| »¯Ñ§Ê½ | CH3COOH | H2CO3 | HClO | |
| µçÀëÆ½ºâ ³£Êý | Ka=1.8¡Á10-5 | Ka1=4.3¡Á10-7 | Ka2=5.6¡Á10-11 | Ka=3.0¡Á10-8 |
·ÖÎö £¨1£©Ëá¸ùÀë×Ó¶ÔÓ¦ËáµÄËáÐÔԽǿ£¬¸ÃÀë×ÓµÄË®½â³Ì¶ÈԽС£¬ÈÜÒºµÄpHԽС£»
£¨2£©0.1mol/LµÄCH3COOHÈÜÒº¼ÓˮϡÊ͹ý³ÌÖУ¬ÇâÀë×ÓÓë´×Ëá¸ùÀë×ÓÎïÖʵÄÁ¿Ôö´ó£¬´×Ëá·Ö×ÓÎïÖʵÄÁ¿¼õС£¬Å¨¶È¼õС£¬ËáÐÔ¼õÈõ£¬Ë®µÄÀë×Ó»ý³£Êý²»±ä£¬´×ËáµÄµçÀëÆ½ºâ³£Êý²»±ä£»
£¨3£©¾Ýͼ·ÖÎö£¬¼ÓˮϡÊ͵Ĺý³ÌÖУ¬HXµÄpH±ä»¯±È½Ï¿ì£¬ËµÃ÷HXµÄËáÐԱȴ×ËáÇ¿£»
£¨4£©±ê×¼×´¿öÏ£¬½«1.12L CO2ͨÈë100mL 1mol•L-1µÄNaOHÈÜÒºÖУ¬1.12L CO2µÄÎïÖʵÄÁ¿Îª£º$\frac{1.12L}{22.4L/mol}$=0.05mol£¬ÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿Îª£º1mol•L-1¡Á0.1L=0.1mol£¬¶þÕßÇ¡ºÃÍêÈ«·´Ó¦Éú³É̼ËáÄÆ£¬ÈÜÒºÖдæÔÚµçºÉÊØºãºÍÎïÁÏÊØºã£»
£¨5£©CH3COOHÓëCH3COONaµÄ»ìºÏÈÜÒºÖУ¬´æÔÚµçºÉÊØºã£ºc£¨Na+£©+c£¨H+£©=c£¨OH-£©+c£¨CH3COO-£©£¬¸ù¾ÝÈÜÒºÖеĵçºÉÊØºãºÍÎïÁÏÊØºãÀ´¼ÆË㣮
½â´ð ½â£º£¨1£©ËÄÖÖÈÜÒºµÄÈÜÖʶ¼ÊÇÇ¿¼îÈõËáÑΣ¬Ë®½â³Ì¶È´óСΪ£ºCO32-£¾ClO-£¾HCO3-£¾CH3COO-£¬Ë®½â¾ùÏÔ¼îÐÔ£¬Ë®½â³Ì¶ÈÔ½´ó£¬¼îÐÔԽǿ£¬ËùÒÔ¼îÐÔ˳ÐòÊÇ£ºNa2CO3£¾NaClO£¾NaHCO3£¾CH3COONa£¬
¼´pHÓÉСµ½´óµÄÅÅÁÐ˳ÐòΪ£ºCH3COONa£¼NaHCO3£¼NaClO£¼Na2CO3£¬¼´£ºa£¼d£¼c£¼b£¬
¹Ê´ð°¸Îª£ºa£¼d£¼c£¼b£»
£¨2£©0.1mol/LµÄCH3COOHÈÜÒº¼ÓˮϡÊ͹ý³ÌÖУ¬ÇâÀë×ÓÓë´×Ëá¸ùÀë×ÓÎïÖʵÄÁ¿Ôö´ó£¬Å¨¶È¼õС£¬ËáÐÔ¼õÈõ£¬
A¡¢ÇâÀë×ÓŨ¶È¼õС£¬¹ÊA´íÎó£»
B¡¢¼ÓˮϡÊ͹ý³ÌÖУ¬ÇâÀë×ÓÎïÖʵÄÁ¿Ôö´ó£¬´×Ëá·Ö×ÓÎïÖʵÄÁ¿¼õС£¬ËùÒÔ$\frac{c£¨{H}^{+}£©}{c£¨C{H}_{3}COOH£©}$Ôö´ó£¬¹ÊBÕýÈ·£»
C¡¢Ë®µÄÀë×Ó»ý³£Êý²»±ä£¬¹ÊC´íÎó£»
D¡¢´×ËáÈÜÒº¼ÓˮϡÊÍʱËáÐÔ¼õÈõ£¬ÇâÀë×ÓŨ¶È¼õСÇâÑõ¸ùÀë×ÓŨ¶ÈÔö´ó£¬ËùÒÔ$\frac{c£¨O{H}^{-}£©}{c£¨{H}^{+}£©}$Ôö´ó£¬¹ÊDÕýÈ·£»
E¡¢´×ËáµÄµçÀëÆ½ºâ³£Êý²»±ä£¬¹ÊE´íÎó£»
¹Ê´ð°¸Îª£ºBD£»
£¨3£©¾Ýͼ·ÖÎö£¬¼ÓˮϡÊ͵Ĺý³ÌÖУ¬HXµÄpH±ä»¯±È½Ï¿ì£¬ËµÃ÷HXµÄËáÐԱȴ×ËáÇ¿£¬HXµÄµçÀëÆ½ºâ³£Êý±È´×Ëá´ó£¬
¹Ê´ð°¸Îª£º´óÓÚ£»
£¨4£©±ê×¼×´¿öÏ£¬½«1.12L CO2ͨÈë100mL 1mol•L-1µÄNaOHÈÜÒºÖУ¬1.12L CO2µÄÎïÖʵÄÁ¿Îª£º$\frac{1.12L}{22.4L/mol}$=0.05mol£¬ÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿Îª£º1mol•L-1¡Á0.1L=0.1mol£¬¶þÕßÇ¡ºÃÍêÈ«·´Ó¦Éú³É̼ËáÄÆ£¬
¢ÙÈÜÒºÖдæÔÚÖÊ×ÓÊØºã£ºc£¨OH-£©=c£¨H+£©+c£¨HCO3-£©+2c£¨H2CO3£©£¬
¹Ê´ð°¸Îª£ºc£¨H+£©+c£¨HCO3-£©£»
¢Ú̼ËáÄÆÈÜÒºÖдæÔÚµçºÉÊØºã£ºc£¨H+£©+c£¨Na+£©=2c£¨CO32-£©+c£¨HCO3-£©+c£¨OH-£©£¬
¹Ê´ð°¸Îª£º2c£¨CO32-£©+c£¨HCO3-£©+c£¨OH-£©£»
£¨5£©CH3COOHÓëCH3COONaµÄ»ìºÏÈÜÒºÖУ¬´æÔÚµçºÉÊØºã£ºc£¨Na+£©+c£¨H+£©=c£¨OH-£©+c£¨CH3COO-£©£¬ËùÒÔc£¨CH3COO-£©-c£¨Na+£©=c£¨H+£©-c£¨OH-£©=10-6mol/L-10-8mol/L=9.9¡Á10-7mol/L£¬
¹Ê´ð°¸Îª£º9.9¡Á10-7£®
µãÆÀ ±¾Ì⿼²éÁËÑεÄË®½â¡¢Èõµç½âÖʵĵçÀë¼°ÆäÓ°Ïì¡¢µçÀëÆ½ºâ³£ÊýµÄÓ¦Óõȣ¬ÌâÄ¿ÄѶÈÖеȣ¬ÊÔÌâ֪ʶµã½Ï¶à¡¢×ÛºÏÐÔ½ÏÇ¿£¬³ä·Ö¿¼²éÁËѧÉúµÄ·ÖÎö¡¢Àí½âÄÜÁ¦¼°Áé»îÓ¦ÓÃËùѧ֪ʶµÄÄÜÁ¦£®
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ¿ªÆôÆ¡¾ÆÆ¿ºó£¬Æ¿ÖÐÁ¢¿Ì·ºÆð´óÁ¿ÅÝÄ | |
| B£® | ½«Ê¢ÓжþÑõ»¯µªºÍËÄÑõ»¯¶þµª»ìºÏÆøµÄÃܱÕÈÝÆ÷ÖÃÓÚÀäË®ÖУ¬»ìºÏÆøÌåÑÕÉ«±ädz- | |
| C£® | ÏòÂÈË®ÖмÓCaCO3ºó£¬ÈÜҺƯ°×ÐÔÔöÇ¿ | |
| D£® | 500¡æ×óÓÒ±ÈÊÒθüÓÐÀûÓںϳÉNH3 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | KSCN | B£® | BaCl2 | C£® | NaOH | D£® | NH3•H2O |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | KOHµÄµç×Óʽ£º | B£® | ¶þÑõ»¯Ì¼µÄ½á¹¹Ê½£ºO=C=O | ||
| C£® | CH4µÄÇò¹÷Ä£ÐÍ£º | D£® | S2¡¥µÄÀë×ӽṹʾÒâͼ£º |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ¶ÆÍÌúÖÆÆ·¶Æ²ãÆÆËðºó£¬ÌúÖÆÆ·±ÈÆÆËðǰ¸üÈÝÒ×ÉúÐâ | |
| B£® | ±ê×¼×´¿öÏ£¬22.4 L Cl2Óë×ãÁ¿NaOHÈÜÒº·´Ó¦£¬×ªÒƵç×ÓÊýΪ2mol | |
| C£® | Ë®µÄÀë×Ó»ý³£ÊýKwËæ×ÅζȵÄÉý¸ß¶øÔö´ó£¬ËµÃ÷Ë®µÄµçÀëÊÇ·ÅÈÈ·´Ó¦ | |
| D£® | Na2CO3ÈÜÒºÖмÓÈëÉÙÁ¿Ca£¨OH£©2¹ÌÌ壬CO32-Ë®½â³Ì¶È¼õС£¬ÈÜÒºµÄpH¼õС |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | ¼ÓÈë¹ýÁ¿°±Ë®£¬Óа×É«³ÁµíÉú³É£¬ÔòÔÈÜÒºÒ»¶¨ÓÐAl3+ | |
| B£® | FeCl3¡¢CuCl2µÄ»ìºÏÈÜÒºÖмÓÈëÌú·Û£¬³ä·Ö·´Ó¦ºóÈÔÓйÌÌåÊ£Ó࣬¼ÓÈëKCSNÈÜÒº¿ÉÄܱä³ÉѪºìÉ« | |
| C£® | пÓëŨÁòËá²»·´Ó¦£¬ÓëÏ¡ÁòËá·´Ó¦Éú³ÉÇâÆø | |
| D£® | ¼ÓÈëNaOHÈÜÒº£¬¼ÓÈȺó²úÉúµÄÆøÌåÄÜʹʪÈóµÄºìɫʯÈïÊÔÖ½±äÀ¶£¬ÔòÔÈÜÒºÒ»¶¨ÓÐNH4+ |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ
| A£® | N¡¢M¾ù²»ÄÜ·¢ÉúÒø¾µ·´Ó¦ | B£® | MÖпÉÄÜûÓм׻ù | ||
| C£® | N¿É·¢ÉúÏûÈ¥·´Ó¦ | D£® | N·Ö×ÓÖк¬Óм׻ù |
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - 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