Question

9£®£¨1£©ÒÑÖª£ºFe£¨s£©+$\frac{1}{2}$O₂£¨g£©=FeO£¨s£©£¬¡÷H=-272.0KJ•mol^-1
2Al£¨s£©+$\frac{3}{2}$O2£¨g£©=Al₂O₃£¨s£©£¬¡÷H=-1675.7KJ•mol^-1
AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ2Al£¨s£©+3FeO£¨s£©¨TAl₂O₃£¨s£©+3Fe£¨s£©¡÷H=-859.7 kJ•mol^-1£®
£¨2£©Ä³¿ÉÄæ·´Ó¦ÔÚ²»Í¬Ìõ¼þϵķ´Ó¦Àú³Ì·Ö±ðΪA¡¢B£¬ÈçͼËùʾ£®
¢Ù¾ÝͼÅжϸ÷´Ó¦ÊÇÎü £¨Ìî¡°Îü¡±»ò¡°·Å¡±£©ÈÈ·´Ó¦£¬µ±·´Ó¦´ïµ½Æ½ºâºó£¬ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȣ¬·´Ó¦ÎïµÄת»¯ÂÊÔö´ó £¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©£»
¢ÚÆäÖÐBÀú³Ì±íÃ÷´Ë·´Ó¦²ÉÓõÄÌõ¼þΪD £¨Ñ¡ÌîÐòºÅ£©£®
A£®Éý¸ßζȠB£®Ôö´ó·´Ó¦ÎïµÄŨ¶È  C£®½µµÍζȠD£®Ê¹Óô߻¯¼Á
£¨3£©1000¡æʱ£¬ÁòËáÄÆÓëÇâÆø·¢ÉúÏÂÁз´Ó¦£ºNa₂SO₄£¨s£©+4H₂£¨g£©£¬Na₂S£¨s£©+4H₂O£¨g£©£¬¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪk=$\frac{{c}^{4}£¨{H}_{2}O£©}{{c}^{4}£¨{H}_{2}£©}$£®
£¨4£©³£ÎÂÏ£¬Èç¹ûÈ¡0.1mol•L^-1HAÈÜÒºÓë0.1mol•L^-1NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©£¬²âµÃ»ìºÏÒºµÄpH=8£®
¢Ù»ìºÏÒºÖÐÓÉË®µçÀë³öµÄOH^-Ũ¶ÈÓë0.1mol•L^-1NaOHÈÜÒºÖÐÓÉË®µçÀë³öµÄOH^-Ũ¶ÈÖ®±ÈΪ10⁷£º1£»
¢ÚÒÑÖªNH₄AÈÜҺΪÖÐÐÔ£¬ÓÖÖª½«HAÈÜÒº¼Óµ½Na₂CO₃ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ÊÔÍƶϣ¨NH₄£©₂CO₃ÈÜÒºµÄpH£¾7£¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ý¸Ç˹¶¨ÂÉÊéдĿ±êÈÈ»¯Ñ§·½³Ìʽ£»
£¨2£©¢ÙÓÉͼ¿ÉÖª£¬·´Ó¦ÎïµÄ×ÜÄÜÁ¿µÍÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬¸Ã·´Ó¦ÎªÎüÈÈ·´Ó¦£»Éý¸ßζÈƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬¾Ý´Ë½â´ð£»
¢ÚÓÉͼ¿ÉÖª£¬·´Ó¦Àú³ÌBÓëAÏà±È£¬¸Ä±ä·´Ó¦Àú³Ì£¬Ó¦ÊÇʹÓô߻¯¼Á£»
£¨3£©»¯Ñ§Æ½ºâ³£Êý£¬ÊÇÖ¸ÔÚÒ»¶¨Î¶ÈÏ£¬¿ÉÄæ·´Ó¦´ïµ½Æ½ºâʱ¸÷Éú³ÉÎïŨ¶ÈµÄ»¯Ñ§¼ÆÁ¿Êý´ÎÃݵij˻ý³ýÒÔ¸÷·´Ó¦ÎïŨ¶ÈµÄ»¯Ñ§¼ÆÁ¿Êý´ÎÃݵij˻ýËùµÃµÄ±ÈÖµ£¬¾Ý´ËÊéд£»
£¨4£©È¡0.1mol•L^-1HAÈÜÒºÓë0.1mol•L^-1NaOHÈÜÒºµÈÌå»ý»ìºÏ£¨»ìºÏºóÈÜÒºÌå»ýµÄ±ä»¯ºöÂÔ²»¼Æ£©£¬²âµÃ»ìºÏÒºµÄpH=8£¬Ôò˵Ã÷HAÊÇÈõËᣬº¬ÓÐÈõ¸ùÀë×ÓµÄÑδٽøË®µçÀ룬
¢Ù¸ù¾ÝÑÎÈÜÒºµÄpH¼ÆËãË®µçÀë³öµÄc£¨OH^-£©£¬¸ù¾Ýc£¨NaOH£©¼°Ë®µÄÀë×Ó»ý³£Êý¼ÆËãË®ÖÐÇâÀë×ÓŨ¶È£¬ÈÜÒºÖÐÇâÀë×ÓŨ¶ÈµÈÓÚË®µçÀë³öµÄc£¨OH^-£©£¬¾Ý´Ë½â´ð£»
¢Ú¸ù¾ÝÌâÒâÖª£¬HAµÄËáÐÔ±È̼ËáÇ¿£¬NH₄AÈÜҺΪÖÐÐÔ£¬ËµÃ÷笠ùÀë×Ӻ͸ÃËá¸ùÀë×ÓË®½â³Ì¶ÈÏàͬ£¬Óɴ˵Ã֪笠ùÀë×ÓË®½â³Ì¶ÈСÓÚ̼Ëá¸ùÀë×Ó£¬ËùÒÔ£¨NH₄£©₂CO₃ÖÐÈÜÒº³Ê¼îÐÔ£®

½â´ð ½â£º£¨1£©¢ÙFe£¨s£©+$\frac{1}{2}$O₂£¨g£©¨TFeO£¨s£©¡÷H₁=-272.0kJ•mol^-1
¢Ú2Al£¨s£©+$\frac{3}{2}$O₂£¨g£©¨TAl₂O₃£¨s£©¡÷H₂=-1675.7kJ•mol^-1
½«·½³Ìʽ¢Ú-¢Ù¡Á3µÃ£º2Al£¨s£©+3FeO£¨s£©¨TAl₂O₃£¨s£©+3Fe£¨s£©¡÷H=-1675.7kJ•mol^-1-£¨3¡Á-272.0kJ•mol^-1£©=-859.7 kJ•mol^-1£¬
¹Ê´ð°¸Îª£º2Al£¨s£©+3FeO£¨s£©¨TAl₂O₃£¨s£©+3Fe£¨s£©¡÷H=-859.7 kJ•mol^-1£»
£¨2£©¢ÙÓÉͼ¿ÉÖª£¬·´Ó¦ÎïµÄ×ÜÄÜÁ¿µÍÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬¸Ã·´Ó¦ÎªÎüÈÈ·´Ó¦£»Éý¸ßζÈƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬·´Ó¦ÎïµÄת»¯ÂÊÔö´ó£»
¹Ê´ð°¸Îª£ºÎü£»Ôö´ó£»
 ¢ÚÓÉͼ¿ÉÖª£¬·´Ó¦Àú³ÌBÓëAÏà±È£¬¸Ä±ä·´Ó¦Àú³Ì£¬×îºó´ïµ½ÏàͬµÄƽºâ״̬£¬¸Ä±äµÄÌõ¼þÓ¦ÊÇʹÓô߻¯¼Á£¬½µµÍ·´Ó¦µÄ»î»¯ÄÜ£»
¹Ê´ð°¸Îª£ºD£»
£¨3£©Na₂SO₄£¨S£©+4H₂£¨g£©¨TNa₂S£¨s£©+4H₂O£¨g£©µÄƽºâ³£Êý±í´ïʽk=$\frac{{c}^{4}£¨{H}_{2}O£©}{{c}^{4}£¨{H}_{2}£©}$£»
¹Ê´ð°¸Îª£º$\frac{{c}^{4}£¨{H}_{2}O£©}{{c}^{4}£¨{H}_{2}£©}$£»
£¨4£©¢ÙÑÎÈÜÒºÖÐc£¨OH^-£©=$\frac{1{0}^{-14}}{1{0}^{-8}}$=10^-6 mol/L£¬0.1mol•L^-1NaOHÈÜÒºÖÐc£¨H⁺£©µÈÓÚÓÉË®µçÀë³öµÄÇâÑõ¸ùÀë×ÓŨ¶È£¬¼´c£¨H⁺£©=c_Ë®µçÀ루OH^-£©=$\frac{1{0}^{-14}}{0.1}$=10^-13 mol/L£¬ËùÒÔ»ìºÏÒºÖÐÓÉË®µçÀë³öµÄOH^-Ũ¶ÈÓë0.1mol•L^-1NaOHÈÜÒºÖÐÓÉË®µçÀë³öµÄOH^-Ũ¶ÈÖ®±È=10^-6 mol/L£º10^-13 mol/L=10⁷£º1£¬
¹Ê´ð°¸Îª£º10⁷£º1£»
¢Ú½«HAÈÜÒº¼Óµ½Na₂CO₃ÈÜÒºÖÐÓÐÆøÌå·Å³ö£¬ËµÃ÷HAµÄËáÐÔ±È̼ËáµÄÇ¿£¬NH₄AÈÜҺΪÖÐÐÔ£¬ËµÃ÷ÏàͬÌõ¼þÏ£¬°±Ë®ºÍHAµÄµçÀë³Ì¶ÈÏàͬ£¬ËùÒÔ£¨NH₄£©₂CO₃ÖÐ笠ùÀë×ÓµÄË®½â³Ì¶ÈСÓÚ̼Ëá¸ùÀë×ÓµÄË®½â³Ì¶È£¬ËùÒÔÈÜÒºµÄpH£¾7£»
¹Ê´ð°¸Îª£º£¾£»

µãÆÀ ±¾Ìâ×ÛºÏÐԽϴ󣬻¯Ñ§·´Ó¦ÄÜÁ¿±ä»¯Í¼Ïó·ÖÎöÅжϣ¬·´Ó¦»î»¯ÄܺÍìʱäµÄÀí½âÓ¦Ó㬵ç½âÖÊÈÜÒºÓйؼÆËã¡¢ÑÎÀàË®½âµÈ£¬ÊǶÔѧÉú×ÛºÏÄÜÁ¦µÄ¿¼²é£¬ÕÆÎÕ»ù´¡Êǹؼü£¬ÌâÄ¿ÄѶÈÖеȣ®