ÌâÄ¿ÄÚÈÝ
¼ä½ÓµâÁ¿·¨²â¶¨µ¨·¯ÖÐͺ¬Á¿µÄÔÀíºÍ·½·¨ÈçÏ£º
ÒÑÖª£ºÔÚÈõËáÐÔÌõ¼þÏ£¬µ¨·¯ÖÐCu2+ÓëI?/SUP>×÷Óö¨Á¿Îö³öI2£¬I2ÈÜÓÚ¹ýÁ¿µÄKIÈÜÒºÖУº
I2£«I£I3££¬ÓÖÖªÑõ»¯ÐÔ£ºFe3+£¾Cu2+£¾I2£¾FeF63££»Îö³öI2¿ÉÓÃcmol/LNa2S2O3±ê×¼ÈÜÒºµÎ¶¨£º2S2O32££«I3£S4O62££«3I££®
²Ù×÷£º×¼È·³ÆÈ¡agµ¨·¯ÊÔÑù(¿ÉÄܺ¬ÉÙÁ¿Fe2(SO4)3)£¬ÖÃÓÚ250mLµâÁ¿Æ¿(´øÄ¥¿ÚÈûµÄ׶ÐÎÆ¿)ÖУ¬¼Ó50mLÕôÁóË®¡¢5mL3mol/LH2SO4ÈÜÒº£¬¼ÓÉÙÁ¿NaF£¬ÔÙ¼ÓÈë×ãÁ¿µÄ10%KIÈÜÒº£¬Ò¡ÔÈ£®¸ÇÉϵâÁ¿Æ¿Æ¿¸Ç£¬ÖÃÓÚ°µ´¦5min£¬³ä·Ö·´Ó¦ºó£¬¼ÓÈë1~2mL0.5%µÄµí·ÛÈÜÒº£¬ÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨µ½À¶É«ÍÊȥʱ£¬¹²ÓÃÈ¥VmL±ê×¼Òº£®
¢ÙʵÑéÖУ¬ÔÚ¼ÓKIÇ°Ðè¼ÓÈëÉÙÁ¿NaF£¬ÍƲâÆä×÷ÓÿÉÄÜÊÇ________£®
¢ÚʵÑéÖмÓÈëÁòËᣬÄãÈÏΪÁòËáµÄ×÷ÓÃÊÇ________£®
¢Û±¾ÊµÑéÖÐÓõâÁ¿Æ¿¶ø²»ÓÃÆÕͨ׶ÐÎÆ¿ÊÇÒòΪ£º________£®
¢ÜÁòËáÍÈÜÒºÓëµâ»¯¼ØÈÜÒº·´Ó¦Éú³É°×É«³Áµí(µâ»¯ÑÇÍ)²¢Îö³öµâ£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º________£®
¢Ý¸ù¾Ý±¾´ÎʵÑé½á¹û£¬¸Ãµ¨·¯ÊÔÑùÖÐÍÔªËصÄÖÊÁ¿·ÖÊý¦Ø(Cu)£½________£®
½âÎö£º
(1)ÑÚ±ÎFe3+£¬·ÀÖ¹Ôì³ÉÆ«´óµÄÎó²î(2·Ö) (2)ÌṩËáÐÔ»·¾³ÒÖÖÆÍÀë×ÓË®½â(2·Ö) (3)·ÀÖ¹¿ÕÆøÖÐÑõÆøÑõ»¯µâ»¯¼Ø;·ÀÖ¹µâÉý»ª(2·Ö) (4)Cu2+£«4I££½£½£½2CuI¡ý£«I2(2·Ö) (5)64Cv/1000a¡Á100£¥(2·Ö) |
