ÌâÄ¿ÄÚÈÝ
Óá°£¾¡±¡¢¡°=¡±¡¢¡°£¼¡±Ìîд£®
£¨1£©³£ÎÂÏ£¬½«0.1mol/LµÄ´×ËáÈÜÒººÍ0.1mol/L´×ËáÄÆÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃÈÜÒºÏÔËáÐÔ£¬Ôò¸Ã»ìºÏÒºÖУºc£¨CH3COOH£© c£¨CH3COO-£©£»2c£¨Na+£© c£¨CH3COO-£©+c£¨CH3COOH£©£»
£¨2£©³£ÎÂÏ£¬½«0.1mol/LµÄHCNÈÜÒººÍ0.1mol/LµÄNaCNÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃ»ìºÏÒºÖÐc£¨HCN£©£¾c£¨CN-£©£¬£¨¼ÙÉè»ìºÏ¹ý³ÌÖÐÈÜÒºÌå»ýµÄ¸Ä±äºöÂÔ²»¼Æ£©£®Ôò¸Ã»ìºÏÒºpH 7£¬c£¨HCN£©+c£¨CN-£© 0.1mol/L£¬c£¨HCN£©-c£¨CN-£© 2c£¨OH-£©£®
£¨1£©³£ÎÂÏ£¬½«0.1mol/LµÄ´×ËáÈÜÒººÍ0.1mol/L´×ËáÄÆÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃÈÜÒºÏÔËáÐÔ£¬Ôò¸Ã»ìºÏÒºÖУºc£¨CH3COOH£©
£¨2£©³£ÎÂÏ£¬½«0.1mol/LµÄHCNÈÜÒººÍ0.1mol/LµÄNaCNÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃ»ìºÏÒºÖÐc£¨HCN£©£¾c£¨CN-£©£¬£¨¼ÙÉè»ìºÏ¹ý³ÌÖÐÈÜÒºÌå»ýµÄ¸Ä±äºöÂÔ²»¼Æ£©£®Ôò¸Ã»ìºÏÒºpH
¿¼µã£ºÀë×ÓŨ¶È´óСµÄ±È½Ï
רÌ⣺µçÀëÆ½ºâÓëÈÜÒºµÄpHרÌâ
·ÖÎö£º£¨1£©³£ÎÂÏ£¬µÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄCH3COOH¡¢CH3COONaÈÜÒº»ìºÏ£¬»ìºÏÈÜÒº³ÊËáÐÔ£¬ËµÃ÷CH3COOHµÄµçÀë³Ì¶È´óÓÚCH3COO-Ë®½â³Ì¶È£¬ÈÜÒºÖдæÔÚÎïÁÏÊØºã£¬½áºÏÎïÁÏÊØºã½â´ðÎÊÌ⣻
£¨2£©³£ÎÂÏ£¬µÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄNaCN¡¢HCN»ìºÏ£¬»ìºÏÈÜÒºÖÐc£¨HCN£©£¾c£¨CN-£©£¬ËµÃ÷HCNµçÀë³Ì¶ÈСÓÚCN-Ë®½â³Ì¶È£¬ÈÜÒº³Ê¼îÐÔ£»»ìºÏÈÜÒºÖдæÔÚÎïÁÏÊØºã¡¢ÖÊ×ÓÊØºã£¬¸ù¾ÝÎïÁÏÊØºã¡¢ÖÊ×ÓÊØºã½â´ðÎÊÌ⣮
£¨2£©³£ÎÂÏ£¬µÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄNaCN¡¢HCN»ìºÏ£¬»ìºÏÈÜÒºÖÐc£¨HCN£©£¾c£¨CN-£©£¬ËµÃ÷HCNµçÀë³Ì¶ÈСÓÚCN-Ë®½â³Ì¶È£¬ÈÜÒº³Ê¼îÐÔ£»»ìºÏÈÜÒºÖдæÔÚÎïÁÏÊØºã¡¢ÖÊ×ÓÊØºã£¬¸ù¾ÝÎïÁÏÊØºã¡¢ÖÊ×ÓÊØºã½â´ðÎÊÌ⣮
½â´ð£º
½â£º£¨1£©³£ÎÂÏ£¬µÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄCH3COOH¡¢CH3COONaÈÜÒº»ìºÏ£¬»ìºÏÈÜÒº³ÊËáÐÔ£¬ËµÃ÷CH3COOHµÄµçÀë³Ì¶È´óÓÚCH3COO-Ë®½â³Ì¶È£¬ÔòÈÜÒºÖдæÔÚc£¨CH3COOH£©£¼c£¨CH3COO-£©£»
Èκεç½âÖÊÈÜÒºÖж¼´æÔÚÎïÁÏÊØºã£¬½áºÏÎïÁÏÊØºãµÃ2c£¨Na+£©=c£¨CH3COO-£©+c£¨CH3COOH£©£¬
¹Ê´ð°¸Îª£º£¼£»=£»
£¨2£©³£ÎÂÏ£¬µÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄNaCN¡¢HCN»ìºÏ£¬»ìºÏÈÜÒºÖÐc£¨HCN£©£¾c£¨CN-£©£¬ËµÃ÷HCNµçÀë³Ì¶ÈСÓÚCN-Ë®½â³Ì¶È£¬ÈÜÒº³Ê¼îÐÔ£¬ÔòpH£¾7£»»ìºÏÈÜÒºÖдæÔÚÎïÁÏÊØºã£¬»ìºÏÈÜÒºÌå»ýÔö´óÒ»±¶£¬ËùÒÔÎïÖÊŨ¶È½µÎªÔÀ´µÄÒ»°ë£¬¸ù¾ÝÎïÁÏÊØºãµÃc£¨HCN£©+c£¨CN-£©=0.1mol/L£»
¸ù¾ÝÖÊ×ÓÊØºãµÃ2c£¨H+£©+c£¨HCN£©=2c£¨OH-£©+c£¨CN-£©£¬Ôòc£¨HCN£©-c£¨CN-£©=2c£¨OH-£©-2c£¨H+£©£¼2c£¨HO-£©£¬
¹Ê´ð°¸Îª£º£¾£»=£»£¼£®
Èκεç½âÖÊÈÜÒºÖж¼´æÔÚÎïÁÏÊØºã£¬½áºÏÎïÁÏÊØºãµÃ2c£¨Na+£©=c£¨CH3COO-£©+c£¨CH3COOH£©£¬
¹Ê´ð°¸Îª£º£¼£»=£»
£¨2£©³£ÎÂÏ£¬µÈÎïÖʵÄÁ¿Å¨¶È¡¢µÈÌå»ýµÄNaCN¡¢HCN»ìºÏ£¬»ìºÏÈÜÒºÖÐc£¨HCN£©£¾c£¨CN-£©£¬ËµÃ÷HCNµçÀë³Ì¶ÈСÓÚCN-Ë®½â³Ì¶È£¬ÈÜÒº³Ê¼îÐÔ£¬ÔòpH£¾7£»»ìºÏÈÜÒºÖдæÔÚÎïÁÏÊØºã£¬»ìºÏÈÜÒºÌå»ýÔö´óÒ»±¶£¬ËùÒÔÎïÖÊŨ¶È½µÎªÔÀ´µÄÒ»°ë£¬¸ù¾ÝÎïÁÏÊØºãµÃc£¨HCN£©+c£¨CN-£©=0.1mol/L£»
¸ù¾ÝÖÊ×ÓÊØºãµÃ2c£¨H+£©+c£¨HCN£©=2c£¨OH-£©+c£¨CN-£©£¬Ôòc£¨HCN£©-c£¨CN-£©=2c£¨OH-£©-2c£¨H+£©£¼2c£¨HO-£©£¬
¹Ê´ð°¸Îª£º£¾£»=£»£¼£®
µãÆÀ£º±¾Ì⿼²éÀë×ÓŨ¶È´óС±È½Ï£¬Ã÷È·ÈÜÒºÖеÄÈÜÖÊÐÔÖÊÊǽⱾÌâ¹Ø¼ü£¬ÖªµÀÈÜÒºËá¼îÐÔÓëËáµÄµçÀëÓëËá¸ùÀë×ÓË®½â³Ì¶ÈÏà¶Ô´óС¹ØÏµ¼´¿É½â´ð£¬×¢Ò⣨2£©ÖÐ[c£¨HCN£©+c£¨CN-£©]µÄ¼ÆË㣬»ìºÏʱÈÜÒºÌå»ý·¢Éú±ä»¯µ¼ÖÂÎïÖÊŨ¶È·¢Éú±ä»¯£¬ÎªÒ×´íµã£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijÎÞÉ«ÈÜÒºÖУ¬³£ÎÂÏÂÓÉË®µçÀë³öµÄc£¨H+£©=1¡Á10-14 mol/L£¬Ôò¸ÃÈÜÒºÖÐÒ»¶¨ÄÜ´óÁ¿¹²´æµÄÀë×ÓÊÇ£¨¡¡¡¡£©
| A¡¢Al3+¡¢Na+¡¢MnO4-¡¢SO42- |
| B¡¢Na+¡¢Ag+¡¢Cl-¡¢NO3-¡¢ |
| C¡¢Na+¡¢K+¡¢Cl-¡¢NO3- |
| D¡¢K+¡¢Ba2+¡¢Cl-¡¢HCO3- |
»¯Ñ§ÓëÉú²ú¡¢Éú»îÃÜÇÐÏà¹Ø£®ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A¡¢Ã÷·¯¾»Ë®ºÍ¼ÒÓÃÏû¶¾¼ÁµÄÏû¶¾ÔÀíÏàͬ |
| B¡¢Ê¹Óú¬ÓÐÂÈ»¯ÄƵÄÈÚÑ©¼Á»á¼Ó¿ìÇÅÁºµÄ¸¯Ê´ |
| C¡¢ÂÁºÏ½ðµÄ¹ã·ºÊ¹ÓÃÊÇÒòΪÈËÃÇÄÜÓý¹Ì¿µÈ»¹Ô¼Á´ÓÑõ»¯ÂÁÖлñÈ¡ÂÁ |
| D¡¢·°µªºÏ½ð¸ÖÊÇ¡°Äñ³²¡±µÄÖ÷Òª²ÄÁÏÖ®Ò»£¬ÆäÈ۵㡢Ӳ¶ÈºÍÇ¿¶È¾ù¸ßÓÚ´¿Ìú |
ÏÂÁÐͼʾÓë¶ÔÓ¦µÄÐðÊöÏà·ûµÄÊÇ£¨¡¡¡¡£©
| A¡¢ ×Ü·´Ó¦ÎªÎüÈÈ·´Ó¦ |
| B¡¢ ʵÑéΪÓÃ0.01 mol?L-1µÄHClÈÜÒºµÎ¶¨20 mL 0.01 mol?L-1µÄNaOHÈÜÒº |
| C¡¢ ¿ÉÒÔ±íʾ·´Ó¦CO£¨g£©+2H2£¨g£©?CH3OH£¨g£©µÄƽºâ³£ÊýKËæÑ¹Ç¿p±ä»¯µÄ¹ØÏµ |
| D¡¢ ¿ÉÒÔ±íʾ³£ÎÂÏÂpH=2µÄHClÈÜÒº¼ÓˮϡÊͱ¶ÊýÓëpHµÄ±ä»¯¹ØÏµ |
ÉèNAΪ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A¡¢ÔÚ³£Î³£Ñ¹Ï£¬2.8 gN2ºÍCOµÄ»ìºÏÆøÌåËùº¬µç×ÓÊýΪ1.4NA | ||
| B¡¢±ê×¼×´¿öÏ£¬22.4 LCCl4º¬ÓеķÖ×ÓÊýĿΪNA | ||
C¡¢1 L O£®1 mol?L-1NaHCO3ÈÜÒºÖк¬ÓÐ0.1NA¸öHCO
| ||
| D¡¢ÔÚµç½â¾«Á¶´Ö͵Ĺý³ÌÖУ¬µ±×ªÒƵç×ÓÊýΪ NAʱ£¬Ñô¼«Èܽâ32 g |
| A¡¢µç³Ø³äµçʱ£¬OH-ÓɼײàÏòÒÒ²àÒÆ¶¯ |
| B¡¢¼×·ÅµçʱΪÕý¼«£¬³äµçʱΪÑô¼« |
| C¡¢·ÅµçʱÕý¼«µÄµç¼«·´Ó¦Ê½ÎªMHn-ne-¨TM+nH+ |
| D¡¢Æû³µÏÂÆÂʱ·¢ÉúͼÖÐʵÏßËùʾµÄ¹ý³Ì |
ÏÂÁÐÀë×Ó·´Ó¦·½³ÌʽÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A¡¢Ð¡ËÕ´òÈÜÒºÓëÑÎËá·´Ó¦£ºCO32-+2H+=CO2¡ü+H2O |
| B¡¢ÂÈ»¯ÂÁÈÜÒºÖмÓÈë¹ýÁ¿°±Ë®£ºAl3++3NH3?H2O=Al£¨OH£©3¡ý+3NH4+ |
| C¡¢ÄÆÓëË®·´Ó¦£ºNa+H2O=Na++OH-+H2¡ü |
| D¡¢FeÈÜÓÚÏ¡ÁòË᣺Fe+3H+=Fe3++H2¡ü |