¼c£¨CH3COO-£©£¬¹ÊB´íÎó£»
C£®WµãËù±íʾµÄÈÜÒºÖÐc£¨CH3COOH£©=c£¨CH3COO-£©£¬ÈÜÒºÖдæÔÚµçºÉÊØºã£ºc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬¹ÊCÕýÈ·£»
D£®ÏòWµãËù±íʾÈÜÒºÖÐͨÈë0.05molHClÆøÌ壬ԭÓÐÆ½ºâ±»´òÆÆ£¬½¨Á¢ÆðÁËÐÂµÄÆ½ºâ£¬ÈÜÒºÖеçºÉÊØºã¹ØÏµÎª£ºc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©+c£¨Cl-£©£»ÎïÁÏÊØºã¹ØÏµÎª£º2c£¨Cl-£©=c£¨CH3COO-£©+c£¨CH3COOH£©=0.1mol•L-1£¬µÃ2c£¨Na+£©+2c£¨H+£©=3c£¨CH3COO-£©+2c£¨OH-£©+c£¨CH3COOH£©£¬c£¨Na+£©=0.05mol/L£¬c£¨CH3COOH£©+c£¨CH3COO-£©=0.1mol/L£¬Ôòc£¨H+£©¨Tc£¨CH3COO-£©+c£¨OH-£©£¬¹ÊD´íÎó£»
¹ÊÑ¡AC£®

µãÆÀ ±¾Ì⿼²éÁËËá¼î»ìºÏµÄ¶¨ÐÔÅжϼ°Àë×ÓŨ¶È´óС±È½Ï£¬ÌâÄ¿ÄѶÈÖеȣ¬Ã÷È·ÈÜÒºÖдæÔÚµÄÑÎÀàË®½â¡¢Ô­×ÓÊØºãºÍÎïÁÏÊØºãÊǽⱾÌâ¹Ø¼ü£¬×¢ÒâÕÆÎÕÈÜÒºËá¼îÐÔÓëÈÜÒºpHµÄ¹ØÏµ£¬Äܹ»ÀûÓõçºÉÊØºã¡¢ÎïÁÏÊØºã¼°ÑεÄË®½âÔ­ÀíÅжϸ÷Àë×ÓŨ¶È´óС£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
20£®°ÂµØÀû»¯Ñ§¼Ò°Ý¶úµÄ¡°¼îÈÜ·¨¡±Óë·¨¹ú»¯Ñ§¼Ò°£Â³µÄ¡°µç½â·¨¡±Ïà½áºÏ£¬µì¶¨ÁËÏÖ´úµç½âÒ±Á¶ÂÁ¹¤Òµ·½·¨µÄ»ù´¡£¬µç½âÂÁµÄÔ­ÁÏÖ÷ÒªÀ´×ÔÂÁÍÁ¿ó£¨Ö÷Òª³É·ÖΪAl2O3¡¢Fe2O3¡¢SiO2µÈ£©£¬»ù±¾Á÷³Ì¼°×°ÖÃͼÈçÏ£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©²Ù×÷¢ÙºÍ²Ù×÷¢Ú¶¼ÊǹýÂË£¨Ìî²Ù×÷Ãû³Æ£©£¬¼ÓÈë¹ýÁ¿ÉÕ¼îÈÜҺʱ·´Ó¦µÄÀë×Ó·½³ÌʽΪAl3++4OH-=AlO2-+2H2O£¬Fe3++3OH-=Fe£¨OH£©3¡ý£®
£¨2£©¼ÓÈë¹ýÁ¿X£¬·¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪNaAlO2+CO2+2H2O=Al£¨OH£©3¡ý+NaHCO3£®
£¨3£©µç½âʱ£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª2O2--4e-=O2¡ü£¬Êµ¼ÊÉú²úÖÐÑô¼«ÐèÒª¶¨ÆÚ¸ü»»£¬Ô­ÒòÊÇÑô¼«²úÉúµÄÑõÆø»áÓëµç¼«²ÄÁÏÖеÄ̼·´Ó¦¶øÊ¹Ñô¼«ËðºÄ£®
£¨4£©ÔÚµç½â²ÛµÄ¸Ö°åºÍÒõ¼«Ì¼ËزÄÁÏÖ®¼äÐèÒª·ÅÖÃa£¨Ñ¡Ìî×ÖĸÐòºÅ£©£®
a¡¢ÄÍ»ð¾øÔµ   b¡¢·À±¬µ¼ÈÈ    c¡¢Äͻ𵼵砠 d¡¢µ¼Èȵ¼µç
£¨5£©µç½âÂÁʱ£¬ÒÔÑõ»¯ÂÁ-±ù¾§Ê¯ÈÜÈÚҺΪµç½âÖÊ£¬ÆäÖÐÒ²³£¼ÓÈëÉÙÁ¿µÄ·ú»¯¸Æ£¬·ú»¯¸ÆµÄ×÷ÓÃÊǰïÖú½µµÍÑõ»¯ÂÁµÄÈÛµã£®ÎªÖÆÈ¡±ù¾§Ê¯£¬¹¤ÒµÉϲÉÓÃÇâÑõ»¯ÂÁ¡¢´¿¼îºÍ·ú»¯ÇâÔÚÒ»¶¨Ìõ¼þϳä·Ö»ìºÏ·´Ó¦ÖÆÈ¡±ù¾§Ê¯£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2Al£¨OH£©3+12HF+3Na2CO3=2Na3AlF6+3CO2¡ü+9H2O£®
£¨6£©ÉÏÊöÁ÷³ÌÉú²ú1mol½ðÊôÂÁºÄµçÄÜÔ¼1.8¡Á106J£®Ä³ÂÁÍÁ¿óÖÐÑõ»¯ÂÁµÄÖÊÁ¿·ÖÊýΪ51%£¬Ôò1¶Ö¸ÃÂÁÍÁ¿óÒ±Á¶½ðÊôÂÁºÄµçÄÜ1.8¡Á1010J£»ÒÑ֪ÿ¸öÂÁÖÆÒ×À­¹ÞÔ¼15g£¬ÆäÖÐÂÁµÄÖÊÁ¿·ÖÊýΪ90%£¬ÓÉÒ»Ö»Ò×À­¹Þ»ØÊÕÉú²ú½ðÊôÂÁºÄµçÄÜÔ¼4.5¡Á104J£¬ÔòÖÆµÃµÈÁ¿½ðÊôÂÁµÄÄܺÄΪÉÏÊöÁ÷³ÌÉú²ú·½·¨µÄ5%£»Í¨¹ý¶Ô¼ÆËãÊý¾ÝµÄ·ÖÎö£¬ÄãÄܵõ½Ê²Ã´½áÂÛÒ×À­¹ÞµÄ»ØÊÕÀûÓÃÓÐÀûÓÚ½ÚÔ¼ÄÜÔ´£®

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

¾«Ó¢¼Ò½ÌÍø