ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿I-²â¶¨µÄ»ù±¾ÔÀíÊǽ«ÆäÑõ»¯³ÉI2£¬ÔÙÓÃNa2S2O3±ê×¼ÈÜÒºÀ´µÎ¶¨¡£Èç¹ûÑùÆ·Öнöº¬ÓÐ΢Á¿I-£¬±ØÐëÓá°»¯Ñ§·Å´ó¡±·´Ó¦½«µâµÄÁ¿¡°·Å´ó¡±£¬È»ºóÔÙ½øÐвⶨ¡£ÏÂÃæÊÇ»¯Ñ§·Å´ó·´Ó¦µÄʵÑé²½Ö裺
I£®½«º¬ÓÐ΢Á¿I-µÄÑùÆ·ÈÜÒºµ÷ÖÁÖÐÐÔ»òÈõËáÐÔ£¬¼ÓÈëäåË®£¬½«I-ÍêÈ«Ñõ»¯³ÉIO£¬Öó·ÐÈ¥µô¹ýÁ¿µÄä壻
¢ò£®È¡ÉÏÊöÈÜÒº£¬¼ÓÈë¹ýÁ¿ÁòËáËữµÄKIÈÜÒº£¬Õñµ´£¬ÈÜÒºÑÕÉ«³Ê×Ø»ÆÉ«£»
¢ó£®½«¢òËùµÃÈÜÒºÖÐÈ«²¿µÄI2ÝÍÈ¡ÖÁCCl4ÖУ»
¢ô£®Ïò·ÖÒººóµÄCCl4ÈÜÒºÖмÓÈëëÂ(N2H4)µÄË®ÈÜÒº£¬²úÉúÎÞ¶¾µÄN2£¬·ÖÈ¥Óлú²ã£»
V£®½«¢ôËùµÃË®ÈÜÒºÖظ´I¡¢¢ò²½Ö裻
¢ö£®ÒÔµí·ÛΪָʾ¼Á£¬ÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨ÖÁÖյ㡣ÒÑÖª£º·´Ó¦Îª£ºI2+2S2O=2I-+S4O£»Na2S2O3ºÍNa2S4O6ÈÜÒºÑÕÉ«¾ùΪÎÞÉ«¡£
(1)IÖÐÑõ»¯¼ÁºÍ»¹Ô¼ÁÎïÖʵÄÁ¿Ö®±ÈΪ______¡£
(2)¢óÖÐʹÓõÄÖ÷Òª²£Á§ÒÇÆ÷ÊÇ______(ÌîÃû³Æ)¡£
(3)¢ôÖз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ______¡£
(4)¢öÖÐÐèÒªÔÚÖÐÐÔ»òÈõËáÐÔ»·¾³ÖнøÐУ¬pH¹ý´ó£¬Na2S2O3Ò×±»I2Ñõ»¯³ÉNa2SO4£¬ÆäÀë×Ó·½³ÌʽΪ_______£»½áºÏÀë×Ó·½³Ìʽ½âÊÍpH²»Ò˹ýСµÄÔÒò£º______¡£VIÖеζ¨ÖÕµãµÄÏÖÏóÊÇ______¡£
(5)ÑùÆ·ÖÐI-×îÖÕËùÏûºÄµÄNa2S2O3µÄÎïÖʵÄÁ¿ÓëÑùÆ·ÖгõʼI-ÎïÖʵÄÁ¿µÄ±ÈÖµ³ÆΪ¡°·Å´ó±¶Êý¡±¡£¾¹ýÒÔÉϲÙ×÷£¬·Å´ó±¶ÊýΪ______¡£
¡¾´ð°¸¡¿3¡Ã1 ·ÖҺ©¶· N2H4+2I2=4I-+N2+4H+ S2O+4I2+10OH-=2SO+8I-+5H2O pH¹ýС£ºS2O+2H+=S+SO2+H2O£»4I-+O2+4H+=2I2+2H2O ÈÜÒºÀ¶É«Ç¡ºÃÏûʧ£¬30Ãë²»»Ö¸´ 36
¡¾½âÎö¡¿
(1)IÖн«º¬ÓÐ΢Á¿I-µÄÑùÆ·ÈÜÒºµ÷ÖÁÖÐÐÔ»òÈõËáÐÔ£¬¼ÓÈëäåË®£¬½«I-ÍêÈ«Ñõ»¯³ÉIO£¬Ñõ»¯¼ÁΪä壬»¹Ô¼ÁΪI-£¬¸ù¾ÝµÃʧµç×ÓÊغã»ò»¯ºÏ¼ÛÉý½µÊغã·ÖÎö¼ÆË㣻
(2)¢óÊǽ«ÈÜÒºÖеÄI2ÝÍÈ¡ÖÁCCl4ÖУ¬¾Ý´Ë·ÖÎö½â´ð£»
(3)¢ôÖÐÏò·ÖÒººóµÄCCl4ÈÜÒºÖмÓÈëëÂ(N2H4)µÄË®ÈÜÒº£¬²úÉúÎÞ¶¾µÄN2£¬Êǵ⽫ëÂÑõ»¯µÄ¹ý³Ì£¬¾Ý´ËÊéд·´Ó¦µÄÀë×Ó·½³Ìʽ£»
(4)¢öÖÐÐèÒªÔÚÖÐÐÔ»òÈõËáÐÔ»·¾³ÖнøÐУ¬pH¹ý´ó£¬Na2S2O3Ò×±»I2Ñõ»¯³ÉNa2SO4£»pH¹ýС£¬S2OÔÚËáÐÔÈÜÒºÖÐÄܹ»·¢ÉúÆ绯·´Ó¦£»ËáÐÔÈÜÒºÖеâÀë×ÓÈÝÒ×±»¿ÕÆøÖеÄÑõÆøÑõ»¯£»¸ù¾Ý·´Ó¦µÄÔÀíÅжϵζ¨ÖÕµãµÄÏÖÏó£»
(5)¸ù¾ÝÉÏÊö²½Ö裬·¢ÉúµÄ·´Ó¦ÎªI-+3Br2+3H2O= IO+6H++6Br-£¬IO+5I-+6H+=3I2+3H2O£¬I2+2S2O=2I-+S4O£¬ÔÚ²½ÖèVÖн«¢ôËùµÃË®ÈÜÒºÖظ´I¡¢¢ò²½Ö裬¾Ý´Ë·ÖÎöÅжÏI-ÎïÖʵÄÁ¿±»·Å´óµÄ±¶Êý¡£
(1)IÖн«º¬ÓÐ΢Á¿I-µÄÑùÆ·ÈÜÒºµ÷ÖÁÖÐÐÔ»òÈõËáÐÔ£¬¼ÓÈëäåË®£¬½«I-ÍêÈ«Ñõ»¯³ÉIO£¬Ñõ»¯¼ÁΪä壬»¹Ô¼ÁΪI-£¬¸ù¾ÝµÃʧµç×ÓÊغã»ò»¯ºÏ¼ÛÉý½µÊغ㣬Ñõ»¯¼ÁºÍ»¹Ô¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈΪ=£¬¹Ê´ð°¸Îª£º£»
(2)¢óÊǽ«ÈÜÒºÖеÄI2ÝÍÈ¡ÖÁCCl4ÖУ¬ÝÍÈ¡ºÍ·ÖÒº¹ý³ÌÖÐʹÓõÄÖ÷Òª²£Á§ÒÇÆ÷ÊÇ·ÖҺ©¶·£¬¹Ê´ð°¸Îª£º·ÖҺ©¶·£»
(3)¢ôÖÐÏò·ÖÒººóµÄCCl4ÈÜÒºÖмÓÈëëÂ(N2H4)µÄË®ÈÜÒº£¬²úÉúÎÞ¶¾µÄN2£¬Êǵ⽫ëÂÑõ»¯µÄ¹ý³Ì£¬·´Ó¦µÄÀë×Ó·½³ÌʽÊÇN2H4+2I2=4I-+N2+4H+£¬¹Ê´ð°¸Îª£ºN2H4+2I2=4I-+N2+4H+£»
(4)¢öÖÐÐèÒªÔÚÖÐÐÔ»òÈõËáÐÔ»·¾³ÖнøÐУ¬pH¹ý´ó£¬Na2S2O3Ò×±»I2Ñõ»¯³ÉNa2SO4£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪS2O+4I2+10OH-=2SO+8I-+5H2O£»pH¹ýС£¬S2OÄܹ»·¢ÉúÆ绯·´Ó¦£¬S2O+2H+=S+SO2+H2O£»ËáÐÔÈÜÒºÖеâÀë×ÓÈÝÒ×±»¿ÕÆøÖеÄÑõÆøÑõ»¯£¬4I-+O2+4H+=2I2+2H2O£¬Òò´ËpHÒ²²»Ò˹ýС£»ÒÔµí·ÛΪָʾ¼Á£¬ÓÃNa2S2O3±ê×¼ÈÜÒºµÎ¶¨I2ÖÁÖյ㣬µÎ¶¨ÖÕµãµÄÏÖÏóΪÈÜÒºÀ¶É«Ç¡ºÃÏûʧ£¬30Ãë²»»Ö¸´£¬¹Ê´ð°¸Îª£ºS2O+4I2+10OH-=2SO+8I-+5H2O£»pH¹ýС£ºS2O+2H+=S+SO2+H2O¡¢4I-+O2+4H+=2I2+2H2O£»ÈÜÒºÀ¶É«Ç¡ºÃÏûʧ£¬30Ãë²»»Ö¸´£»
(5)¸ù¾ÝÉÏÊö²½Ö裬·¢ÉúµÄ·´Ó¦ÎªI-+3Br2+3H2O= IO+6H++6Br-£¬IO+5I-+6H+=3I2+3H2O£¬I2+2S2O=2I-+S4O£¬ÔÚ²½ÖèVÖн«¢ôËùµÃË®ÈÜÒºÖظ´I¡¢¢ò²½Ö裬Òò´ËµÃµ½I-¡«IO3-¡«3I2¡«6I-¡«6IO3-¡«18I2¡«36S2O32-£¬Òò´Ë¾¹ýÒÔÉϲÙ×÷£¬½áºÏµâÔªËØÊغ㣬³õʼI-ÎïÖʵÄÁ¿±»·Å´óÁË36±¶£¬¹Ê´ð°¸Îª£º36¡£
¡¾ÌâÄ¿¡¿Ë®ÌåÉéÎÛȾÒѳÉΪһ¸öؽ´ý½â¾öµÄÈ«ÇòÐÔ»·¾³ÎÊÌ⣬ÎÒ¹ú¿Æѧ¼ÒÑо¿Áã¼ÛÌú»î»¯¹ýÁòËáÄÆ£¨Na2S2O8£©È¥³ý·ÏË®ÖеÄÕýÎå¼ÛÉé[As(¢õ)]£¬Æä»úÖÆÄ£ÐÍÈçͼ¡£
×ÊÁÏ£º
¢ñ.ËáÐÔÌõ¼þÏÂSO ΪÖ÷ÒªµÄ×ÔÓÉ»ù£¬ÖÐÐÔ¼°Èõ¼îÐÔÌõ¼þÏÂSO¡¤ºÍ¡¤OHͬʱ´æÔÚ£¬Ç¿¼îÐÔÌõ¼þÏ¡¤OHΪÖ÷ÒªµÄ×ÔÓÉ»ù¡£
¢ò.Fe2+¡¢Fe3+ÐγÉÇâÑõ»¯Îï³ÁµíµÄpH
Àë×Ó | ¿ªÊ¼³ÁµíµÄpH | ³ÁµíÍêÈ«µÄpH |
Fe2+ | 7.04 | 9.08 |
Fe3+ | 1.87 | 3.27 |
£¨1£©ÉéÓëÁ×ÔÚÔªËØÖÜÆÚ±íÖÐλÓÚͬһÖ÷×壬ÆäÔ×Ó±ÈÁ׶àÒ»¸öµç×Ӳ㡣
¢ÙÉéÔÚÔªËØÖÜÆÚ±íÖеÄλÖÃÊÇ_____¡£
¢ÚÉéËáµÄ»¯Ñ§Ê½ÊÇ_____£¬ÆäËáÐÔ±ÈH3PO4_____£¨Ìî¡°Ç¿¡±»ò¡°Èõ¡±£©¡£
£¨2£©Áã¼ÛÌúÓë¹ýÁòËáÄÆ·´Ó¦£¬¿É³ÖÐøÊÍ·ÅFe2+£¬Fe2+ÓëS2O·´Ó¦Éú³ÉFe3+ºÍ×ÔÓÉ»ù£¬×ÔÓÉ»ù¾ßÓÐÇ¿Ñõ»¯ÐÔ£¬ÀûÓÚÐγÉFe2+ºÍFe3+£¬ÒÔÈ·±£As(¢õ)È¥³ýÍêÈ«¡£
¢ÙÁã¼ÛÌúÓë¹ýÁòËáÄÆ·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_____¡£
¢ÚFe3+ת»¯ÎªFe2+µÄÀë×Ó·½³ÌʽÊÇ_____¡£
¢ÛSO¡¤ºÍH2O·´Ó¦µÄÀë×Ó·½³ÌʽÊÇ_____¡£
£¨3£©²»Í¬pH¶ÔAs(¢õ)È¥³ýÂʵÄÓ°ÏìÈçͼ¡£5minÄÚpH=7ºÍpH=9ʱȥ³ýÂʸߵÄÔÒòÊÇ_____¡£