ÌâÄ¿ÄÚÈÝ
£¨12·Ö£©£¨1£©Ä³Ñо¿ÐÔѧϰС×éÔÚʵÑéÊÒÖÐÅäÖÆ1 mol/LµÄÏ¡ÁòËá±ê×¼ÈÜÒº£¬È»ºóÓÃÆäµÎ¶¨Ä³Î´ÖªÅ¨¶ÈµÄNaOHÈÜÒº¡£ÏÂÁÐÓйØËµ·¨ÖÐÕýÈ·µÄÊÇ______________¡£
A£®ÊµÑéÖÐËùÓõ½µÄµÎ¶¨¹Ü¡¢ÈÝÁ¿Æ¿£¬ÔÚʹÓÃǰ¾ùÐèÒª¼ì©£»
B£®Èç¹ûʵÑéÖÐÐèÓÃ60 mL Ï¡ÁòËá±ê×¼ÈÜÒº£¬ÅäÖÆÊ±Ó¦Ñ¡ÓÃ100 mLÈÝÁ¿Æ¿£»
C£®ÈÝÁ¿Æ¿Öк¬ÓÐÉÙÁ¿ÕôÁóË®£¬»áµ¼ÖÂËùÅä±ê×¼ÈÜÒºµÄŨ¶ÈƫС£»
D£®ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴµÓºó£¬¼´×°Èë±ê׼Ũ¶ÈµÄÏ¡ÁòËᣬÔò²âµÃµÄNaOHÈÜÒºµÄŨ¶È½«Æ«´ó£»
E£®Öк͵ζ¨Ê±£¬ÈôÔÚ×îºóÒ»´Î¶ÁÊýʱ¸©ÊÓ¶ÁÊý£¬Ôòµ¼ÖÂ×îºóʵÑé½á¹ûÆ«´ó
£¨2£©³£ÎÂÏ£¬ÒÑÖª0.1 mol¡¤L£1Ò»ÔªËáHAÈÜÒºÖÐc(OH£)/c(H+)£½1¡Á10£8¡£
¢Ù³£ÎÂÏ£¬0.1 mol¡¤L£1 HAÈÜÒºµÄpH= £»Ð´³ö¸ÃËᣨHA£©ÓëNaOHÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£º £»
¢ÚpH£½3µÄHAÓëpH£½11µÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºÖÐ4ÖÖÀë×ÓÎïÖʵÄÁ¿Å¨¶È´óС¹ØÏµÊÇ£º £»
¢Û0.2 mol¡¤L£1HAÈÜÒºÓë0.1mol¡¤L£1NaOHÈÜÒºµÈÌå»ý»ìºÏºóËùµÃÈÜÒºÖУºc(H+)£«c(HA)£c(OH£)£½ mol¡¤L£1¡££¨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©
£¨3£©t¡æÊ±£¬ÓÐpH=2µÄÏ¡ÁòËáºÍpH=11µÄNaOHÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒº³ÊÖÐÐÔ£¬Ôò¸ÃζÈÏÂË®µÄÀë×Ó»ý³£ÊýKw= ¡£¸ÃζÈÏ£¨t¡æ£©£¬½«100 mL 0.1 mol¡¤L-1µÄÏ¡H2SO4ÈÜÒºÓë100 mL 0.4 mol¡¤L-1µÄNaOHÈÜÒº»ìºÏºó£¨ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ£©£¬ÈÜÒºµÄpH= ¡£