20¡æʱÏò20mL 0.1mol¡¤L£­1´×ËáÈÜÒºÖв»¶ÏµÎÈë0.1mol¡¤L£­1NaOH(aq)£¬ÈÜÒºpH±ä»¯ÈçͼËùʾ¡£´Ë¹ý³ÌÀïÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ¹Øϵ´íÎóµÄÊÇ
  

A£®aµã£ºc(Na+)£¾c(CH3COO£­)£¾c(H+)£¾c(OH£­
B£®bµã£ºc(Na+) = c(CH3COO£­)£¾c(H+) = c(OH£­
C£®cµã£ºc(H+) = c(CH3COOH) + c(OH£­
D£®dµã£ºc(Na+)£¾c(CH3COO£­)£¾c(OH£­)£¾c(H+

CD

½âÎöÊÔÌâ·ÖÎö£ºA£®aµãʱ´×Ëá¹ýÁ¿£¬ÈÜҺΪCH3COOHºÍCH3COONaµÄ»ìºÏÎÈÜÒº³ÊËáÐÔ£¬ËùÒÔc£¨H+£©£¾c£¨OH-£©£¬ÈÜÒº³ÊµçÖÐÐÔ£¬c£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨Cl-£©+c£¨OH-£©£¬ËùÒÔc£¨Na+£©£¾c£¨CH3COO-£©£¬¹ÊAÕýÈ·£» B£®¸ù¾ÝÈÜÒºµçºÉÊغã¿ÉÖªÈÜÒºÖÐÓ¦´æÔÚc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬ÈÜÒº³ÊÖÐÐÔ£¬Ó¦ÓÐc£¨H+£©=c£¨OH-£©£¬Ôòc£¨Na+£©=c£¨CH3COO-£©£¬¹ÊBÕýÈ·£» C£®cµãʱ£¬ÈÜÒº³Ê¼îÐÔ£¬Ó¦ÓÐc£¨H+£©£¼c£¨OH-£©£¬¹ÊC´íÎó£» D£®dµãΪNaOHºÍCH3COONaµÄ»ìºÏÎÈÜÒº³Ê¼îÐÔ£¬ÓÉÓÚCH3COO-´æÔÚ΢ÈõµÄË®½â£¬ÔòÓÐc£¨Na+£©£¾c£¨OH-£©£¾c£¨CH3COO-£©£¾c£¨H+£©£¬¹ÊD´íÎó£®¹ÊÑ¡CD£®
¿¼µã£º±¾Ì⿼²éËá¼î»ìºÏµÄÅжϺÍÀë×ÓŨ¶È´óС±È½Ï£¬´ðÌâʱעÒâa¡¢b¡¢c¡¢dµãÈÜÒºµÄ×é³É£¬°ÑÎÕÈõµç½âÖʵĵçÀëºÍÑÎÀàË®½âµÄÌص㣬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏòÈý·Ý¾ùΪ100mL0.5mol¡¤L-1NaHCO3ÈÜÒºÖУ¬·Ö±ð¼ÓÈëÉÙÁ¿±ù´×Ëá¡¢Ca(OH)2¹ÌÌå¡¢NaAlO2¹ÌÌ壨ºöÂÔÈÜÒºÌå»ý±ä»¯£©£¬ÔòÈý·ÝÈÜÒºÖÐc(CO32£­)µÄ±ä»¯ÒÀ´ÎΪ

A£®¼õС¡¢¼õС¡¢Ôö´óB£®¼õС¡¢Ôö´ó¡¢Ôö´ó
C£®Ôö´ó¡¢¼õС¡¢¼õСD£®¼õС¡¢Ôö´ó¡¢¼õС

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

Ò»¶¨Î¶ÈÏ£¬Ïòº¬ÓÐAgCl(s)µÄ±¥ºÍAgClÈÜÒºÖмÓË®£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨  £©

A£®AgClµÄÈܽâ¶ÈÔö´ó B£®AgClµÄÈܽâ¶ÈÔö´ó£¬Ksp²»±ä
C£®C£¨Ag+£©Ôö´ó D£®AgClµÄÈܽâ¶È¡¢Ksp¾ù²»±ä

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬ÔÚµÈÌå»ý¢ÙpH=0µÄÁòËá¡¢¢Ú0.01mol/L NaOHÈÜÒº¡¢¢ÛpH=10µÄ´¿¼îÈÜÒº¡¢¢ÜpH=5µÄNH4ClÈÜÒºÖУ¬Ë®µçÀë³Ì¶ÈµÄ´óС˳ÐòÊÇ  

A£®¢Ù£¾¢Ú£¾¢Û£¾¢Ü B£®¢Ú£¾¢Ù£¾¢Ü£¾¢Û C£®¢Û£¾¢Ü£¾¢Ú£¾¢Ù D£®¢Ü£¾¢Û£¾¢Ú£¾¢Ù

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬ÏòÌå»ýΪ10 mL¡¢ÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1 mol/LµÄÑÎËáºÍ´×ËáµÄ»ìºÏÈÜÒºÖеÎÈë0.1 mol£¯L NaOHÈÜÒº¡£ÏÂÁÐ˵·¨´íÎóµÄÊÇ(    )

A£®¼ÓÈëNaOHÈÜÒººó£¬ÈÜÒºÒ»¶¨Âú×㣺 c(H+)+c(Na+)=c(OH-)+c(C1-)+c(CH3COO-)
B£®¼ÓÈë10 mLNaOHÈÜҺʱ£¬ÈÜÒºÂú×ã(ÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ)£º
c(CH3COO-)+c(CH3COOH)="0.05" mol/L
C£®¼ÓÈë15 mLNaOHÈÜҺʱ£¬³ÊËáÐÔ£¬ÈÜÒºÂú×㣺
c(Na+)>c(C1-)>c(CH3COO-)>c(CH3COOH)>c(H+)>c(OH-)
D£®¼ÓÈë×ãÁ¿NaOHÈÜҺʱ£¬ÈÜÒºÖз¢ÉúµÄÖкͷ´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ±íʾΪ£º
H+(aq)+OH-(aq)=H2O(1)  ¦¤H="-57.3" KJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬0.1 mol/L°±Ë®ÈÜÒºÖУ½1¡Á10-8£¬ÏÂÁÐÐðÊö´íÎóµÄÊÇ

A£®¸ÃÈÜÒºÖÐÇâÀë×ÓµÄŨ¶È£ºc(H+ ) £½ 1¡Á10-9 mol/L
B£®0.1 mol/L°±Ë®ÈÜÒºÓë0.1 mol/L HClÈÜÒºµÈÌå»ý»ìºÏºóËùµÃÈÜÒºÖУº
c(NH4+ ) + c(H+ ) £½ c(Cl- ) + c(OH-)
C£®0.1 mol/LµÄ°±Ë®ÈÜÒºÓë0.05 mol/L H2SO4ÈÜÒºµÈÌå»ý»ìºÏºóËùµÃÈÜÒºÖУº
c(NH4+ ) + c(NH3) + c(NH3¡¤H2O) £½ 2c(SO42-)
D£®Å¨¶È¾ùΪ0.1 mol/LµÄNH3¡¤H2OºÍNH4ClÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈôÈÜÒº³Ê¼îÐÔ£¬Ôò
c(NH4+ )£¾ c (NH3¡¤H2O) £¾ c(Cl-) £¾ c(OH-) £¾ c(H+ )

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÊÒÎÂÏ£¬ÓÐpH=3µÄÑÎËá¡¢ÁòËá¡¢´×Ëᣨ¼ÙÉèHAcµÄµçÀë¶ÈΪ1%£©ÈýÖÖÏàͬÌå»ýµÄÈÜÒº¡£ÒÔÏÂÐðÊö´íÎóµÄÊÇ

A£®²â¶¨Æäµ¼µçÐÔÄÜÏàͬ
B£®Óë×ãÁ¿µÄп·Û·´Ó¦µÄÆðʼËÙÂÊÏàͬ
C£®Óë×ãÁ¿µÄп·Û·´Ó¦²úÉúÇâÆøµÄÌå»ý±ÈΪ1¡Ã1¡Ã100
D£®ÓëͬŨ¶ÈÇâÑõ»¯ÄÆÈÜÒº·´Ó¦£¬ÏûºÄÇâÑõ»¯ÄÆÈÜÒºµÄÌå»ýΪ1¡Ã2¡Ã100

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏÂpHµÄCH3COOHÏ¡ÈÜÒºÓëpHµÄNaOHÏ¡ÈÜÒºµÈÌå»ý»ìºÏ£¬ÏÂÁÐÅжϴíÎó
µÄÊÇ£¨  £©

A£®Èô»ìºÏºópH£¬Ôò
B£®·´Ó¦¹ý³ÌÖУ¬CH3COOHµÄµçÀë¶ÈÔö´ó
C£®Èô»ìºÏºópH£¬Ôòc£¼c
D£®Èô»ìºÏºóCH3COOHÓëNaOHÇ¡ºÃÍêÈ«·´Ó¦£¬Ôò

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÔÚÏàͬÌõ¼þÏÂËáÐÔÇ¿Èõ˳ÐòΪ£ºCH3COOH>H2CO3>HCN>HCO3¡ª¡£ÏÂÁбíÊö²»ÕýÈ·µÄÊÇ

A£®25¡æʱ£¬µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ¸÷ÈÜÒºpH¹ØϵΪ£ºpH(Na2CO3)£¾pH(NaCN)£¾pH(NaHCO3)£¾pH(CH3COONa)
B£®a mol¡¤L£­1 HCNÈÜÒºÓëbmol¡¤L£­1 NaOHÈÜÒºµÈÌå»ý»ìºÏºó£¬ËùµÃÈÜÒºÖУºc(Na+)£½c(CN£­)£¬ÔòaÒ»¶¨´óÓÚb
C£®0.1mol¡¤L£­1µÄNa2CO3ÈÜÒºÖУºc(OH£­)£½c(H+)£«c(HCO3£­)£«c(H2CO3)
D£®25¡æʱ£¬pH£½4.75£¬Å¨¶È¾ùΪ0.1mol¡¤L£­1µÄCH3COOH¡¢CH3COONa»ìºÏÈÜÒº£ºc()+c()<c(CH3COOH)+c(H+)

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸