ÌâÄ¿ÄÚÈÝ

16£®K¡¢Ka»òKb¡¢KW¡¢Ksp·Ö±ð±íʾ»¯Ñ§Æ½ºâ³£Êý¡¢µçÀë³£Êý¡¢Ë®µÄÀë×Ó»ý¡¢ÈܶȻý³£Êý£¬ÏÂÁйØÓÚÕâЩ³£ÊýµÄ˵·¨ÖУ¬ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®»¯Ñ§Æ½ºâ³£ÊýµÄ´óСÓëζȡ¢Å¨¶È¡¢Ñ¹Ç¿Óйأ¬Óë´ß»¯¼ÁÎÞ¹Ø
B£®ÒÑÖª£ºK1 £¨H2CO3£©£¾Ka £¨HClO£©£¾K2£¨H2CO3£©£¬ÏòNaClOÈÜÒºÖÐͨÈëÉÙÁ¿CO2µÄ»¯Ñ§·½³ÌÊÇ£º2NaClO+CO2+H2O=Na2CO3+2HClO
C£®25¡æʱ£¬pH=4µÄÑÎËáÖУ¬KW=10-20
D£®³£ÎÂÏ£¬Ksp£¨CaSO4£©=9¡Á10-6£¬Ïò100mL±¥ºÍCaSO4ÈÜÒºÖмÓ400mL 0.01mol/LNa2SO4ÈÜÒº£¬ÎÞ³ÁµíÎö³ö

·ÖÎö A£®»¯Ñ§Æ½ºâ³£ÊýÖ»ÓëζÈÓйأ»
B£®ÒÑÖªK1£¨H2CO3£©£¾Ka£¨HClO£©£¾K2£¨H2CO3£©£¬ËùÒÔËáÐÔH2CO3£¾HClO£¾HCO3-£¬¹Ê²úÎïÖ»ÄÜÊÇ̼ËáÇâÄÆ£»
C£®Ë®µÄÀë×Ó»ýÖ»ÓëζÈÓйأ¬25¡æʱ£¬KW=10-14£»
D£®CaSO4±¥ºÍÈÜÒºc£¨Ca2+£©=c£¨SO42-£©=3¡Á10-3mol/L£¬¼ÓÈë400mL 0.01mol/LNa2SO4ÈÜÒººó£¬c£¨Ca2+£©=0.6¡Á10-3mol/L£¬c£¨SO42-£©=8.6¡Á10-3mol/L£¬±È½ÏQcÓëKsp£®

½â´ð ½â£ºA£®»¯Ñ§Æ½ºâ³£ÊýÖ»ÓëζÈÓйأ¬ÓëŨ¶È¡¢Ñ¹Ç¿¡¢´ß»¯¼ÁÎ޹أ¬¹ÊA´íÎó£»
B£®ËáÐÔH2CO3£¾HClO£¾HCO3-£¬ÉÙÁ¿CO2ͨÈë´ÎÂÈËáÄÆÈÜÒºÖз´Ó¦Éú³É̼ËáÇâÄƺʹÎÂÈËᣬ»¯Ñ§·½³ÌʽΪ£ºCO2+H2O+NaClO¨TNaHCO3+HClO£¬¹ÊB´íÎó£»
C£®Ë®µÄÀë×Ó»ýÖ»ÓëζÈÓйأ¬25¡æʱ£¬KW=10-14£¬pH=4µÄÑÎËáÖУ¬KWÈÔȻΪ10-14£¬¹ÊC´íÎó£»
D£®CaSO4±¥ºÍÈÜÒºc£¨Ca2+£©=c£¨SO42-£©=3¡Á10-3mol/L£¬¼ÓÈë400mL 0.01mol/LNa2SO4ÈÜÒººó£¬»ìºÏÒºÖÐc£¨Ca2+£©=$\frac{0.1L¡Á3.0¡Á1{0}^{-3}mol/L}{0.1L+0.4L}$=6.0¡Á10-4 mol/L£¬c£¨SO42-£©=$\frac{0.1L¡Á3.0¡Á1{0}^{-3}mol/L+0.4L¡Á0.01mol/L}{0.1L+0.4L}$=8.6¡Á10-3 mol/L£¬
ÈÜÒºÖÐc£¨Ca2+£©•c£¨ SO42- £©=5.16¡Á10-6£¼Ksp£¨CaSO4£©=9.0¡Á10-6£¬ËùÒÔ»ìºÏÒºÖÐÎÞ³ÁµíÎö³ö£¬¹ÊDÕýÈ·£®
¹ÊÑ¡£ºD£®

µãÆÀ ±¾Ì⿼²é»¯Ñ§Æ½ºâ³£Êý¡¢µçÀë³£Êý¡¢ÈܶȻýµÈ£¬ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕ¸÷ÖÖ³£Êý±í´ïʽµÄÊéдÓëÓ°ÏìÒòËØ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
8£®Ä³Ð¡×éÀûÓÃÈçͼװÖã¬Óñ½ÓëäåÔÚFeBr3´ß»¯×÷ÓÃÏÂÖƱ¸äå±½£º

·´Ó¦¾çÁÒ½øÐУ¬ÉÕÆ¿ÖÐÓдóÁ¿ºì×ØÉ«ÕôÆø£¬×¶ÐÎÆ¿Öе¼¹Ü¿ÚÓа×Îí³öÏÖ£¬ÕôÁóË®Öð½¥±ä³É»ÆÉ«£®·´Ó¦Í£Ö¹ºó°´ ÈçÏÂÁ÷³Ì·ÖÀë²úÆ·£º

ÒÑÖª£º
 ·Ðµã¡æÃܶÈg/cm3ÈܽâÐÔ
äå593.119Ë®ÖÐÈܽâ¶ÈС£¬Ò×ÈÜÓÚÓлúÈܼÁ
±½800.8765ÄÑÈÜÓÚË®£¬ÓëÓлúÈܼÁ»¥ÈÜ
äå±½1561.50ÄÑÈÜÓÚË®£¬ÓëÓлúÈܼÁ»¥ÈÜ
£¨1£©²Ù×÷¢ñΪ¹ýÂË£¬²Ù×÷¢òΪÕôÁó£® 
£¨2£©¡°Ë®Ï´¡±¡¢¡°NaOHÈÜҺϴ¡±ÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷ÊÇ·ÖҺ©¶·¡¢ÉÕ±­£® 
£¨3£©¡°Ë®Ï´¡±Ö÷ҪĿµÄÊdzýÈ¥FeBr3£¬¡°NaOHÈÜҺϴ¡±Ö÷ҪĿµÄÊdzýÈ¥Br2£®
£¨4£©×¶ÐÎÆ¿ÖÐÕôÁóË®±ä»ÆµÄÔ­ÒòÊÇÈܽâÁË´ÓÉÕÆ¿Öлӷ¢³öµÄä壮 
£¨5£©ÒÑÖª±½Óëäå·¢ÉúµÄÊÇÈ¡´ú·´Ó¦£¬ÍƲⷴӦºó׶ÐÎÆ¿ÖÐÒºÌ庬ÓеÄÁ½ÖÖ´óÁ¿Àë×ÓÊÇBr-¡¢H+£¬ÇëÉè¼ÆʵÑé·½°¸ÑéÖ¤ÄãµÄÍƲ⣮£¨ÏÞÑ¡ÊÔ¼Á£ºÃ¾Ìõ¡¢ËÄÂÈ»¯Ì¼¡¢ÂÈË®¡¢äåË®¡¢ÕôÁóË®£©
ÐòºÅʵÑé²½ÖèÔ¤ÆÚÏÖÏó½áÂÛ
1½«×¶ÐÎÆ¿ÖÐÒºÌåתÈë·ÖҺ©¶·£¬¼ÓÈëÊÊÁ¿ËÄÂÈ»¯Ì¼£¬Õñµ´ºó·ÖÒº£®·Ö±ðÈ¡ÉÙÁ¿ÉϲãÎÞÉ«ÈÜÒºÓÚÊÔ¹ÜA¡¢BÖР 
2  ×¶ÐÎÆ¿ÖÐÒºÌ庬´óÁ¿
Br-
3  ×¶ÐÎÆ¿ÖÐÒºÌ庬´óÁ¿
H+ 

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

¾«Ó¢¼Ò½ÌÍø