25¡æ£¬ÓÐc(CH3COOH)£«c(CH3COO£­)="0.1" mol/LµÄÒ»×é´×ËáºÍ´×ËáÄÆ»ìºÏÈÜÒº£¬ÈÜÒºÖÐc(CH3COOH)¡¢c(CH3COO£­)ÓëpHÖµµÄ¹ØϵÈçͼËùʾ¡£ÓйØÀë×ÓŨ¶È¹Øϵ·ÖÎöÕýÈ·µÄÊÇ

A£®pH=5.5ÈÜÒºÖУºc(CH3COO£­) £¾c(CH3COOH)£¾c(H£«)£¾c(OH£­)
B£®Wµã±íʾÈÜÒºÖУºc(Na£«)£«c(H£«)=c(CH3COO£­)£«c(OH£­)
C£®pH=3.5ÈÜÒºÖУºc(Na£«)£­c(OH£­)£«c(CH3COOH)="0.1" mol/L
D£®ÏòWµãËù±íʾÈÜÒºÖÐͨÈë0.05molHClÆøÌå(ÈÜÒºÌå»ý±ä»¯¿ÉºöÂÔ)£ºc(H£«)=c(CH3COOH)£«c(OH£­)

AB

½âÎöÊÔÌâ·ÖÎö£ºA£®ÓÉͼ¿ÉÖª£¬pH=5.5 µÄÈÜÒº£¬ÏÔËáÐÔ£¬ÇÒc£¨CH3COO-£©£¼c£¨CH3COOH£©£¬ÏÔÐÔÀë×Ó´óÓÚÒþÐÔÀë×Ó£¬Ôòc£¨CH3COOH£©£¾c£¨CH3COO-£©£¾c£¨H+£©£¾c£¨OH-£©£¬¹ÊAÕýÈ·£»B£®WµãÓɵçºÉÊغã¿ÉÖª£¬c£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬¹ÊBÕýÈ·£»C£®ÓɵçºÉÊغ㼰c£¨CH3COOH£©+c£¨CH3COO-£©=0.1mol?L-1¿ÉÖª£¬c£¨Na+£©+c£¨H+£©-c£¨OH-£©+c£¨CH3COOH£©==0.1mol/L£¬¹ÊC´íÎó£»D£®WµãΪµÈÁ¿µÄ´×ËáºÍ´×ËáÄƵĻìºÏÒº£¬1.0 L ÈÜÒºÖÐͨÈë0.05 mol HCl ÆøÌåµÃµ½0.05molNaClºÍ0.1molHAc£¬ÓɵçºÉÊغã¿ÉÖª£¬c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬¹ÊD´íÎó£»¹ÊÑ¡AB¡£
¿¼µã£º±¾ÌâÖ÷Òª¿¼²éÀë×ÓŨ¶È´óСµÄ±È½Ï£¬Ã÷È·ÑÎÀàË®½â¡¢ÖÊ×ÓÊغãÊǽâ´ð±¾ÌâµÄ¹Ø¼ü¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬ÓÐÏÂÁÐËÄÖÖÈÜÒº£º

¢Ù
¢Ú
¢Û
¢Ü
0.1mol/L
NH3¡¤H2OÈÜÒº
pH=11
NaOHÈÜÒº
pH=1
H2SO4ÈÜÒº
pH=3
CH3COOHÈÜÒº
 
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ(    )
A£®¢ÙÓë¢ÛµÈÌå»ý»ìºÏ£¬ÈÜÒºpH£¼7
B£®¢ÚÓë¢ÜµÈÌå»ý»ìºÏ£¬ÈÜÒºpH£¾7
C£®½«VmL¢ÜµÄÈÜҺŨËõµ½10¡ª2VmLºó£¬pHÓë¢ÛÏàͬ
D£®¢ÚÓë¢Û»ìºÏ£¬ÈôÈÜÒºpH£½7£¬ÔòV(NaOH) : V(H2SO4) £½100 : 1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬ÏÂÁÐÈÜÒºÖеÄ΢Á£Å¨¶È¹ØϵÕýÈ·µÄÊÇ£¨  £©

A£®ÐÂÖÆÂÈË®ÖмÓÈë¹ÌÌåNaOH£ºc(Na£«)=c(Cl£­)+c(ClO£­)+c(OH£­)
B£®pH=8.3µÄNaHCO3ÈÜÒº£ºc(Na£«)£¾c(HCO3£­)£¾c(CO32£­)£¾c(H2CO3)
C£®pH=11µÄ°±Ë®ÓëpH=3µÄÑÎËáµÈÌå»ý»ìºÏ£ºc(Cl£­)=c(NH4£«) £¾c(OH£­)=c(H£«)
D£®0.2mol¡¤L£­1CH3COOHÈÜÒºÓë0.1mol¡¤L£­1NaOHÈÜÒºµÈÌå»ý»ìºÏ£º2c(H£«)£­2c(OH£­)=c(CH3COO£­)£­c(CH3COOH)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÒͼÊÇÓÃ0.1000 mol¡¤L-1 NaOHÈÜÒºµÎ¶¨20.00 mLδ֪Ũ¶ÈÑÎËá(·Ó̪×öָʾ¼Á)µÄµÎ¶¨ÇúÏß¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®Ë®µçÀë³öµÄÇâÀë×ÓŨ¶È£ºa£¾b
B£®ÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.0100 mol¡¤L-1
C£®Ö¸Ê¾¼Á±äɫʱ£¬ËµÃ÷ÑÎËáÓëNaOHÇ¡ºÃÍêÈ«·´Ó¦
D£®µ±µÎ¼ÓNaOHÈÜÒº10.00 mLʱ£¬¸Ã»ìºÏÒºµÄpH£½1+lg3

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏ£¬Ïò20 mL x mol¡¤L£­1 CH3COOHÈÜÒºÖÐÖðµÎ¼ÓÈëµÈÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº£¬»ìºÏÒºµÄpHËæNaOHÈÜÒºµÄÌå»ý£¨V£©±ä»¯¹ØϵÈçͼËùʾ£¨ºöÂÔζȱ仯£©¡£ÏÂÁÐ˵·¨Öв»ÕýÈ·µÄÊÇ

A£®ÉÏÊö CH3COOHÈÜÒºÖУºc(H+)£½1¡Á10£­3 mol¡¤L£­1
B£®Í¼ÖÐV1 £¾20 mL
C£®aµã¶ÔÓ¦µÄÈÜÒºÖУºc (CH3COO£­)£½c (Na+£©
D£®¼ÓÈëNaOHÈÜÒºÌå»ýΪ20 mLʱ£¬ÈÜÒºÖУºc (CH3COOH) + c (H+)£½c (OH£­£©+c(CH3COO£­)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®µç½âÖÊÈÜÒºÄܵ¼µç£¬ÊÇÒòΪÔÚͨµçʱµç½âÖʵçÀë²úÉúÁË×ÔÓÉÒƶ¯µÄÀë×Ó
B£®±¥ºÍʯ»ÒË®ÖмÓÈëÉÙÁ¿CaO£¬»Ö¸´ÖÁÊÒκóÈÜÒºµÄpHÖµ²»±ä
C£®½«AlCl3ÈÜÒººÍNa2SO3ÈÜÒº·Ö±ðÕô¸É²¢×ÆÉÕ£¬µÃµ½Al2O3ºÍNa2SO3
D£®pHÏàͬµÄCH3COONaÈÜÒº¡¢C6H5ONaÈÜÒº¡¢Na2CO3ÈÜÒº¡¢NaOHÈÜÒº£¬ÔòÈÜҺŨ¶È´óС¹Øϵ£º
c(CH3COONa)> c(Na2CO3)> c(C6H5ONa)>c(NaOH)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚ³£ÎÂÏ£¬0.1000mol¡¤L-1Na2CO3ÈÜÒº25mL ÓÃ0.1000mol¡¤L-1ÑÎËáµÎ¶¨£¬ÆäµÎ¶¨ÇúÏßÈçͼ¡£¶ÔµÎ¶¨¹ý³ÌÖÐËùµÃÈÜÒºÖÐÏà¹ØÀë×ÓŨ¶È¼äµÄ¹Øϵ£¬ÏÂÁÐÓйØ˵·¨ÕýÈ·µÄÊÇ

A£®aµã£ºc£¨CO32-£©=c(HCO3-)>c(OH-
B£®bµã£º5c(Cl-)>4c(HCO3-)+4c(CO32-
C£®cµã£ºc(OH-)=c(H+)+c(HCO3-)+2c(H2CO3
D£®dµã£ºc(H+)=c(CO32-)+c(HCO3-)+c(OH-

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÊÒÎÂÏ£¬ÏÂÁйØÓÚpH¾ùΪ9¡¢Ìå»ý¾ùΪ10 mLµÄNaOHÈÜÒººÍCH3COONaÈÜÒºµÄ˵·¨£¬ÕýÈ·µÄÊÇ

A£®Á½ÖÖÈÜÒºÖеÄc£¨Na+£©ÏàµÈ
B£®·Ö±ð¼ÓÈȵ½ÏàͬζÈʱ£¬CH3COONaÈÜÒºµÄpHС
C£®·Ö±ð¼ÓˮϡÊ͵½100 mLʱ£¬Á½ÖÖÈÜÒºµÄpHÒÀÈ»ÏàµÈ
D£®Á½ÈÜÒºÖÐÓÉË®µçÀë³öµÄc£¨OH-£©Ö®±ÈΪ10-9/10-5

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºµ¥Ñ¡Ìâ

³£ÎÂÏÂÓÐÌå»ýÏàͬµÄËÄÖÖÈÜÒº£º¢ÙpH=3µÄCH3COOHÈÜÒº ¢ÚpH=3µÄÑÎËá ¢ÛpH=11µÄ°±Ë®¢ÜpH=11µÄNaOHÈÜÒº¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®Èô½«ËÄÖÖÈÜҺϡÊÍ100±¶£¬ÈÜÒºpH´óС˳Ðò£º¢Û£¾¢Ü£¾¢Ù£¾¢Ú
B£®¢ÛºÍ¢Ü·Ö±ðÓõÈŨ¶ÈµÄÁòËáÈÜÒºÖкͣ¬ÏûºÄÁòËáÈÜÒºµÄÌå»ý£º¢Û=¢Ü
C£®¢ÙÓë¢Ú·Ö±ðÓë×ãÁ¿Ã¾·Û·´Ó¦£¬Éú³ÉH2µÄÁ¿£º¢Ù<¢Ú
D£®¢ÚºÍ¢Û»ìºÏ£¬ËùµÃ»ìºÏÈÜÒºµÄpH´óÓÚ7

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸