ÌâÄ¿ÄÚÈÝ

ÏÂÁÐÐðÊö´íÎóµÄÊÇ (¡¡¡¡)

A£®°Ña L 0.1 mol/L µÄCH3COOHÈÜÒºÓëb L 0.1 mol/LµÄ KOHÈÜÒº»ìºÏ£¬ËùµÃÈÜ
ÒºÖÐÒ»¶¨´æÔÚ£ºc (K+)+ c (H+) =" c" (CH3COO£­) + c (OH£­)
B£®ÂÈË®ÖУºc(Cl£­)£¾c(H+)£¾c(OH£­)£¾c(ClO£­)
C£®°Ñ0.1 mol/L µÄNaHCO3ÈÜÒºÓë0.3 mol/L µÄBa(OH)2ÈÜÒºµÈÌå»ý»ìºÏ£¬ËùµÃÈÜ
ÒºÖÐÒ»¶¨´æÔÚ£ºc (OH£­) >c (Ba2+)>c (Na+)> c (H£«)
D£®³£ÎÂÏ£¬ÔÚpH=3µÄCH3COOHÈÜÒººÍpH=11µÄNaOHÈÜÒºÖУ¬Ë®µÄµçÀë³Ì¶ÈÏàͬ

B

½âÎöÊÔÌâ·ÖÎö£ºA¡¢µçºÉÊغã,ÕýÈ·£»B¡¢c(ClO£­) £¾c(OH£­)£¬OH£­ÊÇË®µçÀë²úÉú£¬ºÜС£¬´íÎó£»C¡¢HCO3£­+OH-= CO32-+ H2O£¬Ba2++ CO32-=" Ba" CO3¡ý, c(OH£­)="0.5" ,c (Ba2+)="0.2," c (Na+)="0.1," c (H£«)ÓÉË®µçÀë²úÉúµÄ£¬ºÜС£¬ËùÒÔc (OH£­) >c (Ba2+)>c (Na+)> c (H£«)£¬ÕýÈ·£»D¡¢CH3COOHÈÜÒºc (H+)µÈÓÚpH=11µÄNaOHÈÜÒºµÄc(OH£­)£¬Îª10-3£¬Kw =10-14 £¬ÁíÍâµÄH+  ºÍOH£­ ÏàµÈ£¬ÊÇË®µçÀë²úÉúµÄ£¬ÕýÈ·¡£
¿¼µã£º¿¼²éÈÜÒºÖеÄÀë×ÓŨ¶È±È½ÏµÈÏà¹Ø֪ʶ¡£ 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø