ÌâÄ¿ÄÚÈÝ

(10·Ö)ÔÚ25¡æÊ±,Ïò100mLº¬ÂÈ»¯Çâ14.6gµÄÑÎËáÈÜÒºÀï·ÅÈë5.60g´¿Ìú·Û(²»¿¼ÂÇ·´Ó¦Ç°ºóÈÜÒºÌå»ýµÄ±ä»¯),·´Ó¦¿ªÊ¼ÖÁ2minÄ©,ÊÕ¼¯µ½1.12L(±ê×¼×´¿ö)ÇâÆø,ÔÚ´ËÖ®ºó,ÓÖ¾­¹ý4min,Ìú·ÛÍêÈ«Èܽâ,Ôò:
(1)ÔÚǰ2minÄÚÓÃFeCl2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊÊÇ____________
(2)ÔÚºó4minÄÚÓÃHCl±íʾµÄƽ¾ù·´Ó¦ËÙÂÊÊÇ______________
(3)ǰ2minÓëºó4minÏà±È,·´Ó¦ËÙÂʽϿìµÄÊÇ__________________________________,ÆäÔ­ÒòÊÇ__________________________________________¡£
(4)·´Ó¦½áÊøºó,ÈÜÒºÖÐCl-µÄÎïÖʵÄÁ¿Å¨¶ÈΪ________________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÎïÖÊÔÚË®ÖпÉÄÜ´æÔÚµçÀëÆ½ºâ¡¢ÑεÄË®½âƽºâºÍ³ÁµíµÄÈÜ½âÆ½ºâ£¬ËüÃǶ¼¿É¿´×÷»¯Ñ§Æ½ºâ£®Çë¸ù¾ÝËùѧµÄ֪ʶ»Ø´ð£º
£¨1£©AΪ0.1mol/LµÄ£¨NH4£©2SO4ÈÜÒº£¬ÔÚ¸ÃÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡Ë³ÐòΪ
c£¨NH4+£©£¾c£¨SO42-£©£¾c£¨H+£©£¾c£¨OH-£©
c£¨NH4+£©£¾c£¨SO42-£©£¾c£¨H+£©£¾c£¨OH-£©
£®
£¨2£©BΪ0.1mol/L NaHCO3ÈÜÒº£¬Çë·ÖÎöNaHCO3ÈÜÒºÏÔ¼îÐÔµÄÔ­Òò£º
HCO
 
-
3
µÄË®½â³Ì¶È´óÓÚÆäµçÀë³Ì¶È£¬ÈÜÒºÖÐc£¨OH-£©£¾c£¨H+£©£¬¹ÊÈÜÒºÏÔ¼îÐÔ
HCO
 
-
3
µÄË®½â³Ì¶È´óÓÚÆäµçÀë³Ì¶È£¬ÈÜÒºÖÐc£¨OH-£©£¾c£¨H+£©£¬¹ÊÈÜÒºÏÔ¼îÐÔ
£®
£¨3£©CΪFeCl3ÈÜÒº£¬ÊµÑéÊÒÖÐÅäÖÆFeCl3ÈÜҺʱ³£¼ÓÈë
ÑÎËá
ÑÎËá
ÈÜÒºÒÔÒÖÖÆÆäË®½â£¬Èô°ÑBºÍCÈÜÒº»ìºÏ£¬½«²úÉúºìºÖÉ«³ÁµíºÍÎÞÉ«ÆøÌ壬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ
Fe3++3HCO
 
-
3
¨TFe£¨OH£©3¡ý+3CO2¡ü
Fe3++3HCO
 
-
3
¨TFe£¨OH£©3¡ý+3CO2¡ü
£®
£¨4£©DΪº¬ÓÐ×ãÁ¿AgCl¹ÌÌåµÄ±¥ºÍÈÜÒº£¬AgClÔÚË®ÖдæÔÚ³ÁµíÈÜ½âÆ½ºâ£ºAgCl£¨s£©?Ag+£¨aq£©+Cl-£¨aq£©£¬ÔÚ25¡æÊ±£¬ÂÈ»¯ÒøµÄKsp=1.8¡Á10-10£®ÏÖ½«×ãÁ¿ÂÈ»¯Òø·Ö±ð·ÅÈ룺
¢Ù100mLÕôÁóË®ÖУ»¢Ú100mL 0.2mol/L AgNO3ÈÜÒºÖУ»¢Û100mL 0.1mol/LÂÈ»¯ÂÁÈÜÒºÖУ»¢Ü100mL 0.1mol/LÑÎËáÖУ¬³ä·Ö½Á°èºó£¬ÏàͬζÈÏÂc£¨Ag+£©ÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
£©¢Ú£¾¢Ù£¾¢Ü£¾¢Û
£©¢Ú£¾¢Ù£¾¢Ü£¾¢Û
£¨ÌîдÐòºÅ£©£»¢ÚÖÐÂÈÀë×ÓµÄŨ¶ÈΪ
9¡Á10-10
9¡Á10-10
mol/L£®
£¨5£©EΪCuSO4ÈÜÒº£¬ÒÑÖª25¡æÊ±£¬Ksp[Cu£¨OH£©2]=2¡Á10-20£®ÒªÊ¹0.2mol?L-1 EÈÜÒºÖÐCu2+³Áµí½ÏΪÍêÈ«£¨Ê¹Cu2+Ũ¶È½µÖÁÔ­À´µÄǧ·ÖÖ®Ò»£©£¬ÔòÓ¦ÏòÈÜÒºÀï¼ÓÈëNaOHÈÜÒº£¬Ê¹ÈÜÒºµÄpHΪ
6
6
£®
ÓÃÖк͵ζ¨µÄ·½·¨²â¶¨NaOHºÍNa2CO3µÄ»ìºÏÈÜÒºÖÐNaOHµÄº¬Á¿£¬¿ÉÏÈÔÚ»ìºÏÒºÖмÓÈë¹ýÁ¿µÄBaCl2ÈÜÒº£¬Ê¹Na2CO3Íêȫת±ä³ÉBaCO3³Áµí£¬È»ºóÓñê×¼ÑÎËáµÎ¶¨£¨ÒÑÖª¼¸ÖÖËá¼îָʾ¼Á±äÉ«µÄpH·¶Î§£º¢Ù¼×»ù³È3.1¡«4.4  ¢Ú¼×»ùºì4.4¡«6.2  ¢Û·Ó̪8.2¡«10£©£®ÊԻشð£º
£¨1£©ÔÚ25¡æÊ±£¬1LË®ÖÐÔ¼ÄÜÈܽâ0.01g̼Ëá±µ£¬Ôò¸ÃζÈÏÂ̼Ëá±µµÄÈܶȻýKsp=
 
£®
£¨2£©Ïò»ìÓÐBaCO3³ÁµíµÄNaOHÈÜÒºÖеÎÈëÑÎËᣬӦѡÓÃ
 
ָʾ¼Á£¬ÀíÓÉÊÇ
 
£»
Åжϵ½´ïµÎ¶¨ÖÕµãµÄʵÑéÏÖÏóÊÇ
 
£®
£¨3£©Îª²â¶¨Ä³ÉÕ¼îÑùÆ·ÖÐNaOHµÄº¬Á¿£¨ÉèÑùÆ·ÖÐÔÓÖÊΪNa2CO3£©£¬Ä³Í¬Ñ§½øÐÐÈçÏÂʵÑ飺׼ȷ³ÆÈ¡5.0gÑùÆ·ÅäÖÆ³É250mLÈÜÒº£¬È»ºó·ÖÈý´Î¸÷È¡ÅäÖÆºÃµÄÉÕ¼îÈÜÒº20.00mLÓÚÈý¸öÓÃÕôÁóˮϴ¾»µÄ×¶ÐÎÆ¿ÖУ¬·Ö±ð¼ÓÈë¹ýÁ¿µÄBaCl2ÈÜÒº£¬²¢Ïò×¶ÐÎÆ¿Öи÷¼ÓÈë1¡«2µÎָʾ¼Á£¬ÓÃŨ¶ÈΪ0.2000mol?L-1µÄÑÎËá±ê×¼Òº½øÐе樣¬Ïà¹ØÊý¾Ý¼Ç¼Èç±í 
ʵÑé±àºÅ V£¨ÉÕ¼îÈÜÒº£©/mL V£¨HCl£©/mL
³õ¶ÁÊý Ä©¶ÁÊý
1 20.00 0.00 31.00
2 20.00 1.00 32.04
3 20.00 1.10 32.18
¢ÙʵÑé3µ½´ïµÎ¶¨ÖÕµãʱËùºÄHClÈÜÒºµÄÌå»ýΪ
 
£»ÒÀ¾Ý±íÖÐÊý¾Ý£¬¼ÆËã³öÉÕ¼îÑùÆ·Öк¬NaOHµÄÖÊÁ¿·ÖÊýΪ
 
%£®£¨Ð¡Êýµãºó±£ÁôÁ½Î»Êý×Ö£©
¢ÚµÎ¶¨Ê±µÄÕýÈ·²Ù×÷ÊÇ
 
£®
¢ÛÏÂÁвÙ×÷»áµ¼ÖÂÉÕ¼îÑùÆ·ÖÐNaOHº¬Á¿²â¶¨ÖµÆ«¸ßµÄÊÇ
 

A£®×¶ÐÎÆ¿ÓÃÕôÁóˮϴºóδÓôý²âÒºÈóÏ´
B£®ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴºóδÓñê×¼ÒºÈóÏ´
C£®Ôڵζ¨Ç°ÓÐÆøÅÝ£¬µÎ¶¨ºóÆøÅÝÏûʧ
D£®µÎ¶¨Ç°Æ½ÊÓ¶ÁÊý£¬µÎ¶¨½áÊø¸©ÊÓ¶ÁÊý£®

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

¾«Ó¢¼Ò½ÌÍø