ÌâÄ¿ÄÚÈÝ

ÏÖÓÐ25 ¡æʱ0.1 mol¡¤L£­1µÄ°±Ë®£¬Çë»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©ÈôÏò°±Ë®ÖмÓÈëÉÙÁ¿ÁòËá粒ÌÌ壬һˮºÏ°±µÄµçÀëƽºâ________(Ìî¡°Ïò×󡱡¢¡°ÏòÓÒ¡±»ò¡°²»¡±)Òƶ¯£»´ËʱÈÜÒºÖÐ________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)¡£
£¨2£©ÈôÏò°±Ë®ÖмÓÈëµÈŨ¶ÈÏ¡´×ËᣬʹÆäÇ¡ºÃÖкͣ¬Ð´³ö·´Ó¦µÄÀë×Ó·½³Ìʽ£º_________________£»ËùµÃÈÜÒºµÄpH________7(Ìî¡°£¾¡±¡¢¡°<¡±»ò¡°£½¡±)£¬
£¨3£©ÈôÏò°±Ë®ÖмÓÈëÏ¡ÁòËáÖÁÈÜÒºµÄpH£½7£¬´Ëʱ[NH4+]£½a mol¡¤L£­1£¬Ôòc(SO42-)£½________¡£
£¨4£©ÈôÏò°±Ë®ÖмÓÈëpH£½1µÄÁòËᣬÇÒ°±Ë®ÓëÁòËáµÄÌå»ý±ÈΪ1¡Ã1£¬ÔòËùµÃÈÜÒºÖи÷Àë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ________________________________________¡£

£¨1£©Ïò×ó ¼õС £¨2£©NH3¡¤H2O£«CH3COOH£½CH3COO£­£«NH4£«£«H2O £½ £¨3£©
£¨4£©c(NH4£«)£¾c(SO42£­)£¾c(H£«)£¾c(OH£­)

½âÎöÊÔÌâ·ÖÎö£º£¨1£©Ïò°±Ë®ÖмÓÈëÉÙÁ¿ÁòËá粒ÌÌ壬ÁòËá淋çÀë³öÀ´µÄ笠ùÀë×Ó»áÒÖÖÆһˮºÏ°±µÄµçÀ룬ƽºâÏò×óÒƶ¯£¬´ËʱÈÜÒºÖмõС £¨2£©NH3¡¤H2O£«CH3COOH£½CH3COO£­£«NH4£«£«H2O £»ÒòΪǡºÃÖкͣ¬ÈÜÒºÏÔÖÐÐÔ£¬ËùµÃÈÜÒºµÄpH£½7 £¨3£©ÒÀ¾ÝµçºÉÊغãÓУºc(NH4£«)£«c(H£«)£½2c(SO42£­)£«c(OH£­)£¬ÓÖÒòΪÈÜÒºµÄpH£½7£¬Òò´ËÓУºc(SO42£­)£½c(NH4£«)£½£»£¨4£©ÈôÏò°±Ë®ÖмÓÈëpH£½1µÄÁòËᣬÇÒ°±Ë®ÓëÁòËáµÄÌå»ý±ÈΪ1¡Ã1£¬ÔòµÃµ½µÄÊÇÁòËá淋ÄÈÜÒº£¬ÒòΪ°±¸ùÀë×ÓµÄË®½â£¬Ê¹µÃÈÜÒºÏÔËáÐÔ£¬Òò´ËÓÐÈçϹØϵ£ºc(NH4£«)£¾c(SO42£­)£¾c(H£«)£¾c(OH£­)
¿¼µã£º¿¼²éÈõµç½âÖʵķ´Ó¦¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ijѧÉúÓûÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄ´×ËáÀ´²â¶¨Î´ÖªÎïÖʵÄÁ¿Å¨¶ÈµÄÇâÑõ»¯ÄÆÈÜҺʱ£¬Ñ¡ÔñÊʵ±µÄָʾ¼Á¡£ÇëÌîдÏÂÁпհףº
£¨1£©Óñê×¼´×ËáµÎ¶¨´ý²âµÄÇâÑõ»¯ÄÆÈÜҺʱ£¬´ÓÏÂÁÐÑ¡ÏîÖÐÑ¡³ö×îÇ¡µ±µÄÒ»Ïî________¡£

 
׶ÐÎÆ¿ÖÐÈÜÒº
µÎ¶¨¹ÜÖÐÈÜÒº
Ñ¡ÓÃָʾ¼Á
Ñ¡Óõζ¨¹Ü
A
¼î
Ëá
ʯÈï
£¨ÒÒ£©
B
Ëá
¼î
¼×»ù³È
£¨¼×£©
C
¼î
Ëá
·Ó̪
£¨¼×£©
D
Ëá
¼î
ʯÈï
£¨ÒÒ£©
 
µÎ¶¨Ê±Ó¦×óÊÖÎÕËáʽµÎ¶¨¹ÜµÄ»îÈû£¬ÓÒÊÖÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦×¢ÊÓ                        ¡£Ö±µ½¼ÓÈëÒ»µÎ´×Ëáºó£¬ÈÜÒºÑÕÉ«ÓÉ       É«±äΪ    É«£¬²¢ÔÚ°ë·ÖÖÓÄÚÈÜÒºÑÕÉ«²»¸Ä±äΪֹ¡£

£¨2£©ÏÂÁвÙ×÷ÖпÉÄÜʹËù²âÇâÑõ»¯ÄÆÈÜÒºµÄŨ¶ÈֵƫµÍµÄÊÇ        ¡£
A.ËáʽµÎ¶¨¹ÜδÓñê×¼´×ËáÈóÏ´¾ÍÖ±½Ó×¢Èë±ê×¼´×Ëá
B.µÎ¶¨Ç°Ê¢·ÅÇâÑõ»¯ÄÆÈÜÒºµÄ׶ÐÎÆ¿ÓÃÕôÁóˮϴ¾»ºóδ¸ÉÔï
C.ËáʽµÎ¶¨¹ÜÔڵζ¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ
D.¶ÁÈ¡´×ËáÌå»ýʱ£¬¿ªÊ¼ÑöÊÓ¶ÁÊý£¬µÎ¶¨½áÊøºó¸©ÊÓ¶ÁÊý
£¨3£©Ä³Ñ§Éú¸ù¾Ý3´ÎʵÑé·Ö±ð¼Ç¼ÓйØÊý¾ÝÈç±í£º
 
´ý²âÇâÑõ»¯ÄÆ
0.100mol/L´×ËáµÄÌå»ý
µÎ¶¨´ÎÊý
ÈÜÒºµÄÌå»ý£¨ml£©
µÎ¶¨Ç°µÄ¿Ì¶È£¨ml£©
µÎ¶¨ºóµÄ¿Ì¶È£¨ml£©
µÚÒ»´Î
25.00
0.00
24.98
µÚ¶þ´Î
25.00
1.56
27.86
µÚÈý´Î
25.00
0.22
25.24
 
ÒÀ¾ÝÉϱíÊý¾Ý¼ÆËã¸ÃÇâÑõ»¯ÄÆÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ             ¡£
£¨4£©ÏÂͼΪÉÏÊö25 mL NaOHÈÜÒºÖÐÖðµÎµÎ¼ÓCH3COOHÈÜÒº¹ý³ÌÖÐÈÜÒºpHµÄ±ä»¯ÇúÏߣ¬Çë»Ø´ð£º

¢ÙBµãÈÜÒº³ÊÖÐÐÔ£¬ÓÐÈ˾ݴËÈÏΪ£¬ÔÚBµãʱNaOHÓëCH3COOHÇ¡ºÃÍêÈ«·´Ó¦£¬ÕâÖÖ¿´·¨ÊÇ·ñÕýÈ·£¿________£¨Ñ¡Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©¡£Èô²»ÕýÈ·£¬Ôò¶þÕßÇ¡ºÃÍêÈ«·´Ó¦µÄµãÊÇÔÚABÇø¼ä»¹ÊÇBDÇø¼äÄÚ£¿________Çø¼ä£®£¨ÈôÕýÈ·£¬´ËÎʲ»´ð£©¡£
¢ÚÔÚDµãʱ£¬ÈÜÒºÖÐc£¨CH3COO£­£©£«c£¨CH3COOH£©________2c£¨Na£«£©¡££¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©

£¨1£©ÊÒÎÂÏÂÔÚpH=12µÄNaCNÈÜÒºÖУ¬ÓÉË®µçÀëµÄc(OH¡ª)Ϊ    mol?L¡ª1¡£
£¨2£©Å¨¶ÈΪ0.1mol?L¡ª1µÄÏÂÁи÷ÎïÖʵÄÈÜÒºÖУ¬c(NH4+)ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ___£¨ÌîÐòºÅ£©¡£
¢ÙNH4Cl ¢ÚNH4HSO4  ¢ÛNH3?H2O ¢ÜCH3COONH4
£¨3£©Ä³¶þÔªËᣨ»¯Ñ§Ê½ÓÃH2A±íʾ£©ÔÚË®ÖеĵçÀë·½³ÌʽÊÇ£º
H2A=H+ +HA¡ª£¬HA¡¥H+ +A2¡ª¡£
¢ÙÔòNa2AÈÜÒºÏÔ____ÐÔ£»NaHAÈÜÒºÏÔ    ÐÔ£¨Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©¡£
¢ÚÈôÓÐ0.1mo1?L¡ª1Na2AµÄÈÜÒº£¬ÆäÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ£º   £¨ÌîÐòºÅ£©¡£

A£®c(Na+)>c(A2¡ª)>c(OH¡ª)>c(HA¡ª)>c(H+
B£®c(Na+)> c(OH¡ª)>c(HA¡ª)> >c(A2¡ª) > c(H+
C£®c(Na+)> c(H+)> c(A2¡ª)> c(OH¡ª)>c(HA¡ª
D£®c(A2¡ª)>c(Na+)> c(OH¡ª) > c(H+)>c(HA¡ª
£¨4£©ÔÚº¬ÓÐCl¡ª¡¢Br¡ª¡¢I¡ªµÄÈÜÒºÖУ¬ÒÑÖªÆäŨ¶È¾ùΪ0.1mo1/L£¬ÒÑÖªAgCl¡¢AgBr¡¢AgIµÄÈܶȻý·Ö±ðΪ1.6¡Á10¡ª10¡¢4.l¡Á10¡ª15¡¢1.5¡Á10¡ª16£¬ÈôÏò»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈëAgNO3ÈÜÒº£¬ÊԻشð£º
¢Ùµ±AgBr³Áµí¿ªÊ¼Îö³öʱ£¬ÈÜÒºÖÐAg+Ũ¶ÈÊÇ              ¡£
¢Úµ±AgC1³Áµí¿ªÊ¼Îö³öʱ£¬ÈÜÒºÖеÄBr¡ª¡¢I¡ªÊÇ·ñÍêÈ«³Áµí                 £¨µ±ÈÜÒºÖÐÀë×ÓŨ¶ÈСÓÚ1.0¡Á10¡ª5mo1/Lʱ£¬ÈÏΪÒѾ­³ÁµíÍêÈ«£¬±¾¿ÕÑ¡Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©¡£

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

¾«Ó¢¼Ò½ÌÍø