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£©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£©
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
£¨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´ïʽµÄÊéдÔÔò£º¹ÌÌå»ò´¿ÒºÌ岻дÈëÆ½ºâ³£Êý±í´ïʽ£¬ÔòÓÉͼ£¨2£©Åж¨X´ú±íµÄÎïÖÊΪR¡£
£¨2£©·´Ó¦£ºE£¨g£©+F£¨s£©
E£¨g£©+F£¨s£©
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
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿