ÑÇ‘zËᣨH3PO3£©ÊǶþÔªËᣬÓë×ãÁ¿NaOHÈÜÒº·´Ó¦Éú³ÉNa2HPO3£®
£¨1£©PCl3Ë®½â¿ÉÖÆÈ¡ÑÇÁ×Ë᣺PCl3+32O¨TH3PO3+______£®
£¨2£©H3PO3ÈÜÒºÖдæÔÚµçÀëƽºâ£ºH3PO3?H++H2PO3-£®
¢ÙijζÈÏ£¬O.1Omol£®L-1 µÄ H3PO3 ÈÜÒº pH=1.6£¬¼´ÈÜÒºÖРc£¨H+£©=2.5x 10-2 mol£®L-1£®Çó¸ÃζÈÏÂÉÏÊöµçÀëƽºâµÄƽºâ³£ÊýK£¬Ð´³ö¼ÆËã¹ý³Ì£®
£¨H3PO3µÄµÚ¶þ²½µçÀëºöÂÔ²»¼Æ£¬½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö£®£©
¢Ú¸ù¾ÝH3PO3µÄÐÔÖÊ¿ÉÍƲâNa2HPO3Ï¡ÈÜÒºµÄpH______7 £¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£®
£¨3£©ÑÇÁ×Ëá¾ßÓÐÇ¿»¹Ô­ÐÔ£¬¿ÉʹµâË®ÍÊÉ«£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______
£¨4£©µç½âNa2HPO3ÈÜÒºÒ²¿ÉµÃµ½ÑÇ÷[ËᣬװÖÃʾÒâͼÈçͼ£º
¢ÙÒõ¼«µÄµç¼«·´Ó¦Ê½Îª______
¢Ú²úÆ·ÊÒÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®

¡¾´ð°¸¡¿·ÖÎö£º£¨1£©¸ù¾ÝË®½â·´Ó¦·½³Ìʽȷ¶¨Éú³ÉÎ
£¨2£©¢ÙµçÀëƽºâ³£ÊýK=£»
¢Ú¸ù¾ÝH3PO3µÄËáÐÔÇ¿ÈõÈ·¶¨Na2HPO3ÈÜÒºµÄËá¼îÐÔ£»
£¨3£©ÑÇÁ×ËáºÍµâ·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬ÑÇÁ×Ëá×÷»¹Ô­¼Á±»Ñõ»¯Éú³ÉÁ×Ëᣬµâ±»»¹Ô­Éú³ÉÇâµâËᣬ¾Ý´Ëд³ö·´Ó¦·½³Ìʽ£»
£¨4£©¢ÙÒõ¼«Éϵõç×Ó·¢Éú»¹Ô­·´Ó¦£»
¢Ú²úÆ·ÊÒÖÐHPO32-ºÍÇâÀë×Ó½áºÏ Éú³ÉÑÇÁ×Ëᣮ
½â´ð£º½â£º£¨1£©PCl3Ë®½â¿ÉÖÆÈ¡ÑÇÁ×ËáºÍÑÎËᣬˮ½â·½³ÌʽΪ£ºPCl3+32O?H3PO3+3HCl£¬¹Ê´ð°¸Îª£ºHCl£»
£¨2£©¢ÙH3PO3=H++H2PO3-
ÆðʼŨ¶È         0.10  0   0
·´Ó¦Å¨¶È  2.5×10-2 2.5×10-2 2.5×10-2
ƽºâŨ¶È0.10-2.5×10-2 2.5×10-2 2.5×10-2
µçÀëƽºâ³£ÊýK==mol/L=8.3×10-3mol/L£¬¹Ê´ð°¸Îª8.3×10-3mol/L£»
¢ÚH3PO3ÊÇÈõËᣬNa2HPO3ÊÇÇ¿¼îÈõËáÑΣ¬ËùÒÔÆäË®ÈÜÒº³Ê¼îÐÔ£¬¼´pH£¾7£¬¹Ê´ð°¸Îª£º£¾£»
£¨3£©ÑÇÁ×ËáºÍµâ·¢ÉúÑõ»¯»¹Ô­·´Ó¦£¬ÑÇÁ×Ëá×÷»¹Ô­¼Á±»Ñõ»¯Éú³ÉÁ×Ëᣬµâ±»»¹Ô­Éú³ÉÇâµâËᣬ·´Ó¦·½³ÌʽΪ£ºH3PO3+I2+H2O=2HI+H3PO4£¬¹Ê´ð°¸Îª£ºH3PO3+I2+H2O=2HI+H3PO4£»
£¨4£©¢ÙÒõ¼«ÉÏÇâÀë×ӵõç×Ó·¢Éú»¹Ô­·´Ó¦£¬µç¼«·´Ó¦Ê½Îª2H++2e-=H2¡ü£¬¹Ê´ð°¸Îª£º2H++2e-=H2¡ü£»
¢Ú²úÆ·ÊÒÖÐHPO32-ºÍÇâÀë×Ó½áºÏÉú³ÉÑÇÁ×Ëᣬ·´Ó¦Àë×Ó·½³ÌʽΪ£ºHPO32-+2H+=H3PO3£¬¹Ê´ð°¸Îª£ºHPO32-+2H+=H3PO3£®
µãÆÀ£º±¾ÌâÉæ¼°Ë®½â·´Ó¦¡¢Ñõ»¯»¹Ô­·´Ó¦¡¢µç¼«·´Ó¦Ê½µÄÊéдµÈ֪ʶµã£¬µç¼«·´Ó¦Ê½µÄÊéд¡¢ÓйØƽºâ³£ÊýµÄ¼ÆËãÊǸ߿¼Èȵ㣬ӦÖصãÕÆÎÕ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2013?¹ãÖÝһģ£©ÑÇ‘zËᣨH3PO3£©ÊǶþÔªËᣬÓë×ãÁ¿NaOHÈÜÒº·´Ó¦Éú³ÉNa2HPO3£®
£¨1£©PCl3Ë®½â¿ÉÖÆÈ¡ÑÇÁ×Ë᣺PCl3+3H2O¨TH3PO3+
3HCl
3HCl
£®
£¨2£©H3PO3ÈÜÒºÖдæÔÚµçÀëƽºâ£ºH3PO3?H++H2PO3-£®
¢ÙijζÈÏ£¬0.10mol?L-1 µÄ H3PO3 ÈÜÒº pH=1.6£¬¼´ÈÜÒºÖРc£¨H+£©=2.5¡Á10-2 mol?L-1£®Çó¸ÃζÈÏÂÉÏÊöµçÀëƽºâµÄƽºâ³£ÊýK£¬Ð´³ö¼ÆËã¹ý³Ì£®£¨H3PO3µÄµÚ¶þ²½µçÀëºöÂÔ²»¼Æ£¬½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö£®£©
¢Ú¸ù¾ÝH3PO3µÄÐÔÖÊ¿ÉÍƲâNa2HPO3Ï¡ÈÜÒºµÄpH
£¾
£¾
7 £¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£®
£¨3£©ÑÇÁ×Ëá¾ßÓÐÇ¿»¹Ô­ÐÔ£¬¿ÉʹµâË®ÍÊÉ«£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
H3PO3+I2+H2O=2HI+H3PO4
H3PO3+I2+H2O=2HI+H3PO4

£¨4£©µç½âNa2HPO3ÈÜÒºÒ²¿ÉµÃµ½ÑÇ÷[ËᣬװÖÃʾÒâͼÈçͼ£º
¢ÙÒõ¼«µÄµç¼«·´Ó¦Ê½Îª
2H++2e-=H2¡ü
2H++2e-=H2¡ü

¢Ú²úÆ·ÊÒÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ
HPO32-+2H+=H3PO3
HPO32-+2H+=H3PO3
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012-2013ѧÄê¹ã¶«Ê¡¹ãÖÝÊбÏÒµ°à×ۺϲâÊÔ£¨Ò»£©Àí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£º¼ÆËãÌâ

ÑÇ‘zËᣨH3PO3)ÊǶþÔªËᣬÓë×ãÁ¿NaOHÈÜÒº·´Ó¦Éú³ÉNa2HPO3¡£

£¨1£©PCl3Ë®½â¿ÉÖÆÈ¡ÑÇÁ×Ë᣺PCl3+3H2O=H3PO3+_______¡£

£¨2£©H3PO3ÈÜÒºÖдæÔÚµçÀëƽºâ£ºH3PO3H++H2PO3£­¡£

¢ÙijζÈÏ£¬0.10mol•L£­1 µÄ H3PO3 ÈÜÒº pH =1.6£¬¼´ÈÜÒºÖÐ c(H+) =2.5¡Á10£­2 mol•L£­1¡£Çó¸ÃζÈÏÂÉÏÊöµçÀëƽºâµÄƽºâ³£ÊýK£¬Ð´³ö¼ÆËã¹ý³Ì¡££¨H3PO3µÄµÚ¶þ²½µçÀëºöÂÔ²»¼Æ£¬½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö¡££©

¢Ú¸ù¾ÝH3PO3µÄÐÔÖÊ¿ÉÍƲâNa2HPO3Ï¡ÈÜÒºµÄpH________7 (Ìî¡°>¡±¡¢¡°=¡±»ò¡°<¡±£©¡£

£¨3£©ÑÇÁ×Ëá¾ßÓÐÇ¿»¹Ô­ÐÔ£¬¿ÉʹµâË®ÍÊÉ«¡£¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______¡£

£¨4£©µç½âNa2HPO3ÈÜÒºÒ²¿ÉµÃµ½ÑÇ÷[ËᣬװÖÃʾÒâͼÈçÏ£º

¢ÙÒõ¼«µÄµç¼«·´Ó¦Ê½Îª____________________________¡£

¢Ú²úÆ·ÊÒÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ_____________________¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º¹ãÖÝһģ ÌâÐÍ£ºÎÊ´ðÌâ

ÑÇ‘zËᣨH3PO3£©ÊǶþÔªËᣬÓë×ãÁ¿NaOHÈÜÒº·´Ó¦Éú³ÉNa2HPO3£®
£¨1£©PCl3Ë®½â¿ÉÖÆÈ¡ÑÇÁ×Ë᣺PCl3+32O¨TH3PO3+______£®
£¨2£©H3PO3ÈÜÒºÖдæÔÚµçÀëƽºâ£ºH3PO3?H++H2PO3-£®
¢ÙijζÈÏ£¬O.1Omol£®L-1 µÄ H3PO3 ÈÜÒº pH=1.6£¬¼´ÈÜÒºÖРc£¨H+£©=2.5x 10-2 mol£®L-1£®Çó¸ÃζÈÏÂÉÏÊöµçÀëƽºâµÄƽºâ³£ÊýK£¬Ð´³ö¼ÆËã¹ý³Ì£®
£¨H3PO3µÄµÚ¶þ²½µçÀëºöÂÔ²»¼Æ£¬½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö£®£©
¢Ú¸ù¾ÝH3PO3µÄÐÔÖÊ¿ÉÍƲâNa2HPO3Ï¡ÈÜÒºµÄpH______7 £¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£®
£¨3£©ÑÇÁ×Ëá¾ßÓÐÇ¿»¹Ô­ÐÔ£¬¿ÉʹµâË®ÍÊÉ«£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ______
£¨4£©µç½âNa2HPO3ÈÜÒºÒ²¿ÉµÃµ½ÑÇ÷[ËᣬװÖÃʾÒâͼÈçͼ£º
¢ÙÒõ¼«µÄµç¼«·´Ó¦Ê½Îª______
¢Ú²úÆ·ÊÒÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ______£®
¾«Ó¢¼Ò½ÌÍø

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸