ÌâÄ¿ÄÚÈÝ

ÒÑÖª£ºCu(OH)2ÊǶþÔªÈõ¼î£»ÑÇÁ×ËᣨH3PO3£©ÊǶþÔªÈõËᣬÓëNaOHÈÜÒº·´Ó¦£¬Éú³ÉNa2HPO3¡£
£¨1£©ÔÚÍ­ÑÎÈÜÒºÖÐCu2£«·¢ÉúË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ____£¬¸Ã·´Ó¦µÄƽºâ³£ÊýΪ____£»£¨ÒÑÖª£º25¡æʱ£¬Ksp[Cu(OH)2]£½2.0¡Á10£­20mol3/L3£©
£¨2£©¸ù¾ÝH3PO3µÄÐÔÖÊ¿ÉÍƲâNa2HPO3Ï¡ÈÜÒºµÄpH______7£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°£½¡±£©¡£³£ÎÂÏ£¬Ïò10mL0.01mol/L H3PO3ÈÜÒºÖеμÓ10ml0.02mol/LNaOHÈÜÒººó£¬ÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ_________£»
£¨3£©µç½âNa2HPO3ÈÜÒº¿ÉµÃµ½ÑÇÁ×ËᣬװÖÃÈçͼ£¨ËµÃ÷£ºÑôĤֻÔÊÐíÑôÀë×Óͨ¹ý£¬ÒõĤֻÔÊÐíÒõÀë×Óͨ¹ý£©

¢ÙÑô¼«µÄµç¼«·´Ó¦Ê½Îª____________________¡£
¢Ú²úÆ·ÊÒÖз´Ó¦µÄÀë×Ó·½³ÌʽΪ____________¡£
£¨13·Ö£©£¨1£©Cu2+£«2H2O Cu(OH)2£«2H+£¨2·Ö£©£»5¡Á10£­9£¨2·Ö£©
£¨2£©£¾£¨2·Ö£©£»c(Na+)£¾c(HPO32£­)£¾c(OH£­)£¾c(H2PO3£­)£¾c(H+)£¨2·Ö£©
£¨3£©¢Ù4OH¨D¨D4e£­£½2H2O£«O2¡ü£¨2·Ö£©
¢ÚHPO32£­£«2H£«£½H3PO3£¨2·Ö£©»òHPO32£­£«H£«£½H2PO3£­¡¢H2PO3£­£«H£«£½H3PO3£¨¸÷1·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©ÇâÑõ»¯Í­ÊÇÈõ¼î£¬ËùÒÔÍ­Àë×Ó¿ÉÒÔË®½â£¬ÈÜÒºÏÔËáÐÔ£¬ÆäË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪCu2+£«2H2O Cu(OH)2£«2H+¡£»¯Ñ§Æ½ºâ³£ÊýÊÇÔÚÒ»¶¨Ìõ¼þÏ£¬µ±¿ÉÄæ·´Ó¦´ïµ½Æ½ºâ״̬ʱ£¬Éú³ÉÎïŨ¶ÈµÄÃÝÖ®»ýºÍ·´Ó¦ÎïŨ¶ÈµÄÃÝÖ®»ýµÄ±ÈÖµ£¬ËùÒԸ÷´Ó¦µÄƽºâ³£ÊýK£½£½=£½£½5¡Á10£­9¡£
£¨2£©H3PO3ÊÇÈõËᣬNa2HPO3ÊÇÇ¿¼îÈõËáÑΣ¬ËùÒÔHPO32£­Ë®½â£¬ÆäË®ÈÜÒº³Ê¼îÐÔ£¬¼´pH£¾7£»Ïò10mL0.01mol/LH3PO3ÈÜÒºÖеμÓ10ml 0.02mol/LNaOHÈÜÒººó£¬¶þÕßÇ¡ºÃ·´Ó¦Éú³ÉNa2HPO3£¬ÈÜҺˮ½âÏÔ¼îÐÔ£¬ËùÒÔÈÜÒºÖÐÀë×ÓŨ¶È´óСΪc(Na+)£¾c(HPO32£­)£¾c(OH£­)£¾c(H2PO3£­)£¾c(H+)¡£
£¨3£©¢Ùµç½â³ØÑô¼«Ê§È¥µç×Ó£¬·¢ÉúÑõ»¯·´Ó¦£¬Òõ¼«µÃµ½µç×Ó·¢Éú»¹Ô­·´Ó¦¡£ËùÒÔ¸ù¾Ý×°ÖÃͼ¿ÉÖª£¬Ñô¼«ÉÏÇâÑõ¸ùÀë×Óʧµç×Ó·¢ÉúÑõ»¯·´Ó¦£¬µç¼«·´Ó¦Ê½Îª4OH¨D¨D4e£­£½2H2O£«O2¡ü¡£
¢ÚÓÉÓÚÑôĤֻÔÊÐíÑôÀë×Óͨ¹ý£¬ÒõĤֻÔÊÐíÒõÀë×Óͨ¹ý£¬ËùÒÔ²úÆ·ÊÒÖÐHPO32£­ºÍÇâÀë×Ó½áºÏÉú³ÉÑÇÁ×Ëᣬ·´Ó¦Àë×Ó·½³ÌʽΪHPO32£­£«2H£«£½H3PO3£¬»òHPO32£­£«H£«£½H2PO3£­¡¢H2PO3£­£«H£«£½H3PO3¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø