ÌâÄ¿ÄÚÈÝ

(1)ijζÈ(t¡æ)ʱ£¬Ë®µÄÀë×Ó»ýΪKw£½1¡Á10£­13£¬Ôò¸Ãζȡ¡¡¡25¡æ£¨Ìî¡°£¾¡±¡°£½¡±»ò¡°£¼¡±£©£¬ÔÚ´ËζÈÏ£¬Ä³ÈÜÒºÖÐÓÉË®µçÀë³öÀ´µÄH£«Å¨¶ÈΪ1¡Á10£­10mol/L£¬Ôò¸ÃÈÜÒºµÄpH¿ÉÄÜΪ¡¡  ¡¡¡¡»ò         ¡£
ÈôζÈΪ25¡æʱ£¬Ìå»ýΪVa¡¢pH£½aµÄH2SO4ÈÜÒºÓëÌå»ýΪVb¡¢pH£½bµÄNaOHÈÜÒº»ìºÏ£¬Ç¡ºÃÖкÍ.ÒÑÖªVa£¾Vb£¬ÇÒa£½0.5 b£¬ÔòaµÄÈ¡Öµ·¶Î§¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
£¨3£©Èô½«´ËζÈ(t¡æ)Ï£¬pH£½11µÄ¿ÁÐÔÄÆÈÜÒºm LÓëpH£½1µÄÏ¡ÁòËáÈÜÒºnL»ìºÏ(¼ÙÉè»ìºÏºóÈÜÒºÌå»ýµÄ΢С±ä»¯ºöÂÔ²»¼Æ)£¬ÊÔͨ¹ý¼ÆËãÌîдÒÔϲ»Í¬Çé¿öʱÁ½ÖÖÈÜÒºµÄÌå»ý±È£¬²¢±È½ÏÈÜÒºÖи÷Àë×ÓµÄŨ¶È´óС¡£
¢ÙÈôËùµÃ»ìºÏҺΪÖÐÐÔ£¬Ôòm¡Ãn£½¡¡¡¡¡¡¡¡£»´ËÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
¢ÚÈôËùµÃ»ìºÏÒºµÄpH£½2£¬Ôòm¡Ãn£½¡¡¡¡¡¡£»´ËÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£

(1)£¾¡¡10   3£¨Ã¿¿Õ1·Ö£©        (2) £¼a£¼7£¨2·Ö£©
(3) ¢Ù10¡Ã1£¨2·Ö£©¡¡c(Na£«)£¾c(SO)£¾c(H£«)£½c(OH£­) £¨È«¶ÔµÃ2·Ö£©
¢Ú9¡Ã2£¨2·Ö£©¡¡c(H£«)£¾c(SO)£¾c(Na£«)£¾c(OH£­) £¨È«¶ÔµÃ2·Ö£©

ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
»Ø´ðÏÂÁÐÎÊÌâ:
(1)ÓÐÒ»ÖÖÈÜÒº³ýH+¡¢OH-Í⣬»¹ÓÐNa+¡¢SO42-ºÍCl-£¬²âµÃÈýÖÖÀë×ÓµÄŨ¶È·Ö±ðÊÇ0.01 mol¡¤L-1¡¢0.0035 mol¡¤L-1ºÍ0.004 mol¡¤L-1£¬¸ÃÈÜÒºµÄpHΪ         ¡£
(2)Òº°±ÀàËÆÓÚË®µÄµçÀ룬ÊÔд³öÒº°±µÄµçÀë·½³Ìʽ                         £»
£¨3£©ÏÂÁÐÀë×Ó·½³ÌʽÊéдÕýÈ·µÄÊÇ              ¡£                           ?
A£®Ê¯»ÒÈéÓëNa2CO3ÈÜÒº»ìºÏ£ºCa2++CO32¡ª=====CaCO3¡ý?
B£®NH4HSO3ÈÜÒºÓë×ãÁ¿µÄNaOHÈÜÒº»ìºÏ¼ÓÈÈ£º
NH4+ + HSO3- +2OH- NH3¡ü+SO32-  +2H2O?
C£®ËáÐÔÌõ¼þÏÂKIO3ÈÜÒºÓëKIÈÜÒº·´Ó¦Éú³ÉI2£º IO32-  +5I-+3H2O====3I2+6OH-
D£®AgNO3ÈÜÒºÖмÓÈë¹ýÁ¿°±Ë®£ºAg++NH3¡¤H2O=====AgOH¡ý+NH4+ 
E£®H2SO4ÓëBa(OH)2ÈÜÒº·´Ó¦£ºBa2++OH-+H++SO42-="====" BaSO4¡ý+H2O?
F£®CuSO4ÈÜÒºÎüÊÕH2SÆøÌ壺Cu2++H2S======CuS¡ý+2H+?
G£®AlCl3ÈÜÒºÖмÓÈë¹ýÁ¿µÄŨ°±Ë®£ºAl3++4NH3¡¤H2O ="====" AlO2-  +4NH 4+ +2H2O
H£®µÈÌå»ý¡¢µÈŨ¶ÈµÄBa(OH)2Ï¡ÈÜÒºÓëNH4HCO3Ï¡ÈÜÒº»ìºÏ£º
Ba2++2OH-+NH4+  +HCO3-="==  " BaCO3¡ý+NH3¡¤H2O+H2O
I£®LiAlH4ÈÜÓÚÊÊÁ¿Ë®ºóµÃµ½ÎÞÉ«ÈÜÒº£¬Ôò»¯Ñ§·½³Ìʽ¿É±íʾΪ£º
LiAlH4+2H2O ="==LiAlO2" +4H2¡ü
£¨4£©Na2O2¡¢HCl¡¢Al2O3ÈýÖÖÎïÖÊÔÚË®ÖÐÍêÈ«·´Ó¦ºó£¬ÈÜÒºÖÐÖ»º¬ÓÐNa+¡¢H+¡¢Cl-¡¢OH-£¬ÇÒÈÜÒº³ÊÖÐÐÔ£¬ÔòNa2O2¡¢HCl¡¢Al2O3µÄÎïÖʵÄÁ¿Ö®±È¿ÉÄÜΪ     ¡£                                     
A.3¡Ã2¡Ã1             B.2¡Ã4¡Ã1?   C.2¡Ã3¡Ã1             D.4¡Ã2¡Ã1
£¨8·Ö£©¼×¡¢ÒÒÁ½Î»Í¬Ñ§Éè¼ÆÓÃʵÑéÈ·¶¨Ä³ËáHAÊÇÈõµç½âÖÊ£¬´æÔÚµçÀëƽºâ£¬ÇҸıäÌõ¼þƽºâ·¢ÉúÒƶ¯¡£ÊµÑé·½°¸ÈçÏ£º

¼×£º¢Ù׼ȷÅäÖÆ0.1mol¡¤L-1µÄHA¡¢HClÈÜÒº¸÷100mL£»
¢ÚÈ¡´¿¶ÈÏàͬ£¬ÖÊÁ¿¡¢´óСÏàµÈµÄпÁ£·ÅÈëÁ½Ö»ÊÔ¹ÜÖУ¬Í¬Ê±¼ÓÈë0.1mol¡¤L-1µÄHA¡¢HClÈÜÒº¸÷100mL£¬°´ÉÏͼװºÃ£¬¹Û²ìÏÖÏó¡£
ÒÒ£º¢ÙÓÃpH¼Æ²â¶¨ÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1mol¡¤L-1µÄHAºÍHClÈÜÒºµÄpH£»
¢ÚÔÙÈ¡0.1mol¡¤L-1µÄHAºÍHClÈÜÒº¸÷2µÎ£¨1µÎԼΪ£©·Ö±ðÏ¡ÊÍÖÁ100mL£¬
ÔÙÓÃpH¼Æ²âÆäpH±ä»¯¡£
£¨1£©ÒÒ·½°¸ÖÐ˵Ã÷HAÊÇÈõµç½âÖʵÄÀíÓÉÊÇ£¬²âµÃ0.1mol¡¤L-1µÄHAÈÜÒºµÄpH    1£¨Ìî¡°>¡±¡°<¡±»ò¡°=¡±£©£»¼×·½°¸ÖУ¬ËµÃ÷HAÊÇÈõµç½âÖʵÄʵÑéÏÖÏóÊÇ     ¡£
A£®×°HClµÄÊÔ¹ÜÖзųöµÄÇâÆøËÙÂÊ´ó
B£®×°HAÈÜÒºµÄÊÔ¹ÜÖзųöÇâÆøµÄËÙÂÊ´ó
C£®Á½¸öÊÔ¹ÜÖвúÉúÆøÌåËÙÂÊÒ»Ñù´ó
£¨2£©ÒÒͬѧÉè¼ÆµÄʵÑéµÚ¢Ú²½£¬ÄÜÖ¤Ã÷¸Ä±äÌõ¼þ½âÖÊƽºâ·¢ÉúÒƶ¯¡£¼ÓˮϡÊÍ£¬ÈõËáHAµÄµçÀë³Ì¶È       £¨Ìî¡°Ôö´ó¡¢¼õС¡¢²»±ä¡±£©
£¨3£©¼×ͬѧΪÁ˽øÒ»²½Ö¤Ã÷Èõµç½âÖʵçÀëƽºâÒƶ¯µÄÇé¿ö£¬Éè¼ÆÈçÏÂʵÑ飺ʹHAµÄµçÀë³Ì¶ÈºÍc£¨H+£©¶¼¼õС£¬c£¨A£©Ôö´ó£¬¿ÉÔÚ0.1mol¡¤L-1µÄHAÈÜÒºÖУ¬Ñ¡Ôñ¼ÓÈë            ÊÔ¼Á¡£
A£®NaA¹ÌÌ壨¿ÉÍêÈ«ÈÜÓÚË®£©        B£®1mol¡¤L-1NaOHÈÜÒº
C£®1mol¡¤L-1H2SO4                                                       D£®2mol¡¤L-1HA
£¨4£©pH=1µÄÁ½ÖÖËáÈÜÒºA¡¢B¸÷1mL£¬·Ö±ð¼ÓˮϡÊ͵½1000mL£¬ÆäpHÓëÈÜÒºÌå»ýVµÄ¹ØϵÈçÓÒͼËùʾ£¬ÔòÏÂÁÐ˵·¨²»ÕýÈ·µÄÓР        

A£®Èôa=4£¬ÔòAÊÇÇ¿ËᣬBÊÇÈõËá
B£®Èô£¬ÔòA¡¢B¶¼ÊÇÈõËá
C£®Á½ÖÖËáÈÜÒºµÄÎïÖʵÄÁ¿³¢ÊÔÒ»¶¨ÏàµÈ
D£®Ï¡Êͺó£¬AÈÜÒºµÄËáÐÔ±ÈBÈÜÒºÈõ

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

¾«Ó¢¼Ò½ÌÍø