£¨10·Ö£©³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓУºNa+¡¢A2-¡¢HA-¡¢H+¡¢OH-£¬´æÔÚµÄ·Ö ×ÓÓÐH2O¡¢H2A¡£¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ð´³öËáH2AµÄµçÀë·½³Ìʽ__________________________¡£

£¨2£©ÈôÈÜÒºMÓÉ10mL 2 mol¡¤L-1NaHAÈÜÒºÓë10mL 2mol¡¤L-1NaOHÈÜÒº»ìºÏ¶øµÃ£¬ÔòÈÜÒºMµÄpH ________7 £¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡Ë³ÐòΪ                           ¡££¨ÈÜÒº»ìºÏʱÌå»ý±ä»¯ºöÂÔ£¬ÏÂͬ£©

£¨3£©ÒÑÖªKsp(BaA)= 1.8¡Á10-10£¬Ïò¢ÆµÄ»ìºÏÈÜÒºÖмÓÈë10mL 1 mol¡¤L-1  BaCl2ÈÜÒº£¬»ìºÏºóÈÜÒºÖеÄBa2+Ũ¶ÈΪ___________ mol¡¤L£­1¡£

£¨4£©ÈôÈÜÒºMÓÉpH=3µÄH2AÈÜÒºV1 mLÓëpH=11µÄNaOHÈÜÒºV2 mL»ìºÏ·´Ó¦¶øµÃ£¬»ìºÏÈÜÒºc(H+)/c(OH-)=104£¬V1ÓëV2µÄ´óС¹ØϵΪ___________£¨Ìî¡°´óÓÚ¡¢µÈÓÚ¡¢Ð¡ÓÚ¡±»ò¡°¾ùÓпÉÄÜ¡±£©¡£

 

¡¾´ð°¸¡¿

£¨10·Ö£©

¢ÅH2AH+ + HA-     HA-H+ + A2-£¨2·Ö,µÚ¶þ¸ö²»Ð´²»¿Û·Ö£©

¢Æ £¾  £¨2·Ö£©   c(Na+)>c(A2-)>c(OH-)>c(HA-)>c(H+)  £¨2·Ö£©

¢Ç 5.4¡Á10-10        £¨2·Ö£©

¢È¾ùÓпÉÄÜ      £¨2·Ö£©

¡¾½âÎö¡¿

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2009?פÂíµê¶þÄ££©³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓУºNa+¡¢A-¡¢H+¡¢OH-£®¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÈôÈÜÒºMÓÉ0.1mol?L-1µÄHAÈÜÒºÓë0.1mol?L-1µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ¶øµÃ£¬ÔòÈÜÒºMµÄpH²»¿ÉÄÜ
СÓÚ
СÓÚ
7£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£®
£¨2£©ÈôÈÜÒºMµÄpH£¾7£¬Ôòc£¨Na+£©
£¾
£¾
c£¨A-£©£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®
£¨3£©³£ÎÂÏ£¬ÈôÈÜÒºMÓÉpH=3µÄHAÈÜÒºV1 mLÓëpH=11µÄNaOHÈÜÒºV2 mL»ìºÏ·´Ó¦¶øµÃ£¬ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ
AD
AD
£¨Ìî×Öĸ£©£®
A£®ÈôÈÜÒºM³ÊÖÐÐÔ£¬ÔòÈÜÒºMÖÐc£¨H+£©+c£¨OH-£©=2¡Á10-7 mol?L-1
B£®ÈôV1=V2£¬ÔòÈÜÒºMµÄpHÒ»¶¨µÈÓÚ7
C£®ÈôÈÜÒºM³ÊËáÐÔ£¬ÔòV1Ò»¶¨´óÓÚV2
D£®ÈôÈÜÒºM³Ê¼îÐÔ£¬ÔòV1Ò»¶¨Ð¡ÓÚV2£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔÚ³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖУ®
£¨1£©ÈôMÖдæÔÚµÄÀë×ÓÖ»ÓУºNa+¡¢CH3COO-¡¢H+¡¢OH-£¬ÏÂÁйØϵºÏÀíµÄÊÇ
¢Ù¢Ú¢Ü
¢Ù¢Ú¢Ü
£®
¢Ùc£¨Na+£©£¾c£¨CH3COO-£©£¾c£¨OH-£©£¾c£¨H+£©    ¢Úc£¨Na+£©£¾c£¨OH-£©£¾c£¨CH3COO-£©£¾c£¨H+£©
¢Ûc£¨H+£©£¾c£¨OH-£©£¾c£¨Na+£©£¾c£¨CH3COO-£©    ¢Üc£¨Na+£©=c£¨CH3COO-£©ÇÒc£¨OH-£©=c£¨H+£©
£¨2£©ÈôMÓÉÁ½ÖÖÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº»ìºÏ¶ø³É£¬ÆäÖÐÖ»´æÔÚOH-¡¢H+¡¢NH4+¡¢Cl-ËÄÖÖÀë×Ó£¬ÇÒc£¨NH4+£©£¾c£¨Cl-£©£¾c£¨OH-£©£¾c£¨H+£©£¬ÕâÁ½ÖÖÈÜÒºµÄÈÜÖÊÊÇ
NH4ClºÍNH3?H2O
NH4ClºÍNH3?H2O
£®
£¨3£©ÈôMΪ0.01mol/LµÄ°±Ë®£¬ÏòÆäÖмÓÈëµÈÌå»ýpH=2µÄÑÎËáÈÜÒº£¨ºöÂÔÈÜÒºÌå»ý±ä»¯£©£¬ËùµÃ»ìºÏÒºÖУºc£¨NH4+£©+c£¨H+£©-c£¨OH-£©=
0.005mol/L
0.005mol/L
£¨ÌîÊýÖµ£©
£¨4£©ÈôÒÑÖªH2AΪ¶þÔªÈõËᣬÈÜÒºMÓÉÏÂÁÐÈýÖÖÇé¿ö£º¢Ù0.01mol?L-1µÄH2AÈÜÒº¢Ú0.01mol?L-1µÄNaHAÈÜÒº¢Û0.02mol?L-1µÄHClÓë0.04mol?L-1µÄNaHAÈÜÒºµÈÌå»ý»ìºÏÒº£¬ÔòÈýÖÖÇé¿öµÄÈÜÒºÖÐH2A·Ö×ÓŨ¶È×î´óµÄΪ
¢Û
¢Û
£»pHÓÉ´óµ½Ð¡µÄ˳ÐòΪ
¢Û¢Ú¢Û¢Ù
¢Û¢Ú¢Û¢Ù
£®
£¨5£©ÈôMΪ20mLÏ¡ÁòËáºÍÑÎËá»ìºÏÒº£¬Ïò¸Ã»ìºÏËáÈÜÒºÖÐÖðµÎ¼ÓÈëpH=13µÄBa£¨OH£©2ÈÜÒº£¬Éú³ÉBaSO4µÄÎïÖʵÄÁ¿Îª0.001mol£¬µ±¼ÓÈë60mL Ba£¨OH£©2ÈÜҺʱ£¬ÈÜÒºµÄpH=7£¨ºöÂÔÈÜÒºÌå»ý±ä»¯£©£¬ÊÔ¼ÆË㣺×î³õ»ìºÏËáÈÜÒºÖÐc£¨H2SO4£©=
0.05mol/L
0.05mol/L
£¬c£¨HCl£©=
0.05mol/L
0.05mol/L
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓУºNa+¡¢A2-¡¢HA-¡¢H+¡¢OH-£¬´æÔڵķÖ×ÓÓÐH2O¡¢H2A£®¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öËáH2AµÄµçÀë·½³Ìʽ
H2A?H++HA-¡¢HA-?H++A2-
H2A?H++HA-¡¢HA-?H++A2-
£®
£¨2£©ÈôÈÜÒºMÓÉ2mol?L-1NaHAÈÜÒºÓë2mol?L-1NaOHÈÜÒºµÈÌå»ý»ìºÏ¶øµÃ£¬ÔòÈÜÒºMµÄpH
£¾
£¾
7£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡Ë³ÐòΪ
c£¨Na+£©£¾c£¨A2-£©£¾c£¨OH-£©£¾c£¨HA-£©£¾c£¨H+£©
c£¨Na+£©£¾c£¨A2-£©£¾c£¨OH-£©£¾c£¨HA-£©£¾c£¨H+£©
£®
£¨3£©ÈôÈÜÒºMÓÉÏÂÁÐÈýÖÖÇé¿ö£º¢Ù0.01mol?L-1µÄH2AÈÜÒº£»¢Ú0.01mol?L-1µÄNaHAÈÜÒº£»¢Û0.02mol?L-1µÄHClÓë0.04mol?L-1µÄNaHAÈÜÒºµÈÌå»ý»ìºÏÒº£¬ÔòÈýÖÖÇé¿öµÄÈÜÒºÖÐH2A·Ö×ÓŨ¶È×î´óµÄΪ
¢Û
¢Û
£»pHÓÉ´óµ½Ð¡µÄ˳ÐòΪ
¢Ú£¾¢Û£¾¢Ù
¢Ú£¾¢Û£¾¢Ù
£®
£¨4£©ÈôÈÜÒºMÓÉpH=3µÄH2AÈÜÒºV1 mLÓëpH=11µÄNaOHÈÜÒºV2 mL»ìºÏ·´Ó¦¶øµÃ£¬»ìºÏÈÜÒºc£¨H+£©/c£¨OH-£©=104£¬V1ÓëV2µÄ´óС¹ØϵΪ
¾ùÓпÉÄÜ
¾ùÓпÉÄÜ
£¨Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°¾ùÓпÉÄÜ¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓУºNa+¡¢A2-¡¢HA-¡¢H+¡¢OH-£¬´æÔڵķÖ×ÓÓÐH2O¡¢H2A£®¸ù¾ÝÌâÒâ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Ð´³öËáH2AµÄµçÀë·½³Ìʽ
H2A?H++HA-¡¢HA-?H++A2-
H2A?H++HA-¡¢HA-?H++A2-
£®
£¨2£©ÈôÈÜÒºMÓÉ10mL 2mol?L-1NaHAÈÜÒºÓë2mol?L-1NaOHÈÜÒºµÈÌå»ý»ìºÏ¶øµÃ£¬ÔòÈÜÒºMµÄpH
£¾
£¾
7£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡Ë³ÐòΪ
c£¨Na+£©£¾c£¨A2-£©£¾c£¨OH-£©£¾c£¨HA-£©£¾c£¨H+£©
c£¨Na+£©£¾c£¨A2-£©£¾c£¨OH-£©£¾c£¨HA-£©£¾c£¨H+£©
£®ÒÑÖªKsp£¨BaA£©=1.8¡Á10-10£¬Ïò¸Ã»ìºÏÈÜÒºÖмÓÈë10mL 1mol?L-1BaCl2ÈÜÒº£¬»ìºÏºóÈÜÒºÖеÄBa2+Ũ¶ÈΪ
5.4¡Á10-10
5.4¡Á10-10
 mol?L-1£®
£¨3£©ÈôÈÜÒºMÓÉÏÂÁÐÈýÖÖÇé¿ö£º¢Ù0.01mol?L-1µÄH2AÈÜÒº£»¢Ú0.01mol?L-1µÄNaHAÈÜÒº£»¢Û0.02mol?L-1µÄHClÓë0.04mol?L-1µÄNaHAÈÜÒºµÈÌå»ý»ìºÏÒº£¬ÔòÈýÖÖÇé¿öµÄÈÜÒºÖÐH2A·Ö×ÓŨ¶È×î´óµÄΪ
¢Û
¢Û
£»pHÓÉ´óµ½Ð¡µÄ˳ÐòΪ
¢Ú£¾¢Û£¾¢Ù
¢Ú£¾¢Û£¾¢Ù
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

³£ÎÂÏ£¬Ä³Ë®ÈÜÒºMÖдæÔÚµÄÀë×ÓÓУºNa+¡¢A-¡¢H+¡¢OH-£®Èô¸ÃÈÜÒºMÓÉ pH=3µÄHAÈÜÒºV1mLÓëpH=11µÄNaOHÈÜÒºV2mL»ìºÏ·´Ó¦¶øµÃ£¬ÔòÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ
 
£¨Ìî×Öĸ£©£®
A£®ÈôÈÜÒºM³ÊÖÐÐÔ£¬ÔòÈÜÒºMÖÐc£¨H+£©+c£¨OH-£©=2¡Á10-7mol?L-1
B£®ÈôV1=V2£¬ÔòÈÜÒºMµÄpHÒ»¶¨µÈÓÚ7
C£®ÈôÈÜÒºM³ÊËáÐÔ£¬ÔòV1Ò»¶¨´óÓÚV2
D£®ÈôÈÜÒºM³Ê¼îÐÔ£¬ÔòV1Ò»¶¨Ð¡ÓÚV2£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸