ÌâÄ¿ÄÚÈÝ

£¨17·Ö£©»¯Ñ§·´Ó¦Ô­ÀíÔÚ¿ÆÑкÍÉú²úÖÐÓй㷺ӦÓá£
£¨1£©Ò»¶¨Ìõ¼þÏ£¬Ä£Äâij¿óʯÐγɵķ´Ó¦aW+bQ¡úcN+dP+eRµÃµ½Á½¸öͼÏñ¡£

¢Ù¸Ã·´Ó¦µÄ¡÷H            0£¨Ìî¡°>¡±¡¢¡°=¡±»ò¡°<¡±£©¡£
¢ÚijζÈÏ£¬Æ½ºâ³£Êý±í´ïʽΪK =c2£¨X£©£¬ÔòÓÉͼ£¨2£©Åж¨X´ú±íµÄÎïÖÊΪ____¡£
£¨2£©½«EºÍF¼ÓÈëÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºE£¨g£©+F£¨s£©2G£¨g£©¡£ºöÂÔ
¹ÌÌåÌå»ý£¬Æ½ºâʱGµÄÌå»ý·ÖÊý£¨%£©ËæÎ¶ȺÍѹǿµÄ±ä»¯ÈçϱíËùʾ£º

ÔòK£¨915¡æ£©ÓëK£¨810¡æ£©µÄ¹ØÏµÎªK£¨915¡æ£©____K£¨810¡æ£©£¨Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©£¬a¡¢b¡¢fÈýÕߵĴóС¹ØÏµÎª       £¬1000¡æ¡¢3£®0 MPaʱEµÄת»¯ÂÊΪ____¡££¨3£©25¡æÊ±£¬H2CO3 HCO3£­+H+µÄµçÀë³£ÊýKa=4¡Á10¡ª7 mo1¡¤L£­1£¬Ôò¸ÃζÈÏ£¬NaHCO3µÄË®½â³£ÊýKh=           £¬ÇëÓÃÊʵ±µÄÊÔ¹ÜʵÑéÖ¤Ã÷Na2CO3ÈÜÒºÖдæÔÚCO32£­+H2O  HCO3£­+OH£­µÄÊÂʵ               ¡£
£¨1£©¢Ù£¾¡£¢ÚR¡££¨2£©´óÓÚ£»f£¾a£¾b£»69.5%¡££¨3£©2.5¡Á10¡ª8£¬È¡ÉÙÁ¿Ì¼ËáÄÆÈÜÒºÖÃÓÚÊÔ¹ÜÖУ¬µÎ¼ÓÂÈ»¯±µÈÜÒºÓа×É«³ÁµíÉú³É£¬Ö¤Ã÷º¬ÓÐ̼Ëá¸ù£»ÁíÈ¡ÉÙÁ¿Ì¼ËáÄÆÈÜÒºÖÃÓÚÊÔ¹ÜÖУ¬µÎ¼Ó·Ó̪ÈÜÒº±äºì£¬Ö¤Ã÷ÓÐÇâÑõ¸ùÉú³É¡£

ÊÔÌâ·ÖÎö£º£¨1£©¢Ù¸ù¾Ýͼ£¨1£©Öª£¬¸Ã·´Ó¦µÄKÕýËæÎ¶ȵÄÉý¸ß¶øÔö´ó£¬Ôò¸Ã·´Ó¦ÕýÏòΪÎüÈÈ·´Ó¦£¬¡÷H£¾0¡£¢Ú¸ù¾Ýͼ£¨2£©Öª£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ2W+Q¡ú3N+P+2R£»Ä³Î¶ÈÏ£¬Æ½ºâ³£Êý±í´ïʽΪK =c2£¨X£©£¬¸ù¾Ýƽºâ³£Êý±í´ïʽµÄÊéдԭÔò£º¹ÌÌå»ò´¿ÒºÌ岻дÈëÆ½ºâ³£Êý±í´ïʽ£¬ÔòÓÉͼ£¨2£©Åж¨X´ú±íµÄÎïÖÊΪR¡£
£¨2£©·´Ó¦£ºE£¨g£©+F£¨s£© 2G£¨g£©ÕýÏòÎªÆøÌåÌå»ý¼õСµÄ·´Ó¦£¬¼Óѹ£¬Æ½ºâÄæÏòÒÆ¶¯£¬GµÄÌå»ý·ÖÊý£¨%£©¼õС£¬½áºÏÌâ¸øÊý¾Ý·ÖÎö£¬ºáÐÐΪѹǿ¶ÔƽºâÓ°ÏìµÄÊý¾Ý£¬¼´54.0£¾a£¾b£»c£¾75.0£¾d£»e£¾f£¾82.0£»¿ÉµÃa¡¢b¡¢fÈýÕߵĴóС¹ØÏµÎªf£¾a£¾b¡£¸ù¾ÝÉÏÊö·ÖÎöÖª£¬e£¾c£¾54.0,±íÖÐÊý¾Ý×ÝÐÐΪÎÂ¶È¶ÔÆ½ºâµÄÓ°Ï죬ÉýΣ¬GµÄÌå»ý·ÖÊý£¨%£©Ôö´ó£¬Æ½ºâÕýÏòÒÆ¶¯£¬Ôò¸Ã·´Ó¦ÕýÏòΪÎüÈÈ·´Ó¦£¬ÔòK£¨915¡æ£©ÓëK£¨810¡æ£©µÄ¹ØÏµÎªK£¨915¡æ£©´óÓÚK£¨810¡æ£©¡£¸ù¾ÝÌâ¸øÊý¾ÝÖª£¬1000¡æ¡¢3£®0 MPaʱGµÄÌå»ý·ÖÊý£¨%£©Îª82.0£¬¸ù¾Ý·´Ó¦£ºE£¨g£©+F£¨s£© 2G£¨g£©ÀûÓÃÈýÐÐʽ¼ÆË㣬ÉèÆðʼ¼ÓÈëµÄEÎïÖʵÄÁ¿Îª1mol£¬×ª»¯µÄΪx¡£
E£¨g£©+F£¨s£© 2G£¨g£©
Æðʼ£¨mol£©£º1    ×ãÁ¿        0
ת»¯£¨mol£©£ºx                 2x
ƽºâ£¨mol£©£º1¡ªx              2x
Ôò2x/1+x=82.0%£¬½âµÃx=0.695£¬EµÄת»¯ÂÊΪ69.5%¡£
£¨3£©25¡æÊ±£¬H2CO3µçÀë³£ÊýKa=c(HCO3¡ª)c(H£«)/c(H2CO3)£¬NaHCO3µÄË®½â³£ÊýKh=c(H2CO3)c(OH£­)/c(HCO3¡ª)£¬ÔòKa¡¤Kh=Kw£¬Kh=Kw/Ka=10¡ª14/4¡Á10¡ª7=2.5¡Á10¡ª8£¬Ö¤Ã÷Na2CO3ÈÜÒºÖдæÔÚ
CO32£­+H2O        HCO3£­+OH£­µÄÊÂʵΪȡÉÙÁ¿Ì¼ËáÄÆÈÜÒºÖÃÓÚÊÔ¹ÜÖУ¬µÎ¼ÓÂÈ»¯±µÈÜÒºÓа×É«³ÁµíÉú³É£¬Ö¤Ã÷º¬ÓÐ̼Ëá¸ù£»ÁíÈ¡ÉÙÁ¿Ì¼ËáÄÆÈÜÒºÖÃÓÚÊÔ¹ÜÖУ¬µÎ¼Ó·Ó̪ÈÜÒº±äºì£¬Ö¤Ã÷ÓÐÇâÑõ¸ùÉú³É¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ijÑо¿ÐÔѧϰС×éΪÁË̽¾¿´×ËáµÄµçÀëÇé¿ö£¬½øÐÐÁËÈçÏÂʵÑé¡£
ʵÑéÒ»¡¡ÅäÖÆ²¢±ê¶¨´×ËáÈÜÒºµÄŨ¶È
È¡±ù´×ËáÅäÖÆ250 mL 0.2 mol¡¤L£­1µÄ´×ËáÈÜÒº£¬ÓÃ0.2 mol¡¤L£­1µÄ´×ËáÈÜҺϡÊͳÉËùÐèŨ¶ÈµÄÈÜÒº£¬ÔÙÓÃNaOH±ê×¼ÈÜÒº¶ÔËùÅä´×ËáÈÜÒºµÄŨ¶È½øÐб궨¡£»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÅäÖÆ250 mL 0.2 mol¡¤L£­1´×ËáÈÜҺʱÐèÒªÓõ½µÄ²£Á§ÒÇÆ÷ÓÐÁ¿Í²¡¢ÉÕ±­¡¢²£Á§°ô¡¢__________________ºÍ______________¡£
(2)Ϊ±ê¶¨Ä³´×ËáÈÜÒºµÄ׼ȷŨ¶È£¬ÓÃ0.200 0 mol¡¤L£­1µÄNaOHÈÜÒº¶Ô20.00 mL´×ËáÈÜÒº½øÐе樣¬¼¸´ÎµÎ¶¨ÏûºÄNaOHÈÜÒºµÄÌå»ýÈçÏ£º
ʵÑéÐòºÅ
1
2
3
4
ÏûºÄNaOHÈÜÒºµÄÌå»ý(mL)
20.05
20.00
18.80
19.95
 
Ôò¸Ã´×ËáÈÜÒºµÄ׼ȷŨ¶ÈΪ____________________¡£(±£ÁôСÊýµãºóËÄλ)
ʵÑé¶þ¡¡Ì½¾¿Å¨¶È¶Ô´×ËáµçÀë³Ì¶ÈµÄÓ°Ïì
ÓÃpH¼Æ²â¶¨25¡æÊ±²»Í¬Å¨¶ÈµÄ´×ËáÈÜÒºµÄpH£¬½á¹ûÈçÏ£º
´×ËáÈÜҺŨ¶È(mol¡¤L£­1)
0.001 0
0.010 0
0.020 0
0.100 0
0.200 0
pH
3.88
3.38
3.23
2.88
2.73
 
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¸ù¾Ý±íÖÐÊý¾Ý£¬¿ÉÒԵóö´×ËáÊÇÈõµç½âÖʵĽáÂÛ£¬ÄãÈÏΪµÃ³ö´Ë½áÂÛµÄÒÀ¾ÝÊÇ______________________________________
(2)´Ó±íÖеÄÊý¾Ý£¬»¹¿ÉÒԵóöÁíÒ»½áÂÛ£ºËæ×Å´×ËáÈÜҺŨ¶ÈµÄ¼õС£¬´×ËáµÄµçÀë³Ì¶È________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)¡£
ʵÑéÈý¡¡Ì½¾¿Î¶ȶԴ×ËáµçÀë³Ì¶ÈµÄÓ°Ïì
ÇëÄãÉè¼ÆÒ»¸öʵÑéÍê³É¸Ã̽¾¿£¬Çë¼òÊöÄãµÄʵÑé·½°¸
________________________________________

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

¾«Ó¢¼Ò½ÌÍø