Ũ¶È¾ùΪ0.1mol?L-1µÄÈýÖÖÈÜÒº£º¢ÙCH3COOHÈÜÒº£¬¢ÚÇâÑõ»¯ÄÆÈÜÒº¡¡¢Û´×ËáÄÆÈÜÒº£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
·ÖÎö£ºA£®¢ÙºÍ¢ÚµÈÌå»ý¡¢µÈŨ¶È»ìºÏºó£¬Ç¡ºÃÉú³É´×ËáÄÆ£»
B£®¢ÙºÍ¢ÛµÈÌå»ý»ìºÏºóÈÜÒºÏÔËáÐÔ£¬µçÀë´óÓÚË®½â£¬Ôòc£¨H+£©£¾c£¨OH-£©£¬½áºÏµçºÉÊغãʽc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©·ÖÎö£»
C£®¢ÚºÍ¢ÛµÈÌå»ý»ìºÏºóÈÜÒºÈÜÒºÏÔ¼îÐÔ£¬c£¨OH-£©Å¨¶È½Ï´ó£¬ÒÖÖÆË®µÄµçÀ룻
D£®´×ËáÄÆË®½â´Ù½øË®µÄµçÀ룬¢Ù¢Ú¾ùÒÖÖÆË®µÄµçÀ룬µ«¢ÚÖÐc£¨OH-£©´óÓÚ¢ÙÖÐc£¨H+£©£¬Ôò¢ÚÒÖÖƳ̶ȴó£®
½â´ð£º½â£ºA£®¢ÙºÍ¢ÚµÈÌå»ý¡¢µÈŨ¶È»ìºÏºó£¬Ç¡ºÃÉú³É´×ËáÄÆ£¬ÓɵçºÉÊغãʽc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©¡¢ÎïÁÏÊغãʽc£¨Na+£©=c£¨CH3COO-£©+c£¨CH3COOH£©¿ÉÖª£¬ÈÜÒºÖдæÔÚc£¨OH-£©=c£¨H+£©+c£¨CH3COOH£©£¬¹ÊAÕýÈ·£»
B£®¢ÙºÍ¢ÛµÈÌå»ý»ìºÏºóÈÜÒºÏÔËáÐÔ£¬µçÀë´óÓÚË®½â£¬Ôòc£¨H+£©£¾c£¨OH-£©£¬ÓɵçºÉÊغãʽc£¨Na+£©+c£¨H+£©=c£¨CH3COO-£©+c£¨OH-£©£¬Ôòc£¨CH3COO-£©£¾c£¨Na+£©£¬µçÀëÏÔËáÐÔ£¬ÇÒµçÀë³Ì¶È²»´ó£¬ÔòÀë×ÓŨ¶ÈΪc£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©£¬¹ÊBÕýÈ·£»
C£®¢ÚºÍ¢ÛµÈÌå»ý»ìºÏºóÈÜÒºÈÜÒºÏÔ¼îÐÔ£¬c£¨OH-£©Å¨¶È½Ï´ó£¬ÒÖÖÆË®µÄµçÀ룬ÔòÓÉË®µçÀë³öµÄc£¨H+£©£¼10-7 mol/L£¬¹ÊCÕýÈ·£»
D£®´×ËáÄÆË®½â´Ù½øË®µÄµçÀ룬¢Ù¢Ú¾ùÒÖÖÆË®µÄµçÀ룬µ«¢ÚÖÐc£¨OH-£©´óÓÚ¢ÙÖÐc£¨H+£©£¬Ôò¢ÚÒÖÖƳ̶ȴó£¬ËùÒÔÓÉË®µçÀë³öµÄc£¨OH-£©Îª¢Û£¾¢Ù£¾¢Ú£¬¹ÊD´íÎó£»
¹ÊÑ¡D£®
µãÆÀ£º±¾Ì⿼²éËá¼î»ìºÏ¼°Àë×ÓŨ¶È´óСµÄ±È½Ï£¬Îª¸ßƵ¿¼µã£¬°ÑÎÕËá¼î»ìºÏºóÈÜÒºÖеÄÈÜÖÊ¡¢ÑÎÀàË®½â¼°µçºÉÊغ㡢ÎïÁÏÊغãΪ½â´ðµÄ¹Ø¼ü£¬×ÛºÏÐÔ½ÏÇ¿£¬²àÖØѧÉú˼άÑÏÃÜÐÔѵÁ·£¬ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ìå»ýÏàͬ£¬Å¨¶È¾ùΪ0.1mol?L-1µÄNaOHÈÜÒº¡¢°±Ë®£¬·Ö±ð¼ÓˮϡÊÍm±¶¡¢n±¶£¬ÈÜÒºµÄpH¶¼±ä³É9£¬ÔòmÓënµÄ¹ØϵΪ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2011?ÁúÑÒÄ£Ä⣩ijÈÜÒºÖÐͬʱ´æÔÚMg2+¡¢Fe2+¡¢Mn2+ºÍAl3+ËÄÖÖ½ðÊôÀë×Ó£¨Å¨¶È¾ùΪ0.1mol/L£©£®ÏÖÓüîµ÷½ÚÈÜÒºpH£¬¸ù¾Ý±í¿ÉÖª£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂÁÐÇúÏßͼ£¨×Ý×ø±êΪ³ÁµíµÄÁ¿£¬ºá×ø±êΪ¼ÓÈëÎïµÄÁ¿£©Óë¶ÔӦѡÏî²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2012?±£¶¨Ò»Ä££©A¡¢B¡¢C¡¢D¡¢EÎåÖÖÈÜÒºµÄÈÜÖÊ·Ö±ðÊÇHCl¡¢CH3COOH¡¢NaOH¡¢NH3?H2O¡¢Na2CO3ÖеÄÒ»ÖÖ£®³£ÎÂϽøÐÐÏÂÁÐʵÑ飺
¢ñ£®Å¨¶È¾ùΪ0.1mol?L-1DÓëCÈÜÒºµÈÌå»ý»ìºÏ£¬ÈÜÒº³ÊËáÐÔ£®
¢ò£®Å¨¶È¾ùΪ0.1mol?L-1 AºÍEÈÜÒºµÄpH£ºA£¼E
¢ó£®½«Ò»¶¨Ìå»ýpH=10µÄAÈÜÒº·Ö±ðÓë0.5L 0.01mol?L-1BÈÜÒº¡¢0.48L0.01mol?L-1CÈÜÒº³ä·Ö·´Ó¦ºóÈÜÒº³ÊÖÐÐÔ£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©CÊÇ
HCl
HCl
ÈÜÒº£¨Ìѧʽ£©£®
£¨2£©½«µÈÌå»ýµÈÎïÖʵÄÁ¿Å¨¶ÈµÄCºÍE·Ö±ðÓë×ãÁ¿µÄÂÁ·Û·´Ó¦£¬×ªÒƵĵç×ÓÊýÖ®±ÈÊÇ
1£º3
1£º3

£¨3£©½«µÈÌå»ýµÈÎïÖʵĕžÅ¨¶ÈµÄBºÍDÈÜÒº»ìºÏºó£¬ËùµÃÈÜÒºµÄPH=7µÄÔ­ÒòÊÇ
´×Ëá¸ùµÄË®½â³Ì¶ÈÓë笠ùÀë×ÓµÄË®½â³Ì¶ÈÏ൱
´×Ëá¸ùµÄË®½â³Ì¶ÈÓë笠ùÀë×ÓµÄË®½â³Ì¶ÈÏ൱

£¨4£©½«CÈÜÒºÖðµÎµÎ¼Óµ½AÈÜÒºÖУ¬²»¶ÏÕñµ´£¬µ±µÎ¼Óµ½A¡¢CµÈÎïÖʵÄÁ¿Ê±£¬ÈÜÒºÖÐËùÓÐÒõÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶ÈÓÉСµ½´óµÄ¹ØϵÊÇ
c£¨CO32-£©£¼c£¨OH-£©£¼c£¨HCO3-£©£¼c£¨Cl-£©
c£¨CO32-£©£¼c£¨OH-£©£¼c£¨HCO3-£©£¼c£¨Cl-£©
£®
£¨5£©ÏàͬÌõ¼þÏ£¬H++OH-¨TH2OµÄ¡÷H=-57.3/mol£¬CH3COOH£¨aq£©ÓëNaOH£¨aq£©·´Ó¦µÄ¡÷H=-55.4KJ/mol£¬ÔòCH3COOH£¨aq£©?CH3COO-£¨aq£©+H+£¨aq£©µÄ¡÷H=
+1.9kJ/mol
+1.9kJ/mol
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨2010?³±ÖݶþÄ££©ÏÂÁÐÓйØÀë×ÓŨ¶ÈµÄ¹ØϵÕýÈ·µÄÊÇ£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸