³£ÎÂÏ£¬½«0.1mol£®L-1£®ÑÎËáÖðµÎ¼ÓÈëµ½20mL0.1mol?L-1°±Ë®ÖУ¬²âµÃÈÜÒºµÄpHËæ¼ÓÈëÑÎËáµÄÌå»ý±ä»¯ÈçͼËùʾ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¢ÙÈÜÒºÖÐc£¨Cl-£©£¾c(NH4+)£¾c£¨OH-£©£¾c£¨H+£©
B£®¢ÚÈÜÒºÖУºc£¨NH 4+£©=c£¨Cl-£©=c£¨OH-£©=c£¨H+£©
C£®¢ÛÈÜÒºÖУºc£¨Cl-£©+c£¨H+£©=c(NH4+)+c£¨OH-£©
D£®µÎ¶¨¹ý³ÌÖпÉÄܳöÏÖ£ºc£¨NH3H?2O£©£¾c(NH4+)£¾c(OH-)£¾c£¨Cl-£©£¾c£¨H+£©
¾«Ó¢¼Ò½ÌÍø
A£®¢ÙÈÜÒºÖеÄÈÜÖÊΪµÈÎïÖʵÄÁ¿Å¨¶ÈµÄ°±Ë®ºÍÂÈ»¯ï§£¬°±Ë®µÄµçÀë³Ì¶È´óÓÚ笠ùÀë×ÓµÄË®½â³Ì¶È£¬ËùÒÔÈÜÒº³Ê¼îÐÔ£¬c£¨OH-£©£¾c£¨H+£©£¬¸ù¾ÝµçºÉÊغãµÄc£¨Cl-£©£¼c£¨NH4+£©£¬¹ÊA´íÎó£»
B£®¢ÚÈÜÒº³ÊÖÐÐÔ£¬c£¨OH-£©=c£¨H+£©£¬¸ù¾ÝµçºÉÊغãµÃc£¨Cl-£©=c£¨NH4+£©£¬ÒòΪˮµÄµçÀë½Ï΢Èõ£¬ËùÒÔc£¨Cl-£©£¾c£¨OH-£©£¬¹ÊB´íÎó£»
C£®¢ÛÖÐÈÜÖÊÊÇÂÈ»¯ï§£¬ÈÜÒº³ÊËáÐÔ£¬c£¨OH-£©£¼c£¨H+£©£¬ÈÜÒº³ÊµçÖÐÐÔ£¬ËùÒÔc£¨NH4+£©+c£¨H+£©=c£¨Cl-£©+c£¨OH-£©£¬ËùÒÔc£¨Cl-£©£¾c£¨NH4+£©£¬Ôòc£¨Cl-£©+c£¨H+£©£¾c£¨NH4+£©+c£¨OH-£©£¬¹ÊC´íÎó£»
D£®µ±ÈÜÒºÖа±Ë®µÄÁ¿Ô¶Ô¶´óÓÚÑÎËáʱ£¬¿ÉÄܳöÏÖc£¨NH3H?2O£©£¾c£¨NH4+£©£¾c£¨OH-£©£¾c£¨Cl-£©£¾c£¨H+£©£¬¹ÊDÕýÈ·£»
¹ÊÑ¡D£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2010?º£ÄÏ£©³£ÎÂÏ£¬½«0.1mol?L-1ÇâÑõ»¯ÄÆÈÜÒºÓë0.06mol?L-1ÁòËáÈÜÒºµÈÌå»ý»ìºÏ£¬¸Ã»ìºÏÈÜÒºµÄpHµÈÓÚ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ³£ÎÂÏ£¬0.1mol/LÒ»ÔªËáHBÈÜÒºµÄPH=3£®ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©HBÔÚË®ÈÜÒºÖеĵçÀë·½³ÌʽΪ
HB?H++B-
HB?H++B-
£¬100mL0.1mol/LÒ»ÔªËáHBÈÜÒººÍ0.1mol/LNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Ê±£¬ËùÐèNaOHÈÜÒºµÄÌå»ýΪ
100
100
mL£®
£¨2£©ÔÚ³£ÎÂÏ£¬0.1mol/LÒ»ÔªËáHBÈÜÒºÖÐÓÉË®µçÀë³öµÄH+ÎïÖʵÄÁ¿Å¨¶ÈÊÇ
10-11mol/L
10-11mol/L
£»Èô½«Î¶ȸı䵽t¡æ£¨´ËʱKw=10-12£©£¬Ôòt
£¾
£¾
25£¨£¼¡¢£¾»ò=£©£¬¸ÃζÈÏÂ0.1mol/LÒ»ÔªËáHBÈÜÒºÖÐÓÉË®µçÀë³öµÄH+ÎïÖʵÄÁ¿Å¨¶È
D
D
£®
A.10-11mol/L      B.10-9mol/L      C.10-6mol/L        D£®ÒÔÉϴ𰸶¼²»ÕýÈ·
£¨3£©ÔÚ³£ÎÂÏ£¬½«0.1mol/LÒ»ÔªËáHBÈÜҺϡÊÍ100±¶ºó£¬ÈÜÒºµÄPH
£¼
£¼
5£¨£¼¡¢£¾»ò=£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨1£©³£ÎÂÏ£¬½«0.1mol?L-1ÑÎËáÈÜÒººÍ0.06mol?L-1ÇâÑõ»¯±µÈÜÒºµÈÌå»ý»ìºÏºó£¬¸Ã»ìºÏÈÜÒºµÄpH=
12
12
£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©
£¨2£©³£ÎÂÏ£¬pH=aµÄ10Ìå»ýµÄijǿËáÓëpH=bµÄ1Ìå»ýµÄijǿ¼î»ìºÏºó£¬ÈÜÒº³ÊÖÐÐÔ£¬ÔòaºÍbÂú×ãµÄ¹Øϵ
a+b=15
a+b=15
£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©
£¨3£©Ä³Î¶Èʱ£¬Ë®µÄÀë×Ó»ýKW=1¡Á10-13£¬Ôò¸ÃζÈ
£¾
£¾
25¡æ£¨Ñ¡Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£¬Èô½«´ËζÈÏÂpH=2µÄÏ¡ÁòËáaLÓëpH=12µÄÇâÑõ»¯ÄÆÈÜÒºbL»ìºÏ£¬ÈôËùµÃ»ìºÏÒºµÄpH=11£¬Ôòa£ºb=
9£º11
9£º11
£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©
£¨4£©³£ÎÂÏ£¬pH=8µÄNH4ClÈÜÒºÖУ¬c£¨Cl-£©-c£¨NH4+£©=
9.9¡Á10-7mol/L
9.9¡Á10-7mol/L
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª´×ËáºÍÑÎËáÊÇÈÕ³£Éú»îÖм«Îª³£¼ûµÄËᣬÔÚÒ»¶¨Ìõ¼þÏ£¬CH3COOHÈÜÒºÖдæÔÚµçÀëƽºâ£º
CH3COOH?CH3COO-+H+¡÷H£¾0£®
£¨1£©³£ÎÂÏ£¬ÔÚ pH=5µÄÏ¡´×ËáÈÜÒºÖУ¬c£¨CH3COO-£©=
£¨10-5-10-9£©mol/L
£¨10-5-10-9£©mol/L
£¨ÁÐʽ£¬²»±Ø»¯¼ò£©£»ÏÂÁз½·¨ÖУ¬¿ÉÒÔʹ0.10mol?L-1 CH3COOHµÄµçÀë³Ì¶ÈÔö´óµÄÊÇ
bcf
bcf
?
a£®¼ÓÈëÉÙÁ¿0.10mol?L-1µÄÏ¡ÑÎËá        b£®¼ÓÈÈCH3COOHÈÜÒº
c£®¼ÓˮϡÊÍÖÁ0.010mol?L-1             d£®¼ÓÈëÉÙÁ¿±ù´×Ëá
e£®¼ÓÈëÉÙÁ¿ÂÈ»¯ÄƹÌÌå                 f£®¼ÓÈëÉÙÁ¿0.10mol?L-1µÄNaOHÈÜÒº
£¨2£©½«µÈÖÊÁ¿µÄпͶÈëµÈÌå»ýÇÒpH¾ùµÈÓÚ3µÄ´×ËáºÍÑÎËáÈÜÒºÖУ¬¾­¹ý³ä·Ö·´Ó¦ºó£¬·¢ÏÖÖ»ÔÚÒ»ÖÖÈÜÒºÖÐÓÐп·ÛÊ£Ó࣬ÔòÉú³ÉÇâÆøµÄÌå»ý£ºV£¨ÑÎËᣩ
£¼
£¼
V£¨´×Ëᣩ£¬·´Ó¦µÄ×î³õËÙÂÊΪ£ºv£¨ÑÎËᣩ
=
=
v£¨´×Ëᣩ£®£¨Ìîд¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©
£¨3£©³£ÎÂÏ£¬ÏòÌå»ýΪVa mL£¬pHΪ3µÄ´×ËáÈÜÒºÖеμÓpH=11µÄNaOHÈÜÒºVb mLÖÁÈÜҺǡºÃ³ÊÖÐÐÔ£¬ÔòVaÓëVbµÄ¹ØϵÊÇ£º
Va£¼Vb
Va£¼Vb
£®
£¨4£©³£ÎÂÏ£¬½«0.1mol/LÑÎËáºÍ0.1mol/L´×ËáÄÆÈÜÒº»ìºÏ£¬ËùµÃÈÜҺΪÖÐÐÔ£¬Ôò»ìºÏÈÜÒºÖи÷Àë×ÓµÄŨ¶È°´ÓÉ´óµ½Ð¡ÅÅÐòΪ
c£¨Na+£©£¾c£¨CH3COO-£©£¾c£¨Cl-£©£¾c£¨H+£©=c£¨OH-£©
c£¨Na+£©£¾c£¨CH3COO-£©£¾c£¨Cl-£©£¾c£¨H+£©=c£¨OH-£©
£®
£¨5£©ÒÑÖª£º90¡æʱ£¬Ë®µÄÀë×Ó»ý³£ÊýΪKw=3.8¡Á10-13£¬ÔÚ´ËζÈÏ£¬½«pH=3µÄÑÎËáºÍpH=11µÄÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏ£¬Ôò»ìºÏÈÜÒºÖеÄc£¨H+£©=
2.05¡Á10-11
2.05¡Á10-11
£¨±£ÁôÈýλÓÐЧÊý×Ö£©mol/L£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³£ÎÂÏ£¬½«0.1mol/L Ba£¨OH£©2ÈÜÒºÓë0.09mol/LÁòËáÈÜÒºµÈÌå»ý»ìºÏ£¬¸Ã»ìºÏÈÜÒºµÄpHÔ¼µÈÓÚ£¨¡¡¡¡£©
A¡¢2B¡¢7C¡¢10D¡¢12

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸