ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨   £©
A£®pH=2ºÍpH=lµÄÏõËáÖÐc£¨H+£©Ö®±ÈΪ10£º1
B£®Ò»ÔªËáÓëÒ»Ôª¼îÇ¡ºÃÍêÈ«·´Ó¦ºóµÄÈÜÒºÖÐÒ»¶¨´æÔÚc£¨H+£©=c£¨OH-£©
C£®KAl£¨SO4£©2ÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ´óС˳ÐòΪc£¨SO2-4£©>c£¨K+£©>c£¨Al3+£©>c£¨H+£©>c£¨OH-£©
D£®³£ÎÂÏ£¬½«pH¾ùΪ12µÄ°±Ë®ºÍNaOHÈÜÒº·Ö±ð¼ÓˮϡÊÍ100±¶ºó£¬NaOHÈÜÒºµÄpH½Ï´ó
C
A:pH=2ºÍpH=lµÄÏõËáÖÐc£¨H+£©Ö®±ÈΪ1£º10£¬Åųý
B£ºËá¼îÖкͺóÈÜÒº²»Ò»¶¨³ÊÖÐÐÔ£¬c£¨H+£©Óëc£¨OH-£©²»Ò»¶¨ÏàµÈ£¬²»ÕýÈ·
C£ºÓÉÓÚÂÁÀë×Ó²¿·ÖË®½â£¬ÈÜÒº³ÊËáÐÔ£¬ÕýÈ·
D£ºÓÉÓÚÈõ¼î´æÔÚµçÀëƽºâ£¬ÔÚÏ¡Ê͹ý³ÌÖпÉÒÔ´Ù½øÆäµçÀ룬¹ÊÏ¡ÊÍÏàͬ±¶Êýºó£¬°±Ë®µÄpH½Ï´ó£¬²»ÕýÈ·
´ð°¸ÎªC
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂʱ£¬½«V1mL c1 mol/LµÄ°±Ë®µÎ¼Óµ½V2mL c2 mol/LµÄÑÎËáÖУ¬ÏÂÁÐÕýÈ·µÄÊÇ    
A£®Èô»ìºÏÈÜÒºµÄpH=7£¬ÔòÈÜÒºÖÐc £¨NH4+£©£¾ c £¨Cl¡ª£©
B£®ÈôV1=V2£¬c1=c2£¬ÔòÈÜÒºÖÐÒ»¶¨´æÔÚc£¨H+£©=c £¨OH¡ª£©£¬³ÊÖÐÐÔ
C£®Èô»ìºÏÈÜÒºµÄpH=7£¬ÔòÒ»¶¨´æÔÚc1V1£¾c2V2 ¹Øϵ
D£®ÈôV1=V2£¬²¢ÇÒ»ìºÏÒºµÄpH£¼7£¬ÔòÒ»¶¨ÊÇÓÉÓÚÑÎËá¹ýÁ¿¶øÔì³ÉµÄ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

½«0£®20mol¡¤L£­1µÄNH3¡¤H2OÈÜÒººÍ0£®10mol¡¤L£­1µÄHClÈÜÒºµÈÌå»ý»ìºÏ£¬³ä·Ö·´Ó¦ºó£¬ÏÂÁÐÓйظÃÈÜÒºÖи÷Á£×ÓŨ¶ÈµÄ¹Øϵ²»ÕýÈ·µÄÊÇ                £¨   £©
A£®c£¨NH4+£©£¾c£¨Cl£­£©£¾c£¨H+£©£¾c£¨OH£­£©
B£®c£¨Na+£©£«c£¨H+£©£½c£¨Cl£­£©£«c£¨OH£­£©
C£®c£¨NH4+£©£«c£¨NH3¡¤H2O£©£½0£®10mol¡¤L£­1
D£®c£¨NH4+£©£¾c£¨Cl£­£©£¾c£¨OH£­£©£¾c£¨H+£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÖÓÐÎïÖʵÄÁ¿Å¨¶ÈÏàͬµÄ7ÖÖÈÜÒº£º(a)NaHSO4£»(b)NaHCO3£»(c)H2SO4£»(d)Al2(SO4)3£»(e)Na2CO3£»(f)Na2SO4£»(g)Ba(OH)2£»pHÓÉСµ½´óµÄ˳ÐòΪ£¨   £©
A£®(c)(a)(b)(d)(f)(e)(g)B£®(c)(a)(d)(f)(b)(e)(g)
C£®(g)(e)(b)(f)(d)(a)(c)D£®(c)(a)(d)(f)(e)(b)(g)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂÁÐÊÇËáÈÜÒºÓë¼îÈÜÒº·¢Éú·´Ó¦µÄ¸÷ÖÖ¹Øϵ£¬ÇëÓá°>¡±¡°<¡±»ò¡°=¡±Ìî¿Õ£º 
(1) È¡0.2 mol/LHXÈÜÒºÓë0.2 mol/L NaOHÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃ»ìºÏÈÜÒºÖÐc(Na+)>c(X£­)¡£
¢Ù»ìºÏÈÜÒºÖÐc(HX)      c(X£­)£»   
¢Ú»ìºÏÈÜÒºÖÐc(HX) + c(X£­)        0.1 mol/L£¨ºöÂÔÈÜÒºÌå»ý±ä»¯£©£»    
¢Û»ìºÏÈÜÒºÖÐÓÉË®µçÀë³öµÄc(OH£­)      0.1 mol/L µÄHXÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)¡£
(2) Èç¹ûÈ¡0.2 mol/LHXÈÜÒºÓë0.1 mol/LNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬²âµÃ»ìºÏÈÜÒºµÄPH>7£¬Ôò˵Ã÷ÔÚÏàͬŨ¶ÈʱHXµÄµçÀë³Ì¶È       NaXµÄË®½â³Ì¶È¡£
(3) ·Ö±ðÈ¡pH£½2µÄÁòËáºÍ´×Ëá¸÷50mL£¬¼ÓÈë×ãÁ¿µÄþ·Û£¬·´Ó¦¿ªÊ¼Ê±£¬·´Ó¦ËÙÂÊÇ°Õß_________ºóÕߣ»³ä·Ö·´Ó¦ºó£¬ÏàͬÌõ¼þϲúÉúÇâÆøµÄÌå»ýÇ°Õß_________ºóÕß¡£
(4) µÈŨ¶ÈµÈÌå»ýµÄÑÎËáºÍ´×Ëᣬ·Ö±ðÓëµÈŨ¶ÈµÄNaOHÈÜÒº·´Ó¦£¬Ç¡ºÃÖкÍʱ£¬ÏûºÄNaOHµÄÌå»ýÇ°Õß__________ºóÕߣ¬ËùµÃÈÜÒºµÄpHÇ°Õß________ºóÕß¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

³£ÎÂÏ£¬ÓÐŨ¶È¾ùΪ0.1 mol/LµÄ4ÖÖÈÜÒº£º¢ÙÑÎË᣻¢ÚÁòË᣻¢Û´×Ë᣻
¢ÜÇâÑõ»¯±µ¡££¨×¢£ººöÂÔÈÜÒº»ìºÏʱÌå»ýµÄ±ä»¯£©
£¨1£©½«ÉÏÊöÈÜÒºÁ½Á½µÈÌå»ý»ìºÏºó£¬ÈÜÒºpH=7µÄ×éºÏÓУ¨ÌîÐòºÅ£©           ¡£
£¨2£©´×Ëá±µÊÇÒ×ÈÜÓÚË®µÄÇ¿µç½âÖÊ¡£½«¢ÛÓë¢ÜµÈÌå»ý»ìºÏ£¬ËùµÃÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨6·Ö£©£¨1£©ÒÑÖªÌìÈ»ÆøºÍË®ÕôÆø·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCH4(g) + H2O(g)  CO(g)+ 3H2(g)  ¦¤H £½+206.4kJ¡¤mol-1£¬ÔòÖƵÃ56g COÐèÒªÎüÊÕµÄÈÈÁ¿Îª              kJ£»
£¨2£©CO¿É¼ÌÐøÓëË®ÕôÆø·´Ó¦£ºCO(g)+H2O(g)CO2(g)+H2(g) ¦¤H £½£­41.0kJ¡¤mol£­1 £¬Èô½«1mol CH4Óë×ãÁ¿Ë®ÕôÆø³ä·Ö·´Ó¦µÃµ½1molCO2£¬¸Ã·´Ó¦µÄìʱ䦤H £½      kJ¡¤mol£­1¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÓйصç½âÖÊÈÜÒºµÄÍÆÀíÖÐÕýÈ·µÄÊÇ
A£®ÊÒÎÂÏ£¬pH£½4µÄ´×ËáºÍpH£½10µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏ£¬ÆäpH> 7
B£®pH£½3µÄÑÎËáÓëpH£½11µÄ°±Ë®µÈÌå»ý»ìºÏºó£¬ÈÜÒºÖУº
c£¨NH4+£©£¾c£¨Cl-£©£¾c£¨OH-£©£¾c£¨H+£©
C£®ÎïÖʵÄÁ¿Å¨¶ÈÏàµÈµÄNH4ClÈÜÒººÍNH4HSO4ÈÜÒºÖУ¬NH4+Ũ¶ÈÏàµÈ
D£®ÔÚÓÐAgI³ÁµíµÄÉÏÇåÒºÖеμÓNaClÈÜÒº£¬¿ÉÒԹ۲쵽°×É«³Áµí²úÉú

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬ÔÚ10 mL 0£®2 mol£¯LijһԪǿËáHAµÄÈÜÒºÖÐÖðµÎ¼ÓÈëa mL 0£®1 mol£¯LµÄ°±Ë®£¬ÏÂÁÐÓйØ˵·¨ÖÐÕýÈ·µÄÊÇ
A£®µ±c£¨£©£½c£¨A£­£©Ê±£¬aÒ»¶¨µÈÓÚ20
B£®µ±c£¨£©>c£¨A£­£©Ê±£¬aÒ»¶¨´óÓÚ20
C£®µ±c£¨H£«£©>c£¨OH£­£©Ê±£¬aÒ»¶¨Ð¡ÓÚ20
D£®µ±pH£½7ʱ£¬c£¨£©£½c£¨A£­£©£½c£¨H£«£©£½c£¨OH£­£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸