ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©³£ÎÂÏ£¬½«Ä³Ò»ÔªËáHAºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºµÄpHÈçÏÂ±í£º

ʵÑé±àºÅ
HAÎïÖʵÄÁ¿Å¨¶È(mol¡¤L-1)
NaOHÎïÖʵÄÁ¿Å¨¶È(mol¡¤L-1)
»ìºÏÈÜÒºµÄpH
¢Ù
0.1
0.1
pH£½9
¢Ú
c
0.2
pH£½7
¢Û
0.2
0.1
pH£¼7
Çë»Ø´ð£º
(1) ´Ó¢Ù×éÇé¿ö·ÖÎö£¬ HAÊÇ             ËᣨѡÌî¡°Ç¿¡±¡¢¡°Èõ¡±£©¡£
(2) ¢Ú×éÇé¿ö±íÃ÷£¬c      0.2 (Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±)¡£
(3) ´Ó¢Û×éʵÑé½á¹û·ÖÎö£¬ËµÃ÷HAµÄµçÀë³Ì¶È______NaAµÄË®½â³Ì¶È(Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±)£¬¸Ã»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ                          ¡£
(4) ¢Ù×éʵÑéËùµÃ»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc(OH£­)£½              mol¡¤L£­1¡£

£¨¹²10·Ö£©
(1)  HAÊÇÈõË᣻(2) £¾£»  (3) £¾  c(A£­) £¾c(Na£«) £¾c(H£«)£¾c(OH£­)  (4)  10 -5

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
³£ÎÂÏ£¬½«Ä³Ò»ÔªËáHAºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄŨ¶ÈºÍ»ìºÏºóËùµÃÈÜÒºµÄpHÈçÏÂ±í£¬Çë»Ø´ð£º
ʵÑé±àºÅ HAÎïÖʵÄÁ¿Å¨¶È
£¨mol?L-1£©
NaOHÎïÖʵÄÁ¿Å¨¶È
£¨mol?L-1£©
»ìºÏÈÜÒºµÄpH
¼× 0.2 0.2 pH=a
ÒÒ c 0.2 pH=7
±û 0.2 0.1 pH£¾7
¶¡ 0.1 0.1 pH=9
£¨1£©²»¿¼ÂÇÆäËü×éµÄʵÑé½á¹û£¬µ¥´Ó¼××éÇé¿ö·ÖÎö£¬ÈçºÎÓÃa £¨»ìºÏÈÜÒºµÄpH£©À´ËµÃ÷HAÊÇÇ¿ËỹÊÇÈõËá
a=7ΪǿËᣬa£¾7ΪÈõËá
a=7ΪǿËᣬa£¾7ΪÈõËá
£®
£¨2£©²»¿¼ÂÇÆäËü×éµÄʵÑé½á¹û£¬µ¥´ÓÒÒ×éÇé¿ö·ÖÎö£¬CÊÇ·ñÒ»¶¨µÈÓÚ0.2
·ñ
·ñ
£¨Ñ¡Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©£®»ìºÏÒºÖÐÀë×ÓŨ¶Èc£¨A-£©Óë c£¨Na+£©µÄ´óС¹ØÏµÊÇ
c£¨A-£©=c£¨Na+£©
c£¨A-£©=c£¨Na+£©
£®
£¨3£©±û×éʵÑé½á¹û·ÖÎö£¬HAÊÇ
Èõ
Èõ
ËᣨѡÌî¡°Ç¿¡±»ò¡°Èõ¡±£©£®¸Ã»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨Na+£©£¾c£¨A-£©£¾c£¨OH-£©£¾c£¨H+£©
c£¨Na+£©£¾c£¨A-£©£¾c£¨OH-£©£¾c£¨H+£©
£®
£¨4£©¶¡×éʵÑéËùµÃ»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨OH-£©=
10-5
10-5
mol?L-1£®Ð´³ö¸Ã»ìºÏÈÜÒºÖÐÏÂÁÐËãʽµÄ¾«È·½á¹û£¨²»Ðè×ö½üËÆ¼ÆË㣩£®c£¨Na+£©-c£¨A-£©=
10-5-10-9
10-5-10-9
mol?L-1      c£¨OH-£©-c£¨HA£©=
10-9
10-9
mol?L-1£®
£¨¢ñ£©³£ÎÂÏ£¬½«Ä³Ò»ÔªËáHA£¨¼×¡¢ÒÒ¡¢±û¡¢¶¡´ú±í²»Í¬µÄÒ»ÔªËᣩºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈºÍ»ìºÏÈÜÒºµÄpHÈçϱíËùʾ£º
ʵÑé
񅧏
HAÎïÖʵÄÁ¿
Ũ¶È/£¨mol?L-1£©
NaOHÎïÖʵÄÁ¿
Ũ¶È/£¨mol?L-1£©
»ìºÏºóÈÜ
񼵀pH
¼× 0.1 0.1 pH=a
ÒÒ 0.12 0.1 pH=7
±û 0.2 0.1 pH£¾7
¶¡ 0.1 0.1 pH=10
£¨1£©´Ó¼××éÇé¿ö·ÖÎö£¬ÈçºÎÅжÏHAÊÇÇ¿ËỹÊÇÈõË᣿
a=7ʱ£¬HAÊÇÇ¿Ë᣻a£¾7ʱ£¬HAÊÇÈõËá
a=7ʱ£¬HAÊÇÇ¿Ë᣻a£¾7ʱ£¬HAÊÇÈõËá
£®
£¨2£©ÒÒ×é»ìºÏÈÜÒºÖÐÁ£×ÓŨ¶Èc£¨A-£©ºÍc£¨Na+£©µÄ´óС¹ØÏµ
C
C
£®
A£®Ç°Õß´ó  B£®ºóÕß´ó
C£®Á½ÕßÏàµÈ  D£®ÎÞ·¨ÅжÏ
£¨3£©´Ó±û×éʵÑé½á¹û·ÖÎö£¬¸Ã»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨Na+£©£¾c£¨A-£©£¾c£¨OH-£©£¾c£¨H+£©
c£¨Na+£©£¾c£¨A-£©£¾c£¨OH-£©£¾c£¨H+£©
£®
£¨4£©·ÖÎö¶¡×éʵÑéÊý¾Ý£¬Ð´³ö»ìºÏÈÜÒºÖÐÏÂÁÐËãʽµÄ¾«È·½á¹û£¨ÁÐʽ£©£º
c£¨Na+£©-c£¨A-£©=
10-4-10-10
10-4-10-10
mol?L-1£®
£¨¢ò£©Ä³¶þÔªËᣨ»¯Ñ§Ê½ÓÃH2B±íʾ£©ÔÚË®ÖеĵçÀë·½³ÌʽÊÇ£ºH2B¨TH++HB-¡¢HB-?H++B2-
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨5£©ÔÚ0.1mol?L-1µÄNa2BÈÜÒºÖУ¬ÏÂÁÐÁ£×ÓŨ¶È¹ØÏµÊ½ÕýÈ·µÄÊÇ
AC
AC
£®
A£®c£¨B2-£©+c£¨HB-£©=0.1mol?L-1
B£®c£¨B2-£©+c£¨HB-£©+c£¨H2B£©=0.1mol?L-1
C£®c£¨OH-£©=c£¨H+£©+c£¨HB-£©
D£®c£¨Na+£©+c£¨H+£©=c£¨OH-£©+c£¨HB-£©

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

¾«Ó¢¼Ò½ÌÍø