16£®ÎïÖÊÔÚË®ÖпÉÄÜ´æÔÚµçÀëƽºâ¡¢ÑεÄË®½âƽºâºÍ³ÁµíµÄÈܽâƽºâ£®Çë¸ù¾ÝËùѧ֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©³£ÎÂÏ£¬ÓÐŨ¶È¾ùΪ1mol•L-1µÄÏÂÁÐ4ÖÖÈÜÒº£º
A£®AlCl3ÈÜÒº   B£®NaHCO3ÈÜÒº   C£®NH4HSO4ÈÜÒº   D£®NaOHÈÜÒº£®
¢ÙÕâËÄÖÖÈÜÒºpHÓÉ´óµ½Ð¡µÄ˳ÐòÊÇDBAC£¨ÌîÐòºÅ£©£»
¢ÚBÈÜÒº¸÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇc£¨Na+£©£¾c£¨HCO3-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨CO32-£©£»
¢ÛµÈÎïÖʵÄÁ¿Å¨¶ÈµÄNH4ClÈÜÒºÓëCÈÜÒºÏà±È½Ï£¬c£¨NH4+£©£ºÇ°Õߣ¼ºóÕߣ¨Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±£©£»
¢ÜÈô½«CºÍD°´ÎïÖʵÄÁ¿Ö®±È1£º2»ìºÏ£¬Ôò·´Ó¦Àë×Ó·½³ÌʽΪNH4++H++2OH-=NH3•H2O+H20£®
£¨2£©ÈÜÒºÖÐijÀë×ÓÎïÖʵÄÁ¿Å¨¶ÈµÍÓÚ1.0¡Á10-5mol•L-1ʱ£¬¿ÉÈÏΪÒѳÁµíÍêÈ«£®ÏÖÏòÒ»¶¨Å¨¶ÈµÄAlCl3ºÍFeCl3µÄ»ìºÏÈÜÒºÖÐÖðµÎ¼ÓÈ백ˮ£¬µ±Fe3+ÍêÈ«³Áµíʱ£¬²â¶¨c£¨Al3+£©=0.2mol•L-1£®´ËʱËùµÃ³ÁµíÖв»º¬ÓУ¨Ìî¡°»¹º¬ÓС±»ò¡°²»º¬ÓС±£©Al£¨OH£©3£®Çëд³ö¹ý³ÌFe3+ÍêÈ«³Áµíʱ£¬c£¨OH-£©3=$\frac{Ksp[Fe£¨OH£©_{3}]}{c£¨F{e}^{3+}£©}$=$\frac{4.0¡Á1{0}^{-38}}{1.0¡Á1{0}^{-5}}$=4.0¡Á10-33£¬c£¨Al3+£©•£¨OH-£©3=0.2¡Á4.0¡Á10-33=8¡Á10-34£¼Ksp[Al£¨OH£©3]£¬ËùÒÔûÓÐÉú³ÉAl£¨OH£©3³Áµí£®£¨ÒÑÖª25¡æʱKap[Fe£¨OH£©3]=4.0¡Á10-38£¬Kap[Al£¨OH£©3=1.1¡Á10-33]£©£®

·ÖÎö £¨1£©A£®AlCl3ΪǿËáÈõ¼îÑΣ¬Ë®½âÏÔËáÐÔ£¬B£®NaHCO3ÈÜÒºÖУ¬Ì¼ËáÇâ¸ùÀë×ÓµÄË®½â³Ì¶È´óÓÚÆäµçÀë³Ì¶È£¬ÈÜÒºÏÔÈõ¼îÐÔ£¬C£®NH4HSO4ÈÜÒºÖУ¬ÁòËáÇâï§Äܹ»ÍêÈ«µçÀ룬ÈÜÒºËáÐÔºÜÇ¿£¬D£®NaOHΪһԪǿ¼î£¬Äܹ»ÍêÈ«µçÀ룬¾Ý´Ë·ÖÎö£»
£¨2£©¾Ý³ÁµíµÄÈܶȻý£¬Fe3+ÍêÈ«³Áµíʱ£¬c£¨OH-£©3=$\frac{Ksp[Fe£¨OH£©_{3}]}{c£¨F{e}^{3+}£©}$¼ÆËãÇâÑõ¸ùÀë×ÓŨ¶È£¬ÔÙ¾ÝÂÁÀë×ÓŨ¶È¼ÆË㣮

½â´ð ½â£º£¨1£©¢ÙA£®AlCl3ΪǿËáÈõ¼îÑΣ¬Ë®½âÏÔËáÐÔ£¬B£®NaHCO3ÈÜÒºÖУ¬Ì¼ËáÇâ¸ùÀë×ÓµÄË®½â³Ì¶È´óÓÚÆäµçÀë³Ì¶È£¬ÈÜÒºÏÔÈõ¼îÐÔ£¬C£®NH4HSO4ÈÜÒºÖУ¬ÁòËáÇâï§Äܹ»ÍêÈ«µçÀ룬ÈÜÒºËáÐÔºÜÇ¿£¬D£®NaOHΪһԪǿ¼î£¬Äܹ»ÍêÈ«µçÀ룬ËÄÖÖÈÜÒºpHÓÉ´óµ½Ð¡µÄ˳ÐòÊÇDBAC£¬¹Ê´ð°¸Îª£ºDBAC£»
¢ÚÄÆÀë×Ó²»Ë®½â£¬Ì¼ËáÇâ¸ùÀë×ÓË®½â£¬ËùÒÔc£¨Na+£©£¾c£¨HCO3-£©£¬Ì¼ËáÇâÄÆË®½â¶øʹÈÜÒº³Ê¼îÐÔ£¬ËùÒÔc£¨OH-£©£¾c£¨H+£©£¬µ«Ë®½âÊÇ΢ÈõµÄ£¬ÈÜÒºÖÐÒõÀë×ÓÖ÷ÒªÒÔ̼ËáÇâ¸ùÀë×Ó´æÔÚ£¬ËùÒÔc£¨HCO3-£©£¾c£¨OH-£©£¬ÈÜÒºÖÐ̼ËáÇâ¸ùÀë×ÓµçÀë³öÇâÀë×ÓºÍ̼Ëá¸ùÀë×Ó¡¢Ë®µçÀë³öÇâÀë×Ó£¬ËùÒÔc£¨H+£©£¾c£¨CO32-£©£¬¹ÊÈÜÒºÖи÷ÖÖÀë×ÓŨ¶È´óС˳ÐòÊÇc£¨Na+£©£¾c£¨HCO3-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨CO32-£©£»
¹Ê´ð°¸Îª£ºc£¨Na+£©£¾c£¨HCO3-£©£¾c£¨OH-£©£¾c£¨H+£©£¾c£¨CO32-£©£»
¢ÛNH4HSO4ÈÜÒºÖУ¬ÁòËáÇâï§Äܹ»ÍêÈ«µçÀ룬ÈÜÒºËáÐÔºÜÇ¿£¬Äܹ»ÒÖÖÆ笠ùÀë×ÓµÄË®½â£¬NH4ClÈÜÒºÖÐÂÈÀë×Ó¶Ô笠ùÀë×ÓµÄË®½âÎÞÓ°Ï죬ËùÒÔµÈÎïÖʵÄÁ¿Å¨¶ÈµÄNH4ClÈÜÒºÓëNH4HSO4ÈÜÒºÏà±È½Ï£¬c£¨NH4+£©ÁòËáÇâï§ÈÜÒºÖд󣬹ʴð°¸Îª£º£¼£»
¢ÜÏàͬŨ¶ÈµÄNH4HSO4ÈÜÒºÓëNaOHÈÜÒº°´ÎïÖʵÄÁ¿Ö®±È1£º2»ìºÏ£¬ï§¸ùÀë×ÓÓëÇâÀë×Ó¶¼Äܹ»ÓëÇâÑõ¸ùÀë×Ó·´Ó¦£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪNH4++H++2OH-=NH3•H2O+H20£¬
¹Ê´ð°¸Îª£ºNH4++H++2OH-=NH3•H2O+H20£»
£¨2£©Fe3+ÍêÈ«³Áµíʱ£¬c£¨OH-£©3=$\frac{Ksp[Fe£¨OH£©_{3}]}{c£¨F{e}^{3+}£©}$=$\frac{4.0¡Á1{0}^{-38}}{1.0¡Á1{0}^{-5}}$=4.0¡Á10-33£¬c£¨Al3+£©•£¨OH-£©3=0.2¡Á4.0¡Á10-33=8¡Á10-34£¼Ksp[Al£¨OH£©3]£¬ËùÒÔûÓÐÉú³ÉAl£¨OH£©3³Áµí£»¹Ê´ËʱËùµÃ³Áµí²»º¬ÓÐAl£¨OH£©3£¬
¹Ê´ð°¸Îª£º²»º¬ÓУ»Fe3+ÍêÈ«³Áµíʱ£¬c£¨OH-£©3=$\frac{Ksp[Fe£¨OH£©_{3}]}{c£¨F{e}^{3+}£©}$=$\frac{4.0¡Á1{0}^{-38}}{1.0¡Á1{0}^{-5}}$=4.0¡Á10-33£¬c£¨Al3+£©•£¨OH-£©3=0.2¡Á4.0¡Á10-33=8¡Á10-34£¼Ksp[Al£¨OH£©3]£¬ËùÒÔûÓÐÉú³ÉAl£¨OH£©3³Áµí£®

µãÆÀ ÌâÄ¿×ÛºÏÐԽϴó£¬Éæ¼°ÈܶȻý¼ÆËãµÈ£¬ÊǶÔ֪ʶµÄ×ÛºÏÔËÓã¬×¢Ò⣨2£©ÖÐÔËÓÃKsp½øÐз¶Î§µÄÈ·¶¨£¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

6£®ÓÃ50mL 0.50mol•L-1ÑÎËáºÍ50mL 0.55mol•L-1 NaOHÈÜÒº·´Ó¦£¬ÊµÑéÖвâµÃÆðʼζÈΪ£¨20.1¡æ£¬ÖÕֹζÈΪ23.4¡æ£¬·´Ó¦ºóÈÜÒºµÄ±ÈÈÈÈÝΪ4.18J•g-1•¡æ-1£¬ÑÎËáºÍNaOHÈÜÒºµÄÃܶȶ¼½üËÆÈÏΪÊÇ1g•cm-3£¬ÔòÖкͷ´Ó¦Éú³É1molˮʱ·ÅÈÈ£¨¡¡¡¡£©
A£®55.2kJB£®391kJC£®336kJD£®1.38kJ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

7£®ÏÂÁйØÓÚ½ºÌåµÄÐðÊöÖУ¬²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®½ºÌåÖзÖÉ¢ÖÊÁ£×ÓµÄÖ±¾¶Ð¡ÓÚ1 nm
B£®½ºÌåÊÇÒ»ÖÖ½éÎÈÌåϵ
C£®ÓÃÂËÖ½ÄÜ·ÖÀ뽺ÌåºÍÐü×ÇÒº
D£®ÀûÓö¡´ï¶ûЧӦ¿ÉÒÔÇø·ÖÈÜÒººÍ½ºÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

4£®ÈÕ³£Éú»îÖо­³£Óõ½µÄÏÂÁÐÎïÖÊÖУ¬ÊôÓÚ´¿¾»ÎïµÄÊÇ£¨¡¡¡¡£©
A£®µ÷ζÓõÄʳ´×B£®²ÍÒûÓõÄÆ¡¾Æ
C£®ÓÃÕôÁóË®ÖƵõĽµÎÂÓõıù¿éD£®³´²ËÓõÄÌú¹ø

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

11£®ÏÂÁи÷ÈÜÒºÖÐNa+Àë×ÓŨ¶È×î´óµÄÊÇ£¨¡¡¡¡£©
A£®4L 0.5mol•L-1NaClÈÜÒºB£®1L 0.3mol•L-1Na2SO4ÈÜÒº
C£®2L 0.15mol•L-1Na2CO3ÈÜÒºD£®0.8L 0.4mol•L-1NaOHÈÜÒº

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º¶àÑ¡Ìâ

1£®ÏÂÁл¯Ñ§Ê½ÖУ¬Ö»±íʾһÖÖÎïÖʵÄÊÇ£¨¡¡¡¡£©
A£®CH2Cl2B£®C2H4Cl2C£®C3H7ClD£®£¨CH3£©3CCH2Cl

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

8£®ÓÃË«ÏßÇűê³öµç×ÓתÒƵķ½ÏòºÍÊýÄ¿£ºKClO3+6HCl¨TKCl+3Cl2¡ü+3H2O£¬Ñõ»¯²úÎïºÍ»¹Ô­²úÎïµÄÖÊÁ¿Ö®±ÈΪ5£º1£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

11£®ÏÂÁÐÓйØÓлúÎïµÄÐÔÖʺÍÓ¦ÓÃÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÌìÈ»Æø¡¢Ê¯ÓͶ¼ÊÇÇå½àµÄ¿ÉÔÙÉúÄÜÔ´
B£®ÒÒÏ©ºÍ1£¬3-¶¡¶þÏ©»¥ÎªÍ¬ÏµÎËüÃǾùÄÜʹBr2Ë®ÍÊÉ«
C£®±½¡¢ÒÒ´¼ºÍÒÒËᶼÄÜ·¢ÉúÈ¡´ú·´Ó¦
D£®Ê¯Ó͵ķÖÁó¿É»ñµÃÒÒÏ©¡¢±ûÏ©µÈ²»±¥ºÍÌþ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

12£®ÔÚÒ»¶¨Ìõ¼þÏÂÄܹ»·¢ÉúË®½â·´Ó¦£¬ÇÒË®½âµÄ×îÖÕ²úÎïÖ»ÓÐÒ»ÖÖµÄÊÇ£¨¡¡¡¡£©
A£®µí·ÛB£®²ÏË¿C£®¶¹ÓÍD£®¾Û±ûÏ©Ëá¼×õ¥

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸