Question

ÖƱ¸Ð¿Ó¡Ë¢µç·°åÊÇÓÃÏ¡ÏõËḯʴп°å£¬²úÉúµÄ·ÏÒº³Æ¡°ÀðåÒº¡±£®¡°ÀðåÒº¡±Öк¬ÏõËáпÍ⣬»¹º¬ÓÐ×ÔÀ´Ë®´øÈëµÄCl^-ºÍFe³⁺£®ÔÚʵÑéÊÒÀÓá°ÀðåÒº¡±ÖÆÈ¡ZnSO₄?7H₂OµÄ¹ý³ÌÈçÏ£º

£¨1£©ÈôÏ¡ÏõËḯʴп°åʱ»¹Ô­²úÎïΪN₂O£¬Ñõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±ÈÊÇ

 

£®
£¨2£©Èô²½Öè¢ÙµÄpH£¾12£¬ÔòZn£¨OH£©₂ÈܽâÉú³ÉËÄôÇ»ùºÏпËáÄÆ£®Ð´³öZn£¨OH£©₂±»ÈܽâµÄÀë×Ó·½³Ìʽ

 

£®
£¨3£©ÂËÒºDÖгýÁ˺¬ÓÐOH^-Àë×ÓÍ⣬»¹º¬ÓеÄÒõÀë×ÓÓÐ

 

£¨ÌîÀë×Ó·ûºÅ£©£®
£¨4£©ÈôÂËÒºEµÄpH=4£¬c£¨Zn²⁺£©=2mol?L^-1£¬c£¨Fe³⁺£©=2.6¡Ál0^-9mol?L^-1£¬ÄÜÇóµÃµÄÈܶȻýÊÇ

 

£¨ÌîÑ¡Ï£®
A£®K_sp[Zn£¨OH£©₂]B£®K_sp[Zn£¨OH£©₂]ºÍK_sp[Fe£¨OH£©₃]C£®K_sp[Fe£¨OH£©₃]
£¨5£©²½Öè¢ÛÒª¿ØÖÆpHÔÚÒ»¶¨·¶Î§£®ÊµÑéÊÒÓÃpHÊÔÖ½²â¶¨ÈÜÒºpHµÄ·½·¨ÊÇ£º

 

£®
£¨6£©ÒÑÖª£º
¢ÙFe£¨OH£©₃£¨s£©?Fe³⁺£¨aq£©+OH^-£¨aq£©¡÷H=aKJ?mol^-1
¢ÚH₂O£¨l£©?H⁺£¨aq£©+OH^-£¨aq£©¡÷H=bKJ?mol^-1
Çëд³öFe³⁺·¢ÉúË®½â·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£®

 

£®
Èô¢ÙµÄÈܶȻý³£ÊýΪK_SP£¬¢ÚµÄÀë×Ó»ý³£ÊýΪK_W£¬Fe³⁺·¢ÉúË®½â·´Ó¦µÄƽºâ³£ÊýK=

 

£®£¨Óú¬K_SP¡¢K_WµÄ´úÊýʽ±íʾ£©

Answer 1

¿¼µã£ºÖƱ¸ÊµÑé·½°¸µÄÉè¼Æ,ÈÈ»¯Ñ§·½³Ìʽ,ÄÑÈܵç½âÖʵÄÈܽâƽºâ¼°³Áµíת»¯µÄ±¾ÖÊ

רÌ⣺ʵÑéÉè¼ÆÌâ

·ÖÎö£º£¨1£©ÈôÏ¡ÏõËḯʴп°åʱ»¹Ô­²úÎïΪN₂O£¬Ôò·´Ó¦µÄ·½³ÌʽΪ4Zn+10HNO₃=4Zn£¨NO₃£©₂+N₂O¡ü+5H₂O£¬¸ù¾Ý·½³ÌʽÅжÏÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±È£»
£¨2£©¸ù¾Ý·´Ó¦ÎïºÍÉú³ÉÎïÈ·¶¨·´Ó¦µÄÀë×Ó·½³Ìʽ£»
£¨3£©¹ÌÌå¾ßÓÐÎü¸½ÐÔ£¬³ÁµíA¿ÉÎü¸½ÈÜÒºÖеÄCl^-¡¢NO₃^-µÈÀë×Ó£¬ÔòÂËÒºÖÐÓ¦º¬ÓÐCl^-¡¢NO₃^-£»
£¨4£©ÂËÒºEµÄpH=4£¬µÃµ½µÄ³ÁµíΪFe£¨OH£©₃£¬Ö»ÄܼÆËãFe£¨OH£©₃µÄÈܶȻý£»
£¨5£©²â¶¨ÈÜÒºpHʱ£¬½«pHÊÔÖ½·ÅÔÚ²£Á§Æ¬ÉÏ£¬Óò£Á§°ôպȡ´ý²âÒºµãÔÚÊÔÖ½Öм䣬´ýÑÕÉ«±ä»¯Îȶ¨ºóÓë±ÈÉ«¿¨¶Ô±ÈÄ㣬¶Á³ö¶ÁÊý¼´¿É£»
£¨6£©ÀûÓøÇ˹¶¨ÂɼÆËã·´Ó¦ÈÈ£¬¿Éд³öFe³⁺·¢ÉúË®½â·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£¬ÀûÓâ٢Úƽºâ³£Êý¿ÉÈ·¶¨Fe³⁺·¢ÉúË®½â·´Ó¦µÄƽºâ³£Êý£®

½â´ð£º ½â£º£¨1£©ÈôÏ¡ÏõËḯʴп°åʱ»¹Ô­²úÎïΪN₂O£¬Ôò·´Ó¦µÄ·½³ÌʽΪ4Zn+10HNO₃=4Zn£¨NO₃£©₂+N₂O¡ü+5H₂O£¬ÓÉ·½³Ìʽ¿ÉÖª·´Ó¦ÖÐHNO₃ΪÑõ»¯¼Á£¬ZnΪ»¹Ô­¼Á£¬µ±4molZn²Î¼Ó·´Ó¦Ê±£¬ÓÐ2molHNO₃±»»¹Ô­£¬ÔòÑõ»¯¼ÁÓ뻹ԭ¼ÁµÄÎïÖʵÄÁ¿Ö®±È1£º2£¬¹Ê´ð°¸Îª£º1£º2£»
£¨2£©Zn£¨OH£©₂ÈܽâÉú³ÉËÄôÇ»ùºÏпËáÄÆ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪZn£¨OH£©₂+2OH^-=[Zn£¨OH£©₄]^2-£¬¹Ê´ð°¸Îª£ºZn£¨OH£©₂+2OH^-=[Zn£¨OH£©₄]^2-£»
£¨3£©¹ÌÌå¾ßÓÐÎü¸½ÐÔ£¬³ÁµíA¿ÉÎü¸½ÈÜÒºÖеÄCl^-¡¢NO₃^-µÈÀë×Ó£¬ÔòÂËÒºÖÐÓ¦º¬ÓÐCl^-¡¢NO₃^-£¬¹Ê´ð°¸Îª£ºCl^-¡¢NO₃^-£»
£¨4£©ÂËÒºEµÄpH=4£¬µÃµ½µÄ³ÁµíΪFe£¨OH£©₃£¬Ö»ÄܼÆËãFe£¨OH£©₃µÄÈܶȻý£¬¹Ê´ð°¸Îª£ºC£»
£¨5£©²â¶¨ÈÜÒºpHʱ£¬½«pHÊÔÖ½·ÅÔÚ²£Á§Æ¬ÉÏ£¬Óò£Á§°ôպȡ´ý²âÒºµãÔÚÊÔÖ½Öм䣬´ýÑÕÉ«±ä»¯Îȶ¨ºóÓë±ÈÉ«¿¨¶Ô±È£¬¶Á³ö¶ÁÊý¼´¿É£¬×¢Òâ²»ÄÜÊÂÏÈʪÈóÊÔÖ½£¬¸ü²»Äܽ«ÊÔÖ½²åÔÚÈÜÒºÖУ¬¹Ê´ð°¸Îª£º½«pHÊÔÖ½·ÅÔÚ²£Á§Æ¬ÉÏ£¬Óò£Á§°ôպȡ´ý²âÒºµãÔÚÊÔÖ½Öм䣬´ýÑÕÉ«±ä»¯Îȶ¨ºóÓë±ÈÉ«¿¨¶Ô±È£¬¶Á³ö¶ÁÊý£»
£¨6£©Fe³⁺·¢ÉúË®½â·´Ó¦µÄ·½³ÌʽΪFe³⁺+3H₂O?Fe£¨OH£©₃+3H⁺£¬
ÒÑÖª¢ÙFe£¨OH£©₃£¨s£©?Fe³⁺£¨aq£©+3OH^-£¨aq£©¡÷H=aKJ?mol^-1
¢ÚH₂O£¨l£©?H⁺£¨aq£©+OH^-£¨aq£©¡÷H=bKJ?mol^-1£¬
ÔòÀûÓøÇ˹¶¨Âɽ«¢Ú¡Á3-¢Ù¿ÉµÃFe³⁺£¨aq£©+3H₂O£¨l£©?Fe£¨OH£©₃£¨aq£©+3H⁺£¨aq£©¡÷H=£¨3b-a£©KJ?mol^-1£¬
ÔòFe³⁺·¢ÉúË®½â·´Ó¦µÄƽºâ³£ÊýK=

Kw3

Ksp

£¬
¹Ê´ð°¸Îª£ºFe³⁺£¨aq£©+3H₂O£¨l£©?Fe£¨OH£©₃£¨aq£©+3H⁺£¨aq£©¡÷H=£¨3b-a£©KJ?mol^-1£»

Kw3

Ksp

£®

µãÆÀ£º±¾Ì⿼²éÎïÖʵÄÖƱ¸£¬Îª¸ß¿¼³£¼ûÌâÐÍ£¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦¡¢ÊµÑéÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦µÄ¿¼²é£¬×¢Òâ°ÑÎÕÌâ¸øÐÅÏ¢£¬´ðÌâʱעÒâÉóÌ⣬ÄѶÈÖеȣ®