ijѧÉúÓÃ0.1mol/LµÄKOH±ê×¼ÈÜÒºµÎ¶¨Î´ÖªÅ¨¶ÈÑÎËᣬÆä²Ù×÷¿É·Ö½âΪÈçϼ¸²½£ºA£®ÒÆÈ¡20mL´ý²âÑÎËáÈÜҺעÈë½à¾»µÄ׶ÐÎÆ¿£¬²¢¼ÓÈë2~3µÎ·Ó̪
B£®Óñê×¼ÈÜÒºÈóÏ´µÎ¶¨¹Ü2~3´Î
C£®°ÑÊ¢Óбê×¼ÈÜÒºµÄ¼îʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚµÎ¶¨¹Ü¼â×ìʹ֮³äÂúÈÜÒº
D£®È¡±ê×¼KOHÈÜҺעÈë¼îʽµÎ¶¨¹ÜÖÁ¿Ì¶È0ÒÔÉÏ2~3cm
E£®µ÷½ÚÒºÃæÖÁ0»ò0ÒÔÏ¿̶ȣ¬¼Ç϶ÁÊý
F£®°Ñ׶ÐÎÆ¿·ÅÔڵζ¨¹ÜµÄÏÂÃ棬Óñê×¼KOHÈÜÒºµÎ¶¨ÖÁÖյ㲢¼ÇÏµζ¨¹ÜÒºÃæµÄ¿Ì¶È¡£   ¾Í´ËʵÑéÍê³ÉÌî¿Õ£º
(1)ÕýÈ·²Ù×÷²½ÖèµÄ˳ÐòÊÇ(ÓÃÐòºÅ×ÖĸÌîд)_____                  _______¡£
£¨2£©Åжϵ½´ïµÎ¶¨ÖÕµãµÄʵÑéÏÖÏóÊÇ£º_________                 ________¡£
£¨3£©²Ù×÷²½ÖèBµÄÒâÒ壺                                           ¡£
£¨4£©ÏÂÃæa~eÊÇÖÐѧ»¯Ñ§ÊµÑéÖг£¼ûµÄ¼¸ÖÖ¶¨Á¿ÒÇÆ÷£ºa Á¿Í²£»b ÈÝÁ¿Æ¿£»c µÎ¶¨¹Ü£»d ÍÐÅÌÌìƽ£»e ζȼÆ
¢Ù£®ÆäÖбêʾ³öÒÇÆ÷ʹÓÃζȵÄÊÇ________________(Ìîд±àºÅ)£»
¢Ú£®ÓÉÓÚ´íÎó²Ù×÷£¬Ê¹µÃµ½µÄÊý¾Ý±ÈÕýÈ·Êý¾ÝƫСµÄÊÇ_____________(Ìîд±àºÅ)£»
A£®ÓÃÁ¿Í²Á¿È¡Ò»¶¨Á¿ÒºÌåʱ£¬¸©ÊÓÒºÃæ¶ÁÊý
B£®Öк͵ζ¨´ïÖÕµãʱ¸©Êӵζ¨¹ÜÄÚÒºÃæ¶ÈÊý
C£®Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¸©ÊÓÒºÃ涨ÈݺóËùµÃÈÜÒºµÄŨ¶È
D£®Óôý²âÈÜÒºÈóϴ׶ÐÎÆ¿
£¨1£©£Â£Ä£Ã£Å£Á£Æ£»¡¡£¨2£©ÈÜÒºÓÉÎÞÉ«±äΪ·ÛºìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊ
£¨3£©·ÀÖ¹±ê׼ҺŨ¶È±äÏ¡  £¨4£©A,B,C       A,B

ÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝËá¼îÖк͵ζ¨Ô­ÀíÒÔ¼°ÊµÑé²Ù×÷×¢ÒâÊÂÏî¿ÉÖª£¬ÕýÈ·µÄ²Ù×÷²½Öè˳ÐòÊǣ£ģãţÁ£Æ¡£
£¨2£©¼îҺʹ·Ó̪ÏÔºìÉ«£¬Ëáʹ·Ó̪ÏÔÎÞÉ«£¬ÔòÅжϵ½´ïµÎ¶¨ÖÕµãµÄʵÑéÏÖÏóÊÇÈÜÒºÓÉÎÞÉ«±äΪ·ÛºìÉ«£¬ÇÒ°ë·ÖÖÓÄÚ²»ÍÊ¡£
£¨3£©±ê×¼ÒºÈóÏ´µÎ¶¨¹ÜµÄÄ¿µÄÊÇ·ÀÖ¹±ê׼ҺŨ¶È±äÏ¡£¬Ôì³ÉÊÔÑéÎó²î¡£
£¨4£©¢ÙÆäÖбêʾ³öÒÇÆ÷ʹÓÃζȵÄÊÇÁ¿Í²¡¢Á¿Æ¿¡¢µÎ¶¨¹Ü£¬´ð°¸Ñ¡abc¡£
¢ÚÓÃÁ¿Í²Á¿È¡Ò»¶¨Á¿ÒºÌåʱ£¬¸©ÊÓÒºÃæ¶ÁÊý£¬Ôò¶ÁÊýƫС£»Öк͵ζ¨´ïÖÕµãʱ¸©Êӵζ¨¹ÜÄÚÒºÃæ¶ÈÊý£¬Ôò¶ÁÊýƫС£»Ê¹ÓÃÈÝÁ¿Æ¿ÅäÖÆÈÜҺʱ£¬¸©ÊÓÒºÃ涨ÈݺóËùµÃÈÜÒºµÄŨ¶È£¬ÔòÈÝÁ¿Æ¿ÖÐÈÜÒºµÄÌå»ýƫС£¬Å¨¶ÈÆ«¸ß£»Óôý²âÈÜÒºÈóϴ׶ÐÎÆ¿£¬ÔòÏûºÄ±ê×¼ÒºµÄÌå»ýÔö¼Ó£¬Å¨¶ÈÆ«¸ß£¬´ð°¸Ñ¡AB¡£
µãÆÀ£º¸ÃÌâÊÇ»ù´¡ÐÔÊÔÌâµÄ¿¼²é£¬²àÖضÔѧÉú»ù±¾ÊµÑé²Ù×÷ÄÜÁ¦µÄÅàÑø¡£ÓÐÀûÓÚÅàÑøѧÉú¹æ·¶ÑϽ÷µÄʵÑéÉè¼ÆÄÜÁ¦£¬ÓÐÀûÓÚµ÷¶¯Ñ§ÉúµÄѧϰÐËȤ£¬¼¤·¢Ñ§ÉúµÄѧϰ»ý¼«ÐÔ¡£¸ÃÌâÖ÷ÒªÊÇÒÔ³£¼ûÒÇÆ÷µÄÑ¡Óá¢ÊµÑé»ù±¾²Ù×÷ΪÖÐÐÄ£¬Í¨¹ýÊÇʲô¡¢ÎªÊ²Ã´ºÍÔõÑù×öÖص㿼²éʵÑé»ù±¾²Ù×÷µÄ¹æ·¶ÐÔºÍ׼ȷ¼°Áé»îÔËÓÃ֪ʶ½â¾öʵ¼ÊÎÊÌâµÄÄÜÁ¦¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

25¡æʱÏÂÁÐÐðÊöÕýÈ·µÄÊÇ               
A£®0.10mol¡¤L£­1µÄÈýÖÖÈÜÒº£º¢ÙNaCl¡¢¢ÚNaOH¡¢¢ÛCH3COONa£¬ÆäpH´óС˳ÐòΪ£º¢Û£¾¢Ú£¾¢Ù
B£®0.10mol¡¤L£­1NaHCO3ÈÜÒºÖУºc(CO32£­)+c(HCO3£­)+c(H2CO3)=0.10mol¡¤L£­1
C£®pH=2µÄÑÎËáºÍpH=12µÄNaOHÈÜÒºÖУ¬Ë®µÄÀë×Ó»ýKwÏàͬ
D£®ÓëÌå»ýÏàͬ¡¢pH=2µÄÑÎËáºÍ´×ËáÍêÈ«·´Ó¦£¬ÐèÒª0.010mol¡¤L£­1 NaOHµÄÌå»ýÏàͬ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÊµÑéÌâ

ÔĶÁÏÂÁÐʵÑéÄÚÈÝ£¬¸ù¾ÝÌâÄ¿ÒªÇó»Ø´ðÎÊÌ⣺
ijѧÉúΪ²â¶¨Î´ÖªÅ¨¶ÈµÄÁòËáÈÜÒº£¬ÊµÑéÈçÏ£ºÓÃ1.00mL´ý²âÁòËáÅäÖÆ100mLÏ¡H2SO4ÈÜÒº£»ÒÔ0.14 mol¡¤L£­1µÄNaOHÈÜÒºµÎ¶¨ÉÏÊöÏ¡H2SO4 25.00mL£¬µÎ¶¨ÖÕֹʱÏûºÄNaOHÈÜÒº15.00mL¡£
£¨1£©¸ÃѧÉúÓñê×¼0.14 mol¡¤L£­1NaOHÈÜÒºµÎ¶¨ÁòËáµÄʵÑé²Ù×÷ÈçÏ£º
A£®ÓÃËáʽµÎ¶¨¹ÜÈ¡Ï¡H2SO4 25.00mL£¬×¢Èë׶ÐÎÆ¿ÖУ¬¼ÓÈëָʾ¼Á¡£
B£®Óôý²â¶¨µÄÈÜÒºÈóÏ´ËáʽµÎ¶¨¹Ü¡£
C£®ÓÃÕôÁóˮϴ¸É¾»µÎ¶¨¹Ü¡£
D£®È¡Ï¼îʽµÎ¶¨¹ÜÓñê×¼µÄNaOHÈÜÒºÈóÏ´ºó£¬½«±ê׼ҺעÈë¼îʽµÎ¶¨¹Ü¿Ì¶È¡°0¡±ÒÔÉÏ2¡«3cm´¦£¬ÔٰѼîʽµÎ¶¨¹Ü¹Ì¶¨ºÃ£¬µ÷½ÚÒºÃæÖÁ¿Ì¶È¡°0¡±»ò¡°0¡±¿Ì¶ÈÒÔÏ¡£
E¡¢¼ì²éµÎ¶¨¹ÜÊÇ·ñ©ˮ¡£
F¡¢Áíȡ׶ÐÎÆ¿£¬ÔÙÖظ´²Ù×÷Ò»´Î¡£
G¡¢°Ñ׶ÐÎÆ¿·ÅÔڵζ¨¹ÜÏÂÃ棬ƿϵæÒ»ÕÅ°×Ö½£¬±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿Ö±ÖÁµÎ¶¨Öյ㣬¼ÇÏµζ¨¹ÜÒºÃæËùÔڿ̶ȡ£
¢ÙµÎ¶¨²Ù×÷µÄÕýȷ˳ÐòÊÇ£¨ÓÃÐòºÅÌîд£©___________________________              ¡£
¢Ú¸ÃµÎ¶¨²Ù×÷ÖÐӦѡÓõÄָʾ¼ÁÊÇ         ____________  ¡£
¢ÛÔÚG²Ù×÷ÖÐÈçºÎÈ·¶¨Öյ㣿       __________________________     ____    ¡£
£¨2£©¼îʽµÎ¶¨¹ÜÓÃÕôÁóË®ÈóÏ´ºó£¬Î´Óñê×¼ÒºÈóÏ´µ¼Öµζ¨½á¹û______________£¨ÌƫС¡±¡¢¡°Æ«´ó¡±»ò¡°ÎÞÓ°Ï족£©¡£
£¨3£©ÅäÖÆ׼ȷŨ¶ÈµÄÏ¡H2SO4ÈÜÒº£¬±ØÐëʹÓõÄÖ÷ÒªÈÝÆ÷ÊÇ       ___________      ¡£
£¨4£©ÈçÓÐ1 mol¡¤L£­1ºÍ0.1 mol¡¤L£­1µÄNaOHÈÜÒº£¬Ó¦ÓÃ________________   _µÄNaOHÈÜÒº£¬Ô­ÒòÊÇ__________________________   __________________________          ¡£
£¨5£©Óñê×¼NaOHÈÜÒºµÎ¶¨Ê±£¬Ó¦½«±ê×¼NaOHÈÜҺעÈë    __£¨Ñ¡Ìî¡°¼×¡±»ò¡°ÒÒ¡±£©ÖС£

£¨6£©¹Û²ì¼îʽµÎ¶¨¹Ü¶ÁÊýʱ£¬ÈôµÎ¶¨Ç°ÑöÊÓ£¬µÎ¶¨ºó¸©ÊÓ£¬Ôò½á¹û»áµ¼Ö²âµÃµÄÏ¡H2SO4ÈÜҺŨ¶È²â¶¨Öµ           £¨Ñ¡Ìî¡°Æ«´ó¡±¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©
£¨7£©¼ÆËã´ý²âÁòËᣨϡÊÍÇ°µÄÁòËᣩÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È£¨¼ÆËã½á¹û±£Áôµ½Ð¡ÊýµãºóµÚ¶þ룩      ___     ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌâ
£¨1£©Ä³Î¶ÈÏ£¬´¿Ë®ÖÐc (H£«)£½2.0¡Á10£­7 mol¡¤L-1£¬Ôò´Ëʱc (OH£­) £½         mol¡¤L£­£±
0.9mol¡¤L£­£±NaOHÈÜÒºÓë0.1mol¡¤L£­£±HClÈÜÒºµÈÌå»ý»ìºÏ£¨²»¿¼ÂÇÈÜÒºÌå»ý±ä»¯£©ºó£¬ÈÜÒºµÄpH£½£ß£ß£ß£ß£ß¡£
£¨2£©Ïò0.020 mol¡¤L-1µÄHCNÈÜÒºÖмÓÈë0.020 mol NaCN¹ÌÌ壬ÈÜÒºpHÔö´ó£¬Ö÷ÒªÔ­ÒòÊÇ                                                            £»ÒÑÖª¸Ã»ìºÏÈÜÒºÖÐc (Na£«) > c (CN£­)£¬Ôòc (HCN)        c (CN£­)£¨Óá°>¡±¡¢¡°<¡±¡¢¡°£½¡±·ûºÅÌî¿Õ£©¡£
£¨3£©Ïò1L 0.10 mol¡¤L-1µÄHCNÈÜÒºÖмÓÈë0.08molNaOH¹ÌÌ壬µÃµ½»ìºÏÈÜÒº£¬Ôò_________ºÍ__________Á½ÖÖÁ£×ÓµÄÎïÖʵÄÁ¿Ö®ºÍµÈÓÚ0.1mol£»Ð´³ö¸Ã»ìºÏÈÜÒºÖдæÔÚµÄËùÓÐƽºâµÄ±í´ïʽ                                                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÃÑÎËá²â¶¨Ì¼ËáÄÆʱ¼È¿ÉÓ÷Ó̪×÷ָʾ¼ÁÓÖ¿ÉÓü׻ù³È(Ò»ÖÖËá¼îָʾ¼Á)×÷ָʾ¼Á£¬ÏÖ·Ö±ðÓ÷Ó̪ºÍ¼×»ù³È×÷ָʾ¼Á£¬ÓÃ0.1000mol/LµÄHClµÎ¶¨20.00mLµÄ´¿¼îÈÜÒº£¬µÎ¶¨ÖÕµãʱ·Ö±ðÓÃÈ¥ÁË20.00mL¡¢40.00mLµÄÑÎËᣬÏÂÁÐ˵·¨Öв»ÕýÈ·µÄÊÇ  (    )
A£®Ó÷Ó̪×÷ָʾ¼ÁʱµÎ¶¨µÄ×Ü·´Ó¦Îª£ºNa2CO3+HCl=NaHCO3+NaCl
B£®Óü׻ù³È×÷ָʾ¼ÁʱµÎ¶¨µÄ×Ü·´Ó¦Îª£ºNa2CO3+2HCl=NaCl+CO2¡ü+H2O
C£®¿ÉÓüîʽµÎ¶¨¹ÜÁ¿È¡ËùÐèÒªµÄNa2CO3ÈÜÒº
D£®ÈôËáʽµÎ¶¨¹ÜûÓÐÓñê×¼ÈÜÒºÈóÏ´£¬ÔòËù²âµÃµÄ̼ËáÄÆÈÜҺŨ¶ÈÆ«µÍ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚ25 ¡æʱ£¬Ä³ÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H£«)£½10£­12mol¡¤L£­1¡£¸ÃÈÜÒºµÄpH¿ÉÄÜΪ
A£®12¡¡¡¡¡¡¡¡B£®7C£®6D£®4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

½«pH£½8µÄNaOHÓëpH£½10µÄNaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒººóc(H£«)×î½Ó½üÓÚ(¡¡¡¡)
A£®(10£­8£«10£­10) mol/LB£®(10£­4£«10£­6) mol/L
C£®(10£­8£«10£­10) mol/L D£®2¡Á10£­10 mol/L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚÈÜÒºÖУ¬ÏÂÁеÈÁ¿¹Øϵ³ÉÁ¢µÄÊÇ£¨   £©
A£®c(OH-)= c(H+)+c(HCO3-)+2c(H2CO3)
B£®2c(Na+)= c(CO32-)+ c(HCO3-)+ c(H2CO3)
C£®c(Na+)+ c(H+)= c(HCO3-) +2c(CO32-)+ c(OH-)
D£®c(Na+)="2" c(CO32-)+ c(HCO3-)+ c(H2CO3)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓТÙCH3COOH¡¢¢ÚHCl¡¢¢ÛH2SO4ÈýÖÖÈÜÒº£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ
A£®pHÏàͬʱ£¬ÆäÎïÖʵÄÁ¿Å¨¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ¢Ù>¢Ú>¢Û
B£®ÎïÖʵÄÁ¿Å¨¶ÈÏàͬʱ£¬ÆäpHÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ¢Ù>¢Ú>¢Û
C£®Öк͵ÈÁ¿µÄÉÕ¼îÈÜÒº£¬ÐèµÈÎïÖʵÄÁ¿Å¨¶ÈµÄ¢Ù¢Ú¢ÛÈýÖÖËáÈÜÒºµÄÌå»ý±ÈΪ2¡Ã2¡Ã1
D£®Ìå»ýºÍÎïÖʵÄÁ¿Å¨¶È¾ùÏàͬµÄ¢Ù¢Ú¢ÛÈýÈÜÒº £¬·Ö±ðÓëͬŨ¶ÈµÄÉÕ¼îÈÜҺǡºÃÍêÈ«·´Ó¦£¬ËùÐèÉÕ¼îÈÜÒºµÄÌå»ý±ÈΪ2¡Ã1¡Ã2

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸