ÌâÄ¿ÄÚÈÝ

16£®Ïò20mLÇâÑõ»¯ÄÆÈÜÒºÖÐÖðµÎ¼ÓÈë0.2mol/L´×ËáÈÜÒº£¬µÎ¶¨ÇúÏßÈçͼËùʾÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©    
A£®¸ÃÇâÑõ»¯ÄÆÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.1 mol/L
B£®ÔÚbµã£¬c£¨Na+£©=c£¨CH3COO-£©
C£®ÔÚdµã£¬ÈÜÒºÖÐËùÓÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨CH3COO-£©£¾c£¨Na+£©£¾c£¨H+£©£¾c£¨OH-£©
D£®ÇâÑõ»¯ÄÆÈÜÒºÓë´×ËáÈÜҺǡºÃÍêÈ«·´Ó¦µÄµãλÓÚÇúÏßb¡¢d¼äµÄijµã

·ÖÎö A£®ÓɼîÈÜÒºµÄpH¼ÆËãÇâÑõ»¯ÄÆÈÜҺŨ¶È£»
B£®bµãʱÈÜÒº³ÊÖÐÐÔ£¬¸ù¾ÝµçºÉÊغã·ÖÎö£»
C£®dµãʱ£¬¸ù¾ÝÈÜÒºµÄËá¼îÐÔ¼°µçºÉÊغã·ÖÎö£»
D£®µ±ÇâÑõ»¯Äƺʹ×ËáÇ¡ºÃ·´Ó¦Ê±¶þÕßµÄÎïÖʵÄÁ¿ÏàµÈ£®

½â´ð ½â£ºA£®µ±Î´¼Ó´×Ëáʱ£¬ÇâÑõ»¯ÄÆÈÜÒºµÄpH=13£¬ÔòÇâÑõ»¯ÄÆÈÜÒºÖÐÇâÑõ¸ùÀë×ÓŨ¶ÈÊÇ0.1mol/L£¬ÇâÑõ»¯ÄÆÊÇÇ¿µç½âÖÊ£¬ËùÒÔÇâÑõ»¯ÄÆÈÜҺŨ¶ÈÊÇ0.1mol/L£¬¹ÊAÕýÈ·£»
B£®bµãʱÈÜÒº³ÊÖÐÐÔ£¬Ôòc £¨H+£©=c £¨OH-£©£¬¸ù¾ÝµçºÉÊغãµÃc £¨Na+£©=c£¨CH3COO-£©£¬¹ÊBÕýÈ·£»
C£®dµãʱ£¬ÈÜÒº³ÊËáÐÔ£¬Ôòc £¨H+£©£¾c £¨OH-£©£¬¸ù¾ÝµçºÉÊغãµÃc £¨CH3COO£©£¾c £¨Na+£©£¬¸ÃµãÈÜÒºÖеÄÈÜÖÊÊÇ´×ËáÄƺʹ×ËᣬÆäÎïÖʵÄÁ¿ÏàµÈ£¬´×ËáÊÇÈõµç½âÖÊ£¬´×ËáÄÆÊÇÇ¿µç½âÖÊ£¬ËùÒÔc £¨Na+£©£¾c £¨H+£©£¬ÔòÀë×ÓŨ¶È´óС˳ÐòÊÇc £¨CH3COO£©£¾c £¨Na+£©£¾c £¨H+£©£¾c £¨OH-£©£¬¹ÊCÕýÈ·£»
D£®µ±ÇâÑõ»¯Äƺʹ×ËáÇ¡ºÃ·´Ó¦Ê±¶þÕßµÄÎïÖʵÄÁ¿ÏàµÈ£¬´×ËáµÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ0.2mol/L£¬ÇâÑõ»¯ÄÆÈÜÒºµÄŨ¶ÈÊÇ0.1mol/L£¬ÒªÊ¹¶þÕßµÄÎïÖʵÄÁ¿ÏàµÈ£¬ÔòÇâÑõ»¯ÄƵÄÌå»ýÓ¦¸ÃÊÇ´×ËáµÄ2±¶£¬´×ËáΪ10mL£¬µ±ÈÜÒº³ÊÖÐÐÔʱ£¬ËáÉÔ΢¹ýÁ¿£¬ËùÒÔÇâÑõ»¯ÄÆÈÜÒºÓë´×ËáÈÜҺǡºÃÍêÈ«·´Ó¦µÄµãλÓÚ0-bÖ®¼ä£¬¹ÊD´íÎó£»
¹ÊÑ¡£ºD£®

µãÆÀ ±¾Ì⿼²éÁËËá¼î»ìºÏÈÜÒº¶¨ÐÔÅжϣ¬¸ù¾ÝÈõµç½âÖʵĵçÀëÌص㡢ÈÜÒºÖеÄÈÜÖʼ°µçºÉÊغãÀ´·ÖÎö½â´ð¼´¿É£¬ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
5£®¢ñ£®A¡¢B¡¢C¡¢XÊÇÖÐѧ»¯Ñ§³£¼ûÎïÖÊ£¬¾ùÓɶÌÖÜÆÚÔªËØ×é³É£¬×ª»¯¹ØϵÈçͼËùʾ£®   ÈôA¡¢B¡¢CµÄÑæÉ«·´Ó¦¾ù³Ê»ÆÉ«£¬Ë®ÈÜÒº¾ùΪ¼îÐÔ£®
¢ÙXÊôÓڵľ§ÌåÀàÐÍÊÇÀë×Ó¾§Ì壮
¢Ú½«4.48L£¨±ê×¼×´¿öÏ£©XͨÈë100mL 3mol/L  AµÄË®ÈÜÒººó£¬ÈÜÒºÖз¢ÉúµÄÀë  ×Ó·´Ó¦·½³ÌʽΪ2CO2+3OH-=CO32-+HCO3-+H2O£®
¢Û×ÔÈ»½çÖдæÔÚB¡¢CºÍH2O°´Ò»¶¨±ÈÀý½á¾§¶ø³ÉµÄ¹ÌÌ壮ȡһ¶¨Á¿¸Ã¹ÌÌåÈÜÓÚË®Åä
³É100mLÈÜÒº£¬²âµÃÈÜÈÜÖнðÊôÑôÀë×ÓµÄŨ¶ÈΪ0.5mol/L£®ÈôÈ¡ÏàͬÖÊÁ¿µÄ¹ÌÌå¼ÓÈÈÖÁºãÖØ£¬Ê£Óà¹ÌÌåµÄÖÊÁ¿Îª2.65 g
¢ò£®ÎÞ»úÑÎAÊÇҽѧÉϳ£ÓõÄÕò¾²´ßÃßÒ©£¬ÓÉÁ½ÖÖÔªËØ×é³É£®½«ÆäÈÜÓÚË®£¬Í¨ÈëÊÊÁ¿»ÆÂÌÉ«ÆøÌåB£¬È»ºóÏò·´Ó¦ºóµÄÈÜÒºÖмÓÈëËÄÂÈ»¯Ì¼²¢Õñµ´¡¢¾²Öã¬ÈÜÒº·Ö²ã£¬Ï²ãÒºÌå³Ê³ÈºìÉ«£®·ÖÒººóÈ¡ÉϲãÈÜÒº£¬¾­ÔªËØ·ÖÎö£¬ÈÜÖÊΪƯ°×·ÛµÄÖ÷Òª³É·ÖÖ®Ò»£¬Íù´ËÈÜҺͨÈëCO2ºÍNH3¿É»ñµÃÄÉÃײÄÁÏEºÍï§Ì¬µª·ÊF£®
£¨1£©ÎÞ»úÑÎAÖÐÑôÀë×ӵĽṹʾÒâͼ
£¨2£©CO2ºÍNH3Á½ÆøÌåÖУ¬Ó¦¸ÃÏÈͨÈëÈÜÒºÖеÄÊÇNH3£¨Ìѧʽ£©£¬Ð´³öÖƱ¸EºÍF µÄÀë×Ó·´Ó¦·½³ÌʽCa2++CO2+2NH3+H2O=CaCO3¡ý+2NH4+£®

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

¾«Ó¢¼Ò½ÌÍø