ÌâÄ¿ÄÚÈÝ

ijÈÜÒº¿ÉÄܺ¬ÓÐCl£­¡¢SO42£­¡¢CO32£­¡¢OH-¡¢NH4+¡¢Al3+¡¢Fe3+ºÍK+¡£È¡¸ÃÈÜÒº100mL£¬¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬¼ÓÈÈ£¬µÃµ½0.02molÆøÌ壬ͬʱ²úÉúºìºÖÉ«³Áµí£»¹ýÂË£¬Ï´µÓ£¬×ÆÉÕ£¬µÃµ½ 1.6g¹ÌÌ壻ÏòÉÏÊöÂËÒºÖмÓ×ãÁ¿BaCl2ÈÜÒº£¬µÃµ½ 6.99g²»ÈÜÓÚÑÎËáµÄ³Áµí¡£
ÏÂÁйØÓÚÔ­ÈÜÒºµÄ˵·¨ÕýÈ·µÄÊÇ
A£®Ô­ÈÜÒºÖÐÖÁÉÙ´æÔÚ3ÖÖÀë×Ó
B£®Ô­ÈÜÒºÖÐÒ»¶¨²»´æÔÚµÄÀë×ÓÊÇ£ºCl£­¡¢CO32£­¡¢OH-
C£®Ô­ÈÜÒºÖпÉÄÜ´æÔÚµÄÀë×ÓÓУºCl£­¡¢Al3+ºÍK+
D£®Ô­ÈÜÒºÖÐÒ»¶¨´æÔÚµÄÀë×ÓÊÇ£ºNH4+¡¢Fe3+¡¢SO42£­ºÍCl£­£¬ÇÒc£¨Cl£­£©¡Ý0.2mol/L
D

ÊÔÌâ·ÖÎö£º¸ù¾Ý¼ÓÈë¹ýÁ¿NaOHÈÜÒº£¬¼ÓÈÈ£¬µÃµ½0.02 molÆøÌ壬˵Ã÷ÓÐNH4+£¬¶øÇÒΪ0.02 mol£¬Í¬Ê±²úÉúºìºÖÉ«³Áµí£¬ËµÃ÷ÓÐFe3+£¬¶øÇÒΪ0.02 mol£¬ÔòûÓÐCO32£­£¬¸ù¾Ý²»ÈÜÓÚÑÎËáµÄ4.66 g³Áµí£¬ËµÃ÷ÓÐSO42£­£¬ÇÒΪ0.02 mol£¬ÔÙ¸ù¾ÝµçºÉÊغã¿ÉÖªÒ»¶¨º¬ÓÐCl£­£¬ÖÁÉÙΪ0.06 mol£¨0.02¡Á3+0.02¡Á2+0.02¡Á2£©£¬DÏîÕýÈ·¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø