ÌâÄ¿ÄÚÈÝ

£¨1£©³£ÎÂÏ£¬½«0.1mol?L-1ÑÎËáÈÜÒººÍ0.06mol?L-1ÇâÑõ»¯±µÈÜÒºµÈÌå»ý»ìºÏºó£¬¸Ã»ìºÏÈÜÒºµÄpH=
12
12
£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©
£¨2£©³£ÎÂÏ£¬pH=aµÄ10Ìå»ýµÄijǿËáÓëpH=bµÄ1Ìå»ýµÄijǿ¼î»ìºÏºó£¬ÈÜÒº³ÊÖÐÐÔ£¬ÔòaºÍbÂú×ãµÄ¹ØÏµ
a+b=15
a+b=15
£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©
£¨3£©Ä³Î¶Èʱ£¬Ë®µÄÀë×Ó»ýKW=1¡Á10-13£¬Ôò¸ÃζÈ
£¾
£¾
25¡æ£¨Ñ¡Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£¬Èô½«´ËζÈÏÂpH=2µÄÏ¡ÁòËáaLÓëpH=12µÄÇâÑõ»¯ÄÆÈÜÒºbL»ìºÏ£¬ÈôËùµÃ»ìºÏÒºµÄpH=11£¬Ôòa£ºb=
9£º11
9£º11
£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©
£¨4£©³£ÎÂÏ£¬pH=8µÄNH4ClÈÜÒºÖУ¬c£¨Cl-£©-c£¨NH4+£©=
9.9¡Á10-7mol/L
9.9¡Á10-7mol/L
£®
·ÖÎö£º£¨1£©0.06mol?L-1ÇâÑõ»¯±µÈÜÒºc£¨OH-£©=0.12mol/L£¬¶þÕßµÈÌå»ý»ìºÏÈÜÒº³Ê¼îÐÔ£¬ÏȼÆËã»ìºÏÈÜÒºÖÐc£¨OH-£©£¬ÔÙ½áºÏÀë×Ó»ý³£Êý¼ÆËã»ìºÏÈÜÒºÖÐc£¨H+£©£¬´Ó¶øÈ·¶¨»ìºÏÈÜÒºµÄpH£»
£¨2£©Ç¿ËáºÍÇ¿¼î»ìºÏÈÜÒº³ÊÖÐÐÔ£¬ËµÃ÷ËáÖÐc£¨H+£©µÈÓÚ¼îÖÐc£¨OH-£©£»
£¨3£©Ë®µÄµçÀëÊÇÎüÈÈ·´Ó¦£¬Éý¸ßζȴٽøË®µçÀ룬ˮµÄÀë×Ó»ý³£ÊýÔö´ó£¬¸ù¾Ý»ìºÏÈÜÒºÖÐc£¨OH-£©=
n(OH-)-n(H+)
V(Ëá)+V(¼î)
¼ÆË㣻
£¨4£©¸ù¾ÝµçºÉÊØºã¼ÆËãc£¨Cl-£©-c£¨NH4+£©£®
½â´ð£º½â£º£¨1£©ÉèËáºÍ¼îµÄÌå»ý¶¼ÊÇ1L£¬0.06mol?L-1ÇâÑõ»¯±µÈÜÒºc£¨OH-£©=0.12mol/L£¬0.1mol?L-1ÑÎËáÈÜÒºÖÐÇâÀë×ÓŨ¶ÈÊÇ0.1mol/L£¬
¶þÕßµÈÌå»ý»ìºÏÈÜÒº³Ê¼îÐÔ£¬»ìºÏÈÜÒºÖÐc£¨OH-£©=
0.12-0.1
2
mol/L
=0.01mol/L£¬»ìºÏÈÜÒºÖÐÇâÀë×ÓŨ¶ÈΪ10-12 mol/L£¬ËùÒÔ»ìºÏÈÜÒºÖÐpH=12£¬
¹Ê´ð°¸Îª£º12£»
£¨2£©Ç¿ËáºÍÇ¿¼î»ìºÏÈÜÒº³ÊÖÐÐÔ£¬ËµÃ÷ËáÖÐc£¨H+£©µÈÓÚ¼îÖÐc£¨OH-£©£¬ËùÒÔ10¡Á10-a=1¡Á10b-14£¬ËùÒÔa+b=15£¬
¹Ê´ð°¸Îª£ºa+b=15£»
£¨3£©Ë®µÄµçÀëÊÇÎüÈÈ·´Ó¦£¬Éý¸ßζȴٽøË®µçÀ룬ˮµÄÀë×Ó»ý³£ÊýÔö´ó£¬Ä³Î¶Èʱ£¬Ë®µÄÀë×Ó»ýKW=1¡Á10-13£¾¡Á10-14£¬ËùÒÔÔò¸Ãζȣ¾25¡æ£»»ìºÏÈÜÒºÖÐÇâÀë×ÓŨ¶ÈÊÇ10-11 mol/L£¬c£¨OH-£©=10-3 mol/L£¬
»ìºÏÈÜÒºÖÐc£¨OH-£©=
n(OH-)-n(H+)
V(Ëá)+V(¼î)
=
10-2mol/L¡Á(b-a)L
(a+b)L
=10-3 mol/L£¬a£ºb=9£º11£¬
¹Ê´ð°¸Îª£º9£º11£»
£¨4£©¸ù¾ÝµçºÉÊØºã¼ÆËãc£¨Cl-£©-c£¨NH4+£©=c£¨H+£©-c£¨OH-£©=10-6mol/L-10-8 mol/L=9.9¡Á10-7mol/L£¬
¹Ê´ð°¸Îª£º9.9¡Á10-7mol/L£®
µãÆÀ£º±¾Ì⿼²éÁËpHµÄ¼òµ¥¼ÆË㣬Ã÷È·»ù±¾µÄpH¼ÆË㹫ʽÊǽⱾÌâ¹Ø¼ü£¬ÄѵãÊÇ£¨4£©ÖÐÂÈÀë×ÓºÍ笠ùÀë×ÓŨ¶ÈµÄ²î£¬¸ù¾ÝµçºÉÊØºãÀ´·ÖÎö½â´ð¼´¿É£¬ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨Ñ¡×öÌ⣩£¨10·Ö£©ÔÚº¬ÓÐÈõµç½âÖʵÄÈÜÒºÖУ¬ÍùÍùÓжà¸ö»¯Ñ§Æ½ºâ¹²´æ¡£
£¨1£©³£ÎÂÏ£¬½«0.2mol/LµÄijһԪËáHAÈÜÒººÍ0.1mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒºpH´óÓÚ7£¬Èô»ìºÏÒºÌå»ýµÈÓÚÁ½ÈÜÒºÌå»ýÖ®ºÍ£¬Ôò»ìºÏÒºÖÐÏÂÁйØÏµÕýÈ·µÄÊÇ____         
A£®c(HA)£¼c(A£­)                      B£®c(HA)Ò»¶¨´óÓÚ0.1mol/L
C£®c(Na+)£½c(HA)+c(A£­)               D£®2c£¨OH£­£©£½2c(H+)£«[c(HA)£­c(A£­)]
£¨2£©³£ÎÂÏÂÔÚ20mL0.1mol/L Na2CO3ÈÜÒºÖÐÖðµÎ¼ÓÈë0.1mol/L HClÈÜÒº40mL£¬ÈÜÒºÖк¬Ì¼ÔªËصĸ÷ÖÖ΢Á££¨CO2Òݳöδ»­£©ÎïÖʵÄÁ¿·ÖÊý£¨×ÝÖá£©ËæÈÜÒºpH±ä»¯µÄ²¿·ÖÇé¿öÈçͼËùʾ¡£

»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÔÚͬһÈÜÒºÖУ¬H2CO3¡¢HCO3£­¡¢CO32£­£¨Ì¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©         ´óÁ¿¹²´æ¡£
¢Úµ±pH=7ʱ£¬ÈÜÒºÖи÷ÖÖÀë×ÓÆäÎïÖʵÄÁ¿Å¨¶ÈµÄ´óС¹ØÏµÊÇ£º                  
                                                  ¡£
¢ÛÒÑÖªÔÚ25¡æÊ±£¬CO32£­Ë®½â·´Ó¦µÄƽºâ³£Êý¼´Ë®½â³£ÊýKh=  =2¡Á10£­4£¬µ±ÈÜÒºÖÐc(HCO3£­)©Uc(CO32£­)=2©U1ʱ£¬ÈÜÒºµÄpH=            ¡£

£¨Ñ¡×öÌ⣩£¨10·Ö£©ÔÚº¬ÓÐÈõµç½âÖʵÄÈÜÒºÖУ¬ÍùÍùÓжà¸ö»¯Ñ§Æ½ºâ¹²´æ¡£

£¨1£©³£ÎÂÏ£¬½«0.2mol/LµÄijһԪËáHAÈÜÒººÍ0.1mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒºpH´óÓÚ7£¬Èô»ìºÏÒºÌå»ýµÈÓÚÁ½ÈÜÒºÌå»ýÖ®ºÍ£¬Ôò»ìºÏÒºÖÐÏÂÁйØÏµÕýÈ·µÄÊÇ____         

A£®c(HA)£¼c(A£­)                      B£®c(HA)Ò»¶¨´óÓÚ0.1mol/L

C£®c(Na+)£½c(HA)+c(A£­)                D£®2c£¨OH£­£©£½2c(H+)£«[c(HA)£­c(A£­)]

£¨2£©³£ÎÂÏÂÔÚ20mL0.1mol/L Na2CO3ÈÜÒºÖÐÖðµÎ¼ÓÈë0.1mol/L HClÈÜÒº40mL£¬ÈÜÒºÖк¬Ì¼ÔªËصĸ÷ÖÖ΢Á££¨CO2Òݳöδ»­£©ÎïÖʵÄÁ¿·ÖÊý£¨×ÝÖá£©ËæÈÜÒºpH±ä»¯µÄ²¿·ÖÇé¿öÈçͼËùʾ¡£

»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙÔÚͬһÈÜÒºÖУ¬H2CO3¡¢HCO3£­¡¢CO32£­£¨Ì¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©          ´óÁ¿¹²´æ¡£

¢Úµ±pH=7ʱ£¬ÈÜÒºÖи÷ÖÖÀë×ÓÆäÎïÖʵÄÁ¿Å¨¶ÈµÄ´óС¹ØÏµÊÇ£º                  

                                                   ¡£

¢ÛÒÑÖªÔÚ25¡æÊ±£¬CO32£­Ë®½â·´Ó¦µÄƽºâ³£Êý¼´Ë®½â³£ÊýKh=  =2¡Á10£­4£¬µ±ÈÜÒºÖÐc(HCO3£­)©Uc(CO32£­)=2©U1ʱ£¬ÈÜÒºµÄpH=             ¡£

 

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

¾«Ó¢¼Ò½ÌÍø