ÌâÄ¿ÄÚÈÝ

ÓÐÒ»»ìºÏÎïµÄË®ÈÜÒº£¬Ö»¿ÉÄܺ¬ÓÐÒÔÏÂÀë×ÓÖеļ¸ÖÖ£ºK+¡¢H+¡¢Ba2+¡¢Mg2+¡¢CO32-¡¢SO42-£¬ÏÖÈ¡Á½·Ý¸ÃÈÜÒº¸÷100ml½øÐÐÈçÏÂʵÑ飺
¢ÙµÚÒ»·Ý¼ÓÈë×ãÁ¿NaHCO3ÈÜÒººó£¬ÊÕ¼¯µ½ÆøÌå0.03mol£®
¢ÚµÚ¶þ·Ý¼ÓÈë×ãÁ¿Ba£¨NO3£©2ÈÜÒº³ä·Ö·´Ó¦ºó¹ýÂ˸ÉÔµÃ³Áµí4.66g£®
¸ù¾ÝÉÏÊöʵÑ黨´ð£º
£¨1£©Ô­ÈÜÒºÖÐÒ»¶¨²»´æÔÚµÄÀë×ÓÊÇ______£®
£¨2£©Ô­ÈÜÒºÖпÉÄÜ´æÔÚµÄÀë×ÓÊÇ______£®
£¨3£©¢ÙÖз¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®
£¨4£©¾­¼ÆË㣬ԭÈÜÒºÖÐÒ»¶¨´æÔÚµÄÒõÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈΪ______£®

½â£º£¨1£©¸ù¾Ý¢ÙÏÖÏóÖª£¬¸ÃÈÜÒºÖÐÒ»¶¨´æÔÚÓë̼ËáÇâ¸ùÀë×Ó·´Ó¦Éú³ÉÆøÌåµÄH+£¬Ò»¶¨²»´æÔÚÓëH+·´Ó¦µÄ̼Ëá¸ùÀë×Ó£»ËùÒÔÈÜÒºÖдæÔÚµÄÒõÀë×ÓΪÁòËá¸ùÀë×Ó£¬Ò»¶¨²»´æÔÚÓëÁòËá¸ùÀë×Ó·´Ó¦µÄBa2+£»
¹Ê´ð°¸Îª£ºCO32-¡¢Ba2+£»
£¨2£©¸ù¾ÝʵÑéÏÖÏóÎÞ·¨È·¶¨ÈÜÒºÖÐÊÇ·ñº¬ÓÐK+¡¢Mg2+£¬¹Ê´ð°¸Îª£ºK+¡¢Mg2+£»
£¨3£©Ì¼ËáÇâ¸ùÀë×ÓÄܺÍËá·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍË®HCO3-+H+=CO2¡ü+H2O£®
¹Ê´ð°¸Îª£ºHCO3-+H+=CO2¡ü+H2O£»
£¨4£©ÉèÁòËá¸ùµÄÎïÖʵÄÁ¿Îªxmol£® Ba2++SO42-=BaSO4 ¡ý
1 mol 233g
xmol 4.66g
x=0.02
ËùÒÔC==0.2mol/L
¹Ê´ð°¸Îª£º0.2mol/L£®
·ÖÎö£º£¨1£©£¨2£©¸ù¾Ý¢ÙÏÖÏóÅжÏÄܺÍ̼ËáÇâ¸ùÀë×Ó·´Ó¦µÄÀë×Ó£¬È»ºó¸ù¾ÝÀë×Ó¹²´æÅжÏÒ»¶¨´æÔÚµÄÀë×Ó¡¢Ò»¶¨²»´æÔÚµÄÀë×Ó¡¢¿ÉÄܺ¬ÓеÄÀë×Ó£»
£¨3£©Ì¼ËáÇâ¸ùÀë×ÓÄܺÍËá·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍË®£»
£¨4£©¸ù¾ÝÁòËá±µÓëÁòËá¸ùÀë×ÓµÄÎïÖʵÄÁ¿¹ØÏµ¼ÆËãÁòËá¸ùµÄÎïÖʵÄÁ¿£¬ÔÙ¸ù¾ÝŨ¶È¹«Ê½¼ÆËãÆäŨ¶È£®
µãÆÀ£º±¾Ì⿼²éÁËÀë×Ó¹²´æ¼°Àë×Ó·½³ÌʽµÄÓйؼÆË㣬ÄѶȲ»´ó£¬Àë×Ó¹²´æÊǸ߿¼µÄÈȵ㣬ҪעÒâÀë×Ó¹²´æÖеÄÒþº¬Ìõ¼þ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
½ñÓÐÒ»»ìºÏÎïµÄË®ÈÜÒº£¬Ö»¿ÉÄܺ¬ÓÐÒÔÏÂÀë×ÓÖеÄÈô¸ÉÖÖ£ºK+¡¢NH4+¡¢Cl-¡¢Mg2+¡¢Ba2+¡¢CO32-¡¢SO42-£¬ÏÖÈ¡Èý·Ý¸÷100mLÈÜÒº½øÐÐÈçÏÂʵÑ飺
µÚÒ»·Ý¼ÓÈëAgNO3ÈÜÒºÓгÁµí²úÉú£®
µÚ¶þ·Ý¼Ó×ãÁ¿NaOHÈÜÒº¼ÓÈȺó£¬ÊÕ¼¯µ½0.08molÆøÌ壮
µÚÈý·Ý¼Ó×ãÁ¿BaCl2ÈÜÒººó£¬µÃµ½¸ÉÔï³Áµí12.54g£¬¾­×ãÁ¿ÑÎËáÏ´µÓ¡¢¸ÉÔïºó£¬³ÁµíÖÊÁ¿Îª4.66g£®£®
¸ù¾ÝÉÏÊöʵÑ飬»Ø´ðÒÔÏÂÎÊÌ⣺
£¨1£©ÓɵÚÒ»·Ý½øÐеÄʵÑéÍÆ¶Ï¸Ã»ìºÏÎïÊÇ·ñÒ»¶¨º¬ÓÐCl-
 

£¨2£©Óɵڶþ·Ý½øÐеÄʵÑéµÃÖª»ìºÏÎïÖÐÓ¦º¬ÓÐ
 
Àë×Ó£¬ÆäÎïÖʵÄÁ¿Å¨¶ÈΪ
 

£¨3£©ÓɵÚÈý·Ý½øÐеÄʵÑé¿ÉÖª12.54g³ÁµíµÄ³É·ÖΪ
 
Çë¼ÆËãÐγɸóÁµíµÄÔ­»ìºÏÎïÖи÷Àë×ÓµÄÎïÖʵÄÁ¿£®£¨ÒªÇóд³ö¼ÆËã¹ý³Ì£©
£¨4£©×ÛºÏÉÏÊöʵÑ飬ÄãÈÏΪÒÔϽáÂÛÕýÈ·µÄÊÇ
 

A£®¸Ã»ìºÏÒºÖÐÒ»¶¨º¬ÓУºK+¡¢NH4+¡¢CO32-¡¢SO42-£¬¿ÉÄܺ¬Cl-£¬ÇÒn£¨K+£©¡Ý0.04mol
B£®¸Ã»ìºÏÒºÖÐÒ»¶¨º¬ÓУºNH4+¡¢CO32-¡¢SO42-£¬¿ÉÄܺ¬K+¡¢Cl-
C£®¸Ã»ìºÏÒºÖÐÒ»¶¨º¬ÓУºNH4+¡¢CO32-¡¢SO42-£¬¿ÉÄܺ¬Mg2+¡¢K+¡¢Cl-
D£®¸Ã»ìºÏÒºÖÐÒ»¶¨º¬ÓУºNH4+¡¢SO42-£¬¿ÉÄܺ¬Mg2+¡¢K+¡¢Cl-£®

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

¾«Ó¢¼Ò½ÌÍø