ÌâÄ¿ÄÚÈÝ

ÏÖÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖÇ¿µç½âÖÊ£¬ËüÃÇÔÚË®ÖпɵçÀë²úÉúÏÂÁÐÀë×Ó£¨¸÷ÖÖÀë×Ó²»Öظ´£©£®
ÑôÀë×ÓH+¡¢NA+¡¢Al3+¡¢Ag+¡¢Ba2+
ÒõÀë×ÓOH-¡¢Cl-¡¢CO32-¡¢NO3-¡¢SO42-
¢ÙA¡¢BÁ½ÈÜÒº³Ê¼îÐÔ£»C¡¢D¡¢EÈÜÒº³ÊËáÐÔ£®
¢ÚAÈÜÒºÓëEÈÜÒº·´Ó¦¼ÈÓÐÆøÌåÓÖÓгÁµí²úÉú£»AÈÜÒºÓëCÈÜÒº·´Ó¦Ö»ÓÐÆøÌå²úÉú£¨³Áµí°üÀ¨Î¢ÈÜÎÏÂͬ£©£®
¢ÛDÈÜÒºÓëÁíÍâËÄÖÖÈÜÒº·´Ó¦¶¼ÄܲúÉú³Áµí£»CÖ»ÄÜÓëD·´Ó¦²úÉú³Áµí£®
ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©·Ö±ðд³öA¡¢B¡¢C¡¢D¡¢EµÄ»¯Ñ§Ê½£ºA
 
¡¢B
 
¡¢C
 
¡¢D
 
¡¢E
 
£»
£¨2£©Ð´³öA¡¢E·´Ó¦µÄÀë×Ó·½³Ìʽ£º
 
£»
£¨3£©½«CÈÜÒºÖðµÎ¼ÓÈëµÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄAÈÜÒºÖУ¬·´Ó¦ºóÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ£º
 
£®
£¨4£©ÒÑÖª£ºNaOH£¨aq£©+HNO3£¨aq£©¨TNaNO3£¨aq£©+H2O£¨l£©£»¡÷H=-Q kJ?mol-1£®Ð´³öBÓëCÏ¡ÈÜÒº·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
 
£®
£¨5£©ÔÚ100mL 0.1mol?L-1 EÈÜÒºÖУ¬ÖðµÎ¼ÓÈë40mL 1.6mol?L-1 NaOHÈÜÒº£¬×îÖյõ½³ÁµíÎïÖʵÄÁ¿Îª
 
mol£®
¿¼µã£º³£¼ûÀë×ӵļìÑé·½·¨,¼¸×éδ֪ÎïµÄ¼ìÑé
רÌ⣺ÎïÖʼìÑé¼ø±ðÌâ
·ÖÎö£º¢ÙA¡¢BÁ½ÈÜÒº³Ê¼îÐÔ£¬½áºÏÀë×ӵĹ²´æ¿ÉÖª£¬Ó¦ÎªBa£¨OH£©2¡¢Na2CO3ÖеÄÒ»ÖÖ£¬C¡¢D¡¢EÈÜÒº³ÊËáÐÔ£¬Ó¦ÎªAgNO3¡¢ÁòËáÂÁ¡¢HClÖеÄÒ»ÖÖ£»
¢ÚAÈÜÒºÓëEÈÜÒº·´Ó¦¼ÈÓÐÆøÌåÓÖÓгÁµí²úÉú£»AÈÜÒºÓëCÈÜÒº·´Ó¦Ö»ÓÐÆøÌå²úÉú£¬ÔòAΪNa2CO3£¬BΪBa£¨OH£©2£¬EΪAl2£¨SO4£©3£¬CΪHCl£»
¢ÛDÈÜÒºÓëÁíÍâËÄÖÖÈÜÒº·´Ó¦¶¼ÄܲúÉú³Áµí£»CÖ»ÄÜÓëD·´Ó¦²úÉú³Áµí£¬ÔòDΪAgNO3£¬È»ºó½áºÏÎïÖʵÄÐÔÖʼ°»¯Ñ§ÓÃÓïÀ´½â´ð£®
½â´ð£º ½â£º¢ÙA¡¢BÁ½ÈÜÒº³Ê¼îÐÔ£¬½áºÏÀë×ӵĹ²´æ¿ÉÖª£¬Ó¦ÎªBa£¨OH£©2¡¢Na2CO3ÖеÄÒ»ÖÖ£¬C¡¢D¡¢EÈÜÒº³ÊËáÐÔ£¬Ó¦ÎªAgNO3¡¢ÁòËáÂÁ¡¢HClÖеÄÒ»ÖÖ£»
¢ÚAÈÜÒºÓëEÈÜÒº·´Ó¦¼ÈÓÐÆøÌåÓÖÓгÁµí²úÉú£»AÈÜÒºÓëCÈÜÒº·´Ó¦Ö»ÓÐÆøÌå²úÉú£¬ÔòAΪNa2CO3£¬BΪBa£¨OH£©2£¬EΪAl2£¨SO4£©3£¬CΪHCl£»
¢ÛDÈÜÒºÓëÁíÍâËÄÖÖÈÜÒº·´Ó¦¶¼ÄܲúÉú³Áµí£»CÖ»ÄÜÓëD·´Ó¦²úÉú³Áµí£¬ÔòDΪAgNO3£¬
£¨1£©ÓÉÒÔÉÏ·ÖÎö¿ÉÖªAΪNa2CO3£¬BΪBa£¨OH£©2£¬CΪHCl£¬DΪAgNO3£¬EΪAl2£¨SO4£©3£¬
¹Ê´ð°¸Îª£ºNa2CO3£»Ba£¨OH£©2£»HCl£»AgNO3£»Al2£¨SO4£©3£»
£¨2£©AΪNa2CO3£¬EΪAl2£¨SO4£©3£¬Ì¼ËáÄÆÓëÁòËáÂÁÈÜÒº·¢Éú˫ˮ½â·´Ó¦Éú³ÉÇâÑõ»¯ÂÁ³ÁµíºÍ¶þÑõ»¯Ì¼ÆøÌ壬·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º3CO32-+2Al3++3H2O=2Al£¨OH£©3¡ý+3CO2¡ü£¬
¹Ê´ð°¸Îª£º3CO32-+2Al3++3H2O=2Al£¨OH£©3¡ý+3CO2¡ü£»
£¨3£©AΪNa2CO3£¬CΪHCl£¬½«HClÈÜÒºÖðµÎ¼ÓÈëµÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄ̼ËáÄÆÈÜÒºÖУ¬·´Ó¦Éú³É̼ËáÇâÄÆ£¬Ì¼ËáÇâ¸ùÀë×Ó²¿·ÖË®½â£¬ÈÜÒºÏÔʾ¼îÐÔ£¬Ôòc£¨OH-£©£¾c£¨H+£©¡¢c£¨Na+£©£¾c£¨C1-£©£¾c£¨HCO3-£©£¬·´Ó¦ºóÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ£ºc£¨Na+£©£¾c£¨C1-£©£¾c£¨HCO3-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨CO32-£©£¨»òc£¨Na+£©£¾c£¨C1-£©£¾c£¨HCO3-£©£¾c£¨OH-£©£¾c£¨H+£©£©£¬
¹Ê´ð°¸Îª£ºc£¨Na+£©£¾c£¨C1-£©£¾c£¨HCO3-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨CO32-£©£¨»òc£¨Na+£©£¾c£¨C1-£©£¾c£¨HCO3-£©£¾c£¨OH-£©£¾c£¨H+£©£©£»
£¨4£©BΪBa£¨OH£©2£¬CΪHCl£¬BÓëCµÄÏ¡ÈÜÒº·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºOH-£¨aq£©+H+£¨aq£©=H2O£¨1£©¡÷H=-a kJ/mol»ò1/2Ba£¨OH£©2£¨aq£©+HC1£¨aq£©=1/2BaC12£¨aq£©+H2O£¨1£©¡÷H=-a kJ/mol»òBa£¨OH£©2£¨aq£©+2HC1£¨aq£©=BaC12£¨aq£©+2H2O£¨1£©¡÷H=-2a kJ/mol£¬
¹Ê´ð°¸Îª£ºOH-£¨aq£©+H+£¨aq£©=H2O£¨1£©¡÷H=-a kJ/mol»ò1/2Ba£¨OH£©2£¨aq£©+HC1£¨aq£©=1/2BaC12£¨aq£©+H2O£¨1£©¡÷H=-a kJ/mol
»òBa£¨OH£©2£¨aq£©+2HC1£¨aq£©=BaC12£¨aq£©+2H2O£¨1£©¡÷H=-2a kJ/mol£»
£¨5£©EΪAl2£¨SO4£©3£¬100mL 0.1mol?L-1 EÈÜÒºÖУºn£¨Al2£¨SO4£©3£©=0.1L¡Á0.1mol/L=0.01mol£¬n£¨NaOH£©=0.04L¡Á1.6mol/L=0.064mol£»
Ôòn£¨Al3+£©=0.02mol£¬n£¨OH-£©=0.064mol£¬
·¢Éú£ºAl3++3OH-=Al£¨OH£©3¡ý£¬
 0.02mol 0.06mol  0.02mol
ËùÒÔ0.004molÇâÑõ¸ùÀë×ÓÈܽâµÄÇâÑõ»¯ÂÁµÄÎïÖʵÄÁ¿Îª£º
   Al£¨OH£©3+OH-=AlO2-+2H2O£¬
0.004mol   0.004mol
¼´ÈܽâÁË0.004molÇâÑõ»¯ÂÁ£¬
Ôò·´Ó¦ºó£¬Éú³É³ÁµíÇâÑõ»¯ÂÁµÄÎïÖʵÄÁ¿Îª£º0.02mol-0.004mol=0.016mol£¬
¹Ê´ð°¸Îª£º0.016mol£®
µãÆÀ£º±¾Ì⿼²é֪ʶµã½Ï¶à£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦µÄ¿¼²é£¬¸ù¾ÝÀë×ӵĹ²´æ¡¢Àë×ÓµÄÐÔÖÊÀ´ÍƶÏÎïÖÊÊǽâ´ð±¾ÌâµÄ¹Ø¼ü£¬×ÛºÏÐÔÇ¿£¬ÄѶȽϴó£¬Ñ§ÉúÐèÊìϤˮ½â¡¢ÈÈ»¯Ñ§·´Ó¦·½³Ìʽ¡¢Àë×Ó·´Ó¦µÈ֪ʶÀ´½â´ð£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø