ÌâÄ¿ÄÚÈÝ

5£®ÏÂÁÐÈÜÒºÖи÷΢Á£µÄŨ¶È¹ØÏµ²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®0.1 mol/LCH3COOHÈÜÒºÖУºc£¨CH3COO-£©+c£¨CH3COOH£©=0.1mol/L
B£®½«0.2 mol/L NaAÈÜÒººÍ0.1 mol/LÑÎËáµÈÌå»ý»ìºÏËùµÃ¼îÐÔÈÜÒºÖУºc£¨Na+£©+c£¨H+£©=c£¨A-£©+c£¨Cl-£©
C£®CH3COONaÈÜÒºÖУºc£¨Na+£©£¾c£¨CH3COO-£©£¾c£¨OH-£©£¾c£¨H+£©
D£®Na2SÈÜÒºÖУºc£¨Na+£©=2£¨c£¨S2-£©+c£¨HS-£©+c£¨H2S£©£©

·ÖÎö A.0.1 mol/LCH3COOHÈÜÒºÖдæÔÚÎïÁÏÊØºã£»
B£®½«0.2 mol/L NaAÈÜÒººÍ0.1 mol/LÑÎËáµÈÌå»ý»ìºÏµÃµ½µÈŨ¶ÈµÄNaCl¡¢NaA¡¢HAµÄ»ìºÏÈÜÒº£¬ÈÜÒºÖдæÔÚµçºÉÊØºãÅжϣ»
C£®´×ËáÄÆÈÜÒºÖд×Ëá¸ùÀë×ÓË®½âÈÜÒºÏÔ¼îÐÔ£»
D£®Na2SÈÜÒºÖдæÔÚÎïÁÏÊØºãn£¨Na£©=2n£¨S£©£»

½â´ð ½â£ºA.0.1 mol/LCH3COOHÈÜÒºÖдæÔÚÎïÁÏÊØºã£ºc£¨CH3COO-£©+c£¨CH3COOH£©=0.1mol/L£¬¹ÊAÕýÈ·£»
B£®½«0.2 mol/L NaAÈÜÒººÍ0.1 mol/LÑÎËáµÈÌå»ý»ìºÏµÃµ½µÈŨ¶ÈµÄNaCl¡¢NaA¡¢HAµÄ»ìºÏÈÜÒº£¬ÈÜÒºÖдæÔÚµçºÉÊØºãc£¨Na+£©+c£¨H+£©=c£¨A-£©+c£¨Cl-£©+c£¨OH-£©£¬¹ÊB´íÎó£»
C£®CH3COONaÈÜÒºÖд×Ëá¸ùÀë×ÓË®½âÈÜÒºÏÔ¼îÐÔ£¬Àë×ÓŨ¶È´óС£ºc£¨Na+£©£¾c£¨CH3COO-£©£¾c£¨OH-£©£¾c£¨H+£©£¬¹ÊCÕýÈ·£»
D£®Na2SÈÜÒºÖдæÔÚÎïÁÏÊØºãn£¨Na£©=2n£¨S£©£¬c£¨Na+£©=2£¨c£¨S2-£©+c£¨HS-£©+c£¨H2S£©£©£¬¹ÊDÕýÈ·£»
¹ÊÑ¡B£®

µãÆÀ ±¾Ì⿼²éÁ˵ç½âÖÊÈÜÒºÖÐÀë×ÓŨ¶È´óС¡¢µçºÉÊØºã¡¢ÎïÁÏÊØºã·ÖÎöÅжϣ¬ÕÆÎÕ»ù´¡ÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø