Question

³£ÎÂÏ£¬ÏÂÁÐÓйش×ËáÈÜÒºµÄÐðÊöÖв»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

A£®pH=5.6µÄCH₃COOHÓëCH₃COONa»ìºÏÈÜÒºÖУºc£¨Na⁺£©£¼c£¨CH₃COO^-£©

B£®Å¨¶È¾ùΪ0.1mol?L^-1µÄCH₃COOHºÍCH₃COONaÈÜÒºµÈÌå»ý»ìºÏºó£ºc£¨CH₃COO^-£©-c£¨CH₃COOH£©=2[c£¨H⁺£©-c£¨OH^-£©]

C£®½«pH=aµÄ´×ËáÏ¡ÊÍΪpH=a+1µÄ¹ý³ÌÖУ¬c£¨CH₃COOH£©/c£¨H⁺£©²»±ä

D£®µÈÌå»ýpH=aµÄ´×ËáÓëpH=bµÄNaOHÈÜҺǡºÃÖкÍʱ£¬a+b=14

Answer 1

·ÖÎö£ºA£®¸ù¾ÝµçºÉÊغãÅжϣ»
B£®¸ù¾ÝÎïÁÏÊغãºÍµçºÉÊغãÅжϣ»
C£®¼ÓˮϡÊÍÈõµç½âÖÊÈÜÒº£¬ÄÜ´Ù½øÈõµç½âÖʵĵçÀ룻
D£®´×ËáµÄŨ¶È£¾10^-a mol/L£¬c£¨NaOH£©=10 ^b-14 mol/L£¬ÒªÊ¹´×ËáºÍÇâÑõ»¯ÄÆÇ¡ºÃ·´Ó¦£¬Ôò´×ËáºÍÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿ÏàµÈ£®

½â´ð£º½â£ºA£®»ìºÏÈÜÒºµÄpH=5.6£¼7£¬ÔòÈÜÒºÖÐc£¨H+£©£¾c£¨OH^-£©£¬ÈÜÒºÖдæÔÚµçºÉÊغ㣬¼´c£¨Na⁺£©+c£¨H+£©=c£¨OH^-£©+c£¨CH₃COO^-£©£¬ËùÒÔc£¨Na⁺£©£¼c£¨CH₃COO^-£©£¬¹ÊAÕýÈ·£»
B£®¸ù¾ÝÎïÁÏÊغãµÃ2c£¨Na⁺£©=c£¨CH₃COO^-£©+c£¨CH₃COOH£©=0.1mol/L¢Ù£¬¸ù¾ÝµçºÉÊغãµÃc£¨Na⁺£©+c£¨H+£©=c£¨CH₃COO^-£©+c£¨OH^-£©¢Ú£¬½«¢Ú¡Á2-¢ÙµÃc£¨CH₃COO^-£©-c£¨CH₃COOH£©=2[c£¨H⁺£©-c£¨OH^-£©]£¬¹ÊBÕýÈ·£»
C£®´×ËáÊÇÈõµç½âÖÊ£¬¼ÓˮϡÊÍ´×ËáÄÜ´Ù½ø´×ËáµçÀ룬ÔòÇâÀë×Ó¸öÊýÔö´ó£¬´×Ëá·Ö×Ó¸öÊý¼õС£¬ËùÒÔc£¨CH₃COOH£©/c£¨H⁺£©±ÈÖµ¼õС£¬¹ÊC´íÎó£»
D£®´×ËáµÄŨ¶È£¾10^-a mol/L£¬c£¨NaOH£©=10 ^b-14 mol/L£¬ÒªÊ¹´×ËáºÍÇâÑõ»¯ÄÆÇ¡ºÃ·´Ó¦£¬Ôò´×ËáºÍÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿ÏàµÈ£¬ÒªÊ¹´×ËáµÄÎïÖʵÄÁ¿µÈÓÚÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿£¬ÇÒÁ½ÕßÌå»ýÏàͬ£¬ÔòÆäŨ¶ÈÏàµÈ£¬ËùÒÔa+b£¾14£¬¹ÊD´íÎó£»
¹ÊÑ¡CD£®

µãÆÀ£º±¾Ì⿼²éÈõµç½âÖʵĵçÀë¡¢Àë×ÓŨ¶È´óСµÄ±È½ÏµÈ֪ʶµã£¬¸ù¾ÝÎïÁÏÊغãºÍµçºÉÊغãÀ´·ÖÎö½â´ð¼´¿É£¬ÄѶȽϴó£¬×¢ÒâBÑ¡ÏîÖУ¬Á½ÖÖÈÜÒº»ìºÏʱ£¬¸÷ÈÜÖʵÄŨ¶È±äΪԭÀ´µÄÒ»°ë£¬ÎªÒ×´íµã£®