½â´ð£º½â£º£¨1£©Òª³ýÈ¥SO
42-£¬Ö»ÓÐÑ¡BaCl
2ÈÜÒº£¬ÈôÑ¡ÓÃBa£¨NO
3£©
2£¬»áÒýÈëеÄÀë×ÓNO
3-£¬ÔÙÑ¡ÓÃNaOHÈÜÒº³ýÈ¥Mg
2+ºÍFe
3+ÈÜÒº£¬×îºóÑ¡ÓÃNa
2CO
3ÈÜÒº³ýÈ¥Ca
2+£¬´Ë´¦²»ÄÜÑ¡ÓÃK
2CO
3ÈÜÒº£¬·ñÔò»áÒýÈëеÄK
+£¬ÔÙÓÃHCl³ýÈ¥¹ýÁ¿µÄCO
32-£¬Na
2CO
3ÈÜÒº²»ÄܼÓÔÚBaCl
2ÈÜҺǰ£¬·ñÔò»áÒýÈëBa
2+£¬
¹Ê´ð°¸Îª£ºBaCl
2¡¢NaOH¡¢Na
2CO
3£»
£¨2£©¸ù¾ÝʵÑéÊÒÌá´¿NaClµÄÁ÷³ÌͼºÍʵÑéÄ¿µÄ¿ÉÖªÕô·¢Å¨ËõÈÜÒº¢òÊdzýÈ¥ÈÜÒºÖеÄCa
2+¡¢Mg
2+¡¢Fe
3+¡¢SO
42-Àë×ӵõ½µÄºý×´ÎËùÒÔËüµÄÖ÷Òª»¯Ñ§³É·ÖÊÇÂÈ»¯ÄÆ£¬
¹Ê´ð°¸Îª£ºNaCl£»
£¨3£©ÅäÖÆ500mL4.00mol?L
-1NaClÈÜÒº£¬²Ù×÷²½ÖèÓмÆËã¡¢³ÆÁ¿¡¢Èܽ⡢ÒÆÒº¡¢Ï´µÓÒÆÒº¡¢¶¨ÈÝ¡¢Ò¡ÔȵȲÙ×÷£¬Ò»°ãÓÃÍÐÅÌÌìƽ³ÆÁ¿£¬ÓÃÒ©³×È¡ÓÃÒ©Æ·£¬ÔÚÉÕ±ÖÐÈܽ⣬Óò£Á§°ô½Á°è£¬¼ÓËÙÈܽ⣬»Ö¸´ÊÒκóתÒƵ½500mLÈÝÁ¿Æ¿ÖУ¬²¢Óò£Á§°ôÒýÁ÷£¬Ï´µÓ2-3´Î£¬²¢½«Ï´µÓÒºÒÆÈëÈÝÁ¿Æ¿ÖУ¬µ±¼ÓË®ÖÁÒºÃæ¾àÀë¿Ì¶ÈÏß1¡«2cmʱ£¬¸ÄÓýºÍ·µÎ¹ÜµÎ¼Ó£¬×îºó¶¨Èݵߵ¹Ò¡ÔÈ£¬ËùÒÔÐèÒªµÄÒÇÆ÷ÓÐÍÐÅÌÌìƽ¡¢ÉÕ±¡¢²£Á§°ô¡¢500mlµÄÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü¡¢Ò©³×£¬Ê¹Óû¹ÐèÒªÍÐÅÌÌìƽ¡¢500mlµÄÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü£®
¹Ê´ð°¸Îª£ºÍÐÅÌÌìƽ¡¢500mlµÄÈÝÁ¿Æ¿¡¢½ºÍ·µÎ¹Ü£»
£¨4£©Èçͼ2µç½â±¥ºÍʳÑÎË®µÄ·´Ó¦£º2NaCl+2H
2O
2NaOH+Cl
2¡ü+H
2¡ü£¬ÂÈÆøÔÚË®ÖÐÓÐÒ»¶¨Á¿µÄÈܽ⣬²úÉúµÄCl
2ÄÜÓëÉú³ÉµÄNaOH·´Ó¦NaCl¡¢NaClOºÍH
2O£¬Ê¹µÃ²¿·ÖµÄCl
2±»ÏûºÄ£¬ËùÒÔͬÑùÌõ¼þÏÂÊÕ¼¯µ½µÄCl
2СÓÚ2L£»bΪµçÔ´Õý¼«£¬·¢Éú2Cl
--2e
-=Cl
2¡ü£¬Èô½«´Ë×°ÖõÄb¿ÚÃÜ·âס£¬Ôòµç½âÒ»¶Îʱ¼äºó£¬2NaCl+2H
2O
2NaOH+Cl
2¡ü+H
2¡ü¢Ù£¬2NaOH+Cl
2=NaCl+NaClO+H
2O¢Ú£¬¢Ù+¢ÚµÃ´ËÏû¶¾ÒºµÄÒ»¸ö×Ü·´Ó¦·½³Ìʽ£ºNaCl+H
2O
NaClO+H
2¡ü£¬
¹Ê´ð°¸Îª£º£¼£»ÂÈÆøÔÚË®ÖÐÓÐÒ»¶¨Á¿µÄÈܽ⣬µç½âÉú³ÉµÄÂÈÆøÓëµç½âÉú³ÉµÄNaOH·¢ÉúÁË·´Ó¦£»NaCl+H
2O
NaClO+H
2¡ü£»
£¨5£©A£®ÑôÀë×Ó½»»»Ä¤½öÔÊÐíNa
+ͨ¹ý£¬µç½âʱ£¬Ñô¼«Éú³ÉÂÈÆø£¬ÏûºÄNaCl£¬ÔòÓ¦ÔÚÑô¼«²¹³äNaCl£¬ËùÒÔBΪÒõ¼«£¬·¢Éú2H
++2e
-=H
2¡ü£¬¹ÊA´íÎó£»
B£®µç½âʱ£¬Ñô¼«Éú³ÉÂÈÆø£¬ÏûºÄNaCl£¬WÊÇÏ¡µÄÂÈ»¯ÄÆÈÜÒº£¬¹ÊBÕýÈ·£»
C£®Òõ¼«Éú³ÉOH
-£¬ÇÒNa
+ÏòÒõ¼«Òƶ¯£¬Ôò²úÆ·ÉÕ¼îÈÜÒº´ÓZ¿Úµ¼³ö£¬¹ÊCÕýÈ·£»
D£®ÑôÀë×Ó½»»»Ä¤½öÔÊÐíNa
+ͨ¹ý£¬µç½âʱ£¬Ñô¼«Éú³ÉÂÈÆø£¬ÏûºÄNaCl£¬ÔòÓ¦ÔÚÑô¼«²¹³äNaCl£¬ËùÒÔAΪÑô¼«£¬aΪµçÔ´Õý¼«£¬·¢Éú£º2Cl
--2e
-=Cl
2¡ü£¬ÔòBΪÒõ¼«£¬bΪµçÔ´¸º¼«£¬·¢Éú2H
++2e
-=H
2¡ü£¬¹ÊD´íÎó£»
¹ÊÑ¡BC£®