ÌâÄ¿ÄÚÈÝ
(1)³£ÎÂʱ£¬ËÄÖÖÈÜÒº£º¢ñ.pH£½4µÄCH3COOHÈÜÒº£»¢ò.pH£½4µÄHClÈÜÒº£»¢ó.pH£½10µÄNaOHÈÜÒº£»¢ô.pH£½10µÄCH3COONaÈÜÒº¡£
¢Ù¢óºÍ¢ôµÄÈÜÒºÖÐË®µçÀëµÄc(H£«)Ũ¶ÈÖ®±È_____________________________________¡£
¢ÚÏÂÁÐÓйØÐðÊöÕýÈ·µÄÊÇ________¡£
A£®¢ò¡¢¢óµÄÈÜÒº·Ö±ðÓë10 gÂÁ·Û·´Ó¦£¬Éú³ÉH2µÄÁ¿¢ó¸ü¶à
B£®½«¢ñºÍ¢óµÈÌå»ý»ìºÏºó£¬ÈÜÒºpHСÓÚ7
C£®ËÄÖÖÈÜÒº¸÷10 mL·Ö±ð¼ÓˮϡÊÍÖÁ100 mLºó£¬ÈÜÒºµÄpH£º¢ó£¾¢ô£¾
¢ñ£¾¢ò
(2)CH3COOHÈÜÒºµÄKa£½1.6¡Á10£5£¬Ôò1.0 mol¡¤L£1µÄCH3COONaÈÜÒº
ÖÐc(OH£)£½________________________________________________________________________¡£
(3)½«CO2ͨÈëNaOHÈÜÒºÖУ¬»Ø´ðÏÂÁÐÎÊÌâ¡£
¢Ùµ±CO2ÓëNaOHÎïÖʵÄÁ¿Ö®±ÈΪ1¡Ã2ʱ£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС˳ÐòΪ________________________________________________________________________¡£
¢Úµ±c(Na£«)£½c(CO
)£«c(HCO
)£«c(H2CO3)ʱ£¬·´Ó¦ºóÈÜÒºÖеÄÈÜÖÊΪ________¡£
(4)ÓÃNaOHÈÜÒºµÎ¶¨´×ËáÈÜÒºµÄ¹ý³ÌÖУ¬×¶ÐÎÆ¿ÖеÄÈÜÒºÆäÀë×ÓŨ¶È¹ØÏµÓÐÈçÏÂʽ×Ó£¬Ôڵζ¨¹ý³ÌÖÐÕâЩʽ×Ó³öÏÖµÄÏȺó˳ÐòΪ________(ÇëÑ¡³öÕýÈ·µÄ˳Ðò)¡£
¢Ùc(Na£«)£¾c(CH3COO£)£¾c(OH£)£¾c(H£«)
¢Úc(CH3COO£)£¾c(Na£«)£¾c(H£«)£¾c(OH£)
¢Ûc(CH3COO£)£¾c(H£«)£½c(Na£«)£¾c(OH£)
¢Üc(Na£«)£¾c(CH3COO£)£½c(OH£)£¾c(H£«)
¢Ýc(CH3COO£)£¾c(H£«)£¾c(Na£«)£¾c(OH£)
¢Þc(Na£«)£½c(CH3COO£)£¾c(OH£)£½c(H£«)
¢ßc(Na£«)£¾c(OH£)£¾c(CH3COO£)£¾c(H£«)
A£®¢ß¢Ü¢Ù¢Þ¢Ú¢Û¢Ý¡¡¡¡¡¡ B£®¢Ý¢Ú¢Û¢Ù¢Þ¢Ü¢ß
C£®¢Û¢Ý¢Ú¢Ù¢Þ¢Ü¢ß D£®¢Ý¢Û¢Ú¢Þ¢Ù¢Ü¢ß
´ð°¸¡¡(1)¢Ù10£6¡¡¢ÚB¡¡(2)2.5¡Á10£5 mol¡¤L£1¡¡
(3)¢Ùc(Na£«)£¾c(CO
)£¾c(OH£)£¾c(HCO
)£¾c(H£«)¡¡¢ÚNaHCO3¡¡(4)D
ÓÐA¡¢BÁ½ÖÖÌþ£¬ÆäÏà¹ØÐÅÏ¢ÈçÏ£º
| A | ¢ÙÍêȫȼÉյIJúÎïÖÐn(CO2)¡Ãn(H2O)£½2¡Ã1 ¢Ú28£¼Mr(A)£¼60 ¢Û²»ÄÜʹäåµÄËÄÂÈ»¯Ì¼ÈÜÒºÍÊÉ« ¢ÜÒ»ÂÈ´úÎïÖ»ÓÐÒ»Öֽṹ |
| B | ¢Ù±¥ºÍÁ´Ìþ£¬Í¨³£Çé¿öÏÂ³ÊÆøÌ¬ ¢ÚÓÐͬ·ÖÒì¹¹Ìå ¢Û¶þäå´úÎïÓÐÈýÖÖ |
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÌþAµÄ×î¼òʽÊÇ________£»
(2)ÌþAµÄ½á¹¹¼òʽÊÇ_____________________________________________________£»
(3)ÌþBµÄÈýÖÖ¶þäå´úÎïµÄ½á¹¹¼òʽΪ_______________________________________£»
(4)ÌþCΪÌþBµÄͬϵÎ³£ÎÂÏÂÎªÆøÌ¬ÇÒÖ»ÓÐÒ»ÖÖÒ»äå´úÎÔòÌþCµÄÒ»äå´úÎïµÄ½á¹¹¼òʽΪ________(ÌîÒ»ÖÖ¼´¿É)¡£