£¨1£©25¡æÊ±£¬Å¨¶ÈΪ0.1 mol¡¤L£­1µÄ6ÖÖÈÜÒº£º¢ÙHCl£¬ ¢ÚCH3OOH£¬ ¢ÛBa(OH)2£¬¢ÜNa 2CO3£¬¢ÝKCl£¬¢ÞNH4ClÈÜÒºpHÓÉСµ½´óµÄ˳ÐòΪ__________________(Ìîд±àºÅ)¡£

£¨2£©25¡æÊ±£¬´×ËáµÄµçÀë³£ÊýKa=1.7¡Á10-5mol/L£¬Ôò¸ÃζÈÏÂCH3COONaµÄË®½âƽºâ³£ÊýKh=          mol ¡¤L-1£¨±£Áôµ½Ð¡Êýµãºóһ룩¡£

£¨3£©25¡æÊ±£¬pH£½3µÄ´×ËáºÍpH£½11µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒº³Ê          £¨Ìî¡°ËáÐÔ¡±£¬¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£© £¬Çëд³öÈÜÒºÖÐÀë×ÓŨ¶È¼äµÄÒ»¸öµÈʽ£º

                                                                         ¡£

£¨4£©25¡æÊ±£¬½«m mol/LµÄ´×ËáºÍn mol/LµÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpH£½7£¬

ÔòÈÜÒºÖÐc(CH3COO£­) + c(CH3COOH)=       £¬mÓënµÄ´óС¹ØÏµÊÇ£í      £î£¨Ìî¡° >¡±¡°£½¡±»ò¡°<¡± £©¡£

£¨5£©25¡æÊ±£¬½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ´×ËáÓ백ˮ»ìºÏºó£¬ÈÜÒºµÄpH£½7 £¬ÔòNH3¡¤H2OµÄµçÀë³£ÊýKa=               ¡£

   (1)  ¢Ù¢Ú¢Þ¢Ý¢Ü¢Û

    (2)  5.9¡Á10-10

(3) ËáÐÔ         c(Na£«) + c(H£«) = c(CH3COO£­) + c(OH£­)¡£

(4)  m/2 mol/L             >

(5)  1.7¡Á10-5mol/L

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©25¡æÊ±£¬Å¨¶ÈΪ0.1mol?L-1µÄ6ÖÖÈÜÒº£º¢ÙHCl£¬¢ÚCH3OOH£¬¢ÛBa£¨OH£©2£¬¢ÜNa 2CO3£¬¢ÝKCl£¬
¢ÞNH4ClÈÜÒºpHÓÉСµ½´óµÄ˳ÐòΪ
 
£¨Ìîд±àºÅ£©£®
£¨2£©25¡æÊ±£¬´×ËáµÄµçÀë³£ÊýKa=1.7¡Á10-5mol/L£¬Ôò¸ÃζÈÏÂCH3COONaµÄË®½âƽºâ³£ÊýKh=
 
mol?L-1£¨±£Áôµ½Ð¡Êýµãºóһ룩£®
£¨3£©25¡æÊ±£¬pH=3µÄ´×ËáºÍpH=11µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒº³Ê
 
£¨Ìî¡°ËáÐÔ¡±£¬¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£¬Çëд³öÈÜÒºÖÐÀë×ÓŨ¶È¼äµÄÒ»¸öµÈʽ£º
 
£®
£¨4£©25¡æÊ±£¬½«m mol/LµÄ´×ËáºÍn mol/LµÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpH=7£¬ÔòÈÜÒºÖÐ
c£¨CH3COO-£©+c£¨CH3COOH£©=
 
£¬mÓënµÄ´óС¹ØÏµÊÇm
 
n£¨Ìî¡°£¾¡±¡°=¡±»ò¡°£¼¡±£©£®
£¨5£©25¡æÊ±£¬½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ´×ËáÓ백ˮ»ìºÏºó£¬ÈÜÒºµÄpH=7£¬ÔòNH3?H2OµÄµçÀë³£ÊýKa=
 
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2015½ìɽ¶«Ê¡Î«·»Êи߶þÉÏѧÆÚÆÚĩͳ¿¼»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©25¡æÊ±£¬Å¨¶ÈΪ0.1 mol¡¤L£­1µÄ6ÖÖÈÜÒº£º¢ÙHCl£¬ ¢ÚCH3OOH£¬ ¢ÛBa(OH)2£¬¢ÜNa 2CO3£¬¢ÝKCl£¬¢ÞNH4ClÈÜÒºpHÓÉСµ½´óµÄ˳ÐòΪ__________________(Ìîд±àºÅ)¡£

£¨2£©25¡æÊ±£¬´×ËáµÄµçÀë³£ÊýKa=1.7¡Á10-5mol/L£¬Ôò¸ÃζÈÏÂCH3COONaµÄË®½âƽºâ³£ÊýKh=          mol ¡¤L-1£¨±£Áôµ½Ð¡Êýµãºóһ룩¡£

£¨3£©25¡æÊ±£¬pH£½3µÄ´×ËáºÍpH£½11µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒº³Ê          £¨Ìî¡°ËáÐÔ¡±£¬¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£© £¬Çëд³öÈÜÒºÖÐÀë×ÓŨ¶È¼äµÄÒ»¸öµÈʽ£º                                 ¡£

£¨4£©25¡æÊ±£¬½«m mol/LµÄ´×ËáºÍn mol/LµÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºµÄpH£½7£¬

ÔòÈÜÒºÖÐc(CH3COO£­) + c(CH3COOH)=       £¬mÓënµÄ´óС¹ØÏµÊÇ£í      £î£¨Ìî¡° £¾¡±¡°£½¡±»ò¡°<¡± £©¡£

£¨5£©25¡æÊ±£¬½«µÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ´×ËáÓ백ˮ»ìºÏºó£¬ÈÜÒºµÄpH£½7 £¬ÔòNH3¡¤H2OµÄµçÀë³£ÊýKa=               ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ÒÑÖª´×ËáºÍÑÎËáÊÇÈÕ³£Éú»îÖм«Îª³£¼ûµÄËᣬÔÚÒ»¶¨Ìõ¼þÏ£¬CH3COOHÈÜÒºÖдæÔÚµçÀëÆ½ºâ£ºCH3COOH  CH3COO£­+H+ £»¦¤H>0

£¨1£©25 ¡æÊ±£¬Å¨¶È¾ùΪ0.1mol/LµÄÑÎËáºÍ´×ËáÈÜÒº£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ                 £»

¢ÙÁ½ÈÜÒºµÄpHÏàͬ           

¢ÚÁ½ÈÜÒºµÄµ¼µçÄÜÁ¦Ïàͬ   

¢ÛÓÉË®µçÀë³öµÄc(OH£­)Ïàͬ    

¢ÜÖк͵ÈÎïÖʵÄÁ¿µÄNaOHÈÜÒº£¬ÏûºÄÁ½ÈÜÒºµÄÌå»ýÏàͬ

£¨2£©25 ¡æÊ±£¬ÏòpH¾ùΪ1µÄÑÎËáºÍ´×ËáÈÜÒºÖзֱð¼ÓË®£¬Ëæ¼ÓË®

Á¿µÄÔö¶à£¬Á½ÈÜÒºpHµÄ±ä»¯ÈçͼËùʾ£¬Ôò·ûºÏÑÎËápH±ä»¯

µÄÇúÏßÊÇ               £»

£¨3£©25 ¡æÊ±£¬ÏòÌå»ýΪVa mL pH=3µÄ´×ËáÈÜÒºÖеμÓpH=11µÄ

NaOHÈÜÒºVb mLÖÁÈÜҺǡºÃ³ÊÖÐÐÔ£¬ÔòVa             Vb£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©

¼ÙÉèÖʼ첿ÃŹ涨ÊÐÊÛ´×ËáŨ¶È²»µÃµÍÓÚ4.8g/100mL£¬Ä³Í¬Ñ§ÓûÓÃÖк͵ζ¨µÄ·½·¨À´²â¶¨Ä³Æ·ÅƵÄʳÓô×ÖеĴ×ËẬÁ¿ÊÇ·ñ´ï±ê¡£ÊµÑé¾ßÌå²½ÖèÈçÏ£º¢ÙÍÐÅÌÌìÆ½³ÆÈ¡Ò»¶¨ÖÊÁ¿NaOH²¢ÅäÖÆ³É500mL NaOHÈÜÒº£»¢ÚÓÃÒÑ֪Ũ¶ÈµÄÑÎËá±ê×¼ÈÜҺ׼ȷ±ê¶¨¸ÃNaOHÈÜÒºµÄŨ¶È£»¢ÛÓÃÉÏÊöÒÑ֪׼ȷŨ¶ÈµÄNaOHÈÜÒº²â¶¨´×ËáµÄŨ¶È¡£

£¨4£©²»Ö±½ÓÓÃÅäÖõÄNaOHÈÜÒºµÎ¶¨ÑùÆ·£¬¶øÒªÓñê×¼ÑÎËáÏȱ궨Ôٵ樵ÄÔ­ÒòÊÇ                                                                          £»

£¨5£©ÈôʵÑé¹ý³ÌÈçÏ£º×¼È·Á¿È¡¸ÃʳÓô×20.00mL£¬ÖÃÓÚ250mL×¶ÐÎÆ¿ÖУ¬ÔٵμӷÓָ̪ʾ¼Á£¬Óñ궨ºÃµÄ0.1000mol/LµÄNaOHÈÜÒºµÎ¶¨£¬·Óָ̪ʾ¼ÁÓÉ        ɫǡºÃ±ä³É__________É«ÇÒ              ¼´ÎªÖյ㡣

ÖØ¸´µÎ¶¨¶à´Î£¬½á¹û¼Ç¼ÈçÏ£º

²â¶¨´ÎÐò

µÚÒ»´Î

µÚ¶þ´Î

µÚÈý´Î

µÚËÄ´Î

V£¨mL£©

19.40

15.10

14.90

15.00

Ôò¸ÃʳÓô×Öд×ËáµÄÎïÖʵÄÁ¿Å¨¶È£½________ mol¡¤L£­1£¬ÊÇ·ñºÏ¸ñ        £¨ÌÊÇ¡±»ò¡±·ñ¡±£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª´×ËáºÍÑÎËáÊÇÈÕ³£Éú»îÖм«Îª³£¼ûµÄËᣬÔÚÒ»¶¨Ìõ¼þÏÂ,CH3COOHÈÜÒºÖдæÔÚµçÀëÆ½ºâ£ºCH3COOH  CH3COO-+H+  ¦¤H>0¡£

£¨1£©25 ¡æÊ±£¬Å¨¶È¾ùΪ0.1mol/LµÄÑÎËáºÍ´×ËáÈÜÒº£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ                 £»

¢ÙÁ½ÈÜÒºµÄpHÏàͬ            

¢ÚÁ½ÈÜÒºµÄµ¼µçÄÜÁ¦Ïàͬ   

¢ÛÓÉË®µçÀë³öµÄc(OH-)Ïàͬ     

¢ÜÖк͵ÈÎïÖʵÄÁ¿µÄNaOHÈÜÒº£¬ÏûºÄÁ½ÈÜÒºµÄÌå»ýÏàͬ

£¨2£©25 ¡æÊ±£¬ÔÚ pH =5µÄÏ¡´×ËáÈÜÒºÖУ¬c(CH3COO-)=                       (ÌîÊý×Ö±í´ïʽ)£»

£¨3£©25 ¡æÊ±£¬ÏòpH¾ùΪ1µÄÑÎËáºÍ´×ËáÈÜÒºÖзֱð¼ÓË®£¬Ëæ¼ÓË®

Á¿µÄÔö¶à£¬Á½ÈÜÒºpHµÄ±ä»¯ÈçͼËùʾ£¬Ôò·ûºÏÑÎËápH±ä»¯

µÄÇúÏßÊÇ               £»

£¨4£©25 ¡æÊ±£¬ÏòÌå»ýΪVa mLpH=3µÄ´×ËáÈÜÒºÖеμÓpH=11µÄ

NaOHÈÜÒºVb mLÖÁÈÜҺǡºÃ³ÊÖÐÐÔ£¬ÔòVaÓëµÄVb¹ØÏµÊÇ

Va       Vb£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£

£¨5£©25 ¡æÊ±£¬ÈôÏò°±Ë®ÖмÓÈëÏ¡ÑÎËáÖÁÈÜÒºµÄpH£½7£¬´Ëʱ[NH4£«]£½a mol/L£¬

Ôò[Cl£­]£½________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸