ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÔÚt¡æʱ£¬AgBrÔÚË®ÖеijÁµíÈܽâƽºâÇúÏßÈçͼËùʾ¡£ÓÖÖªt¡æʱAgClµÄKsp=4¡Á10-10£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨ £©
A£®ÔÚt¡æʱ£¬AgBrµÄKspΪ4.9¡Ál0-13
B£®ÔÚAgBr±¥ºÍÈÜÒºÖмÓÈëNaBr¹ÌÌ壬¿ÉʹÈÜÒºÓÉcµãµ½bµã
C£®Í¼ÖÐaµã¶ÔÓ¦µÄÊÇAgBrµÄ²»±¥ºÍÈÜÒº
D£®ÔÚt¡æʱ£¬AgCl(s)+Br-(aq) AgBr(s)+Cl- (aq)µÄƽºâ³£ÊýK¡Ö816
¡¾´ð°¸¡¿B
¡¾½âÎö¡¿A¡¢¸ù¾ÝÈܶȻýµÄ¶¨Ò壬Ksp=c(Ag£«)¡Ác(Br£)=7¡Á10£7¡Á7¡Á10£7=4.9¡Á10£13£¬¹Ê˵·¨ÕýÈ·£»B¡¢AgBr(s) Ag£«(aq)£«Br£(aq)£¬¼ÓÈëä廯ÄƹÌÌ壬c(Br£)Ôö´ó£¬´ÙʹƽºâÏòÄæ·´Ó¦·½Ïò½øÐУ¬c(Ag£«)¼õС£¬¹Ê˵·¨´íÎó£»C¡¢´ËʱµÄŨ¶ÈÉÌ<Ksp£¬ËµÃ÷´ËÈÜÒºÊôÓÚ²»±¥ºÍÈÜÒº£¬¹Ê˵·¨ÕýÈ·£»D¡¢K=c(Cl£)/c(Br£)=c(Ag£«)¡Ác(Cl£)/[c(Ag£«)¡Ác(Br£)]=Ksp(AgCl)/Ksp(AgBr)=4¡Á10£10/4.9¡Á10£13¡Ö816£¬¹Ê˵·¨ÕýÈ·¡£
![](http://thumb2018.1010pic.com/images/loading.gif)
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿