Question

6£®ÒԷϾÉǦËáµç³ØÖеĺ¬Ç¦·ÏÁÏ£¨Pb¡¢PbO¡¢PbO₂¡¢PbSO₄¼°Ì¿ºÚµÈ£©ÎªÔ­ÁÏ£¬ÖƱ¸´ÖǦ£¬ÊµÏÖǦµÄÔÙÉúÀûÓã®Æ乤×÷Á÷³ÌÈçͼ1Ëùʾ£º

ÒÑÖª£ºK_sp£¨PbSO₄£©=1.6¡Á10^-5£¬K_sp£¨PbCO₃£©=3.3¡Á10^-14£®
£¨1£©¹ý³Ì¢ñÖУ¬ÔÚFe²⁺´ß»¯Ï£¬PbºÍPbO₂·´Ó¦Éú³ÉPbSO₄µÄ»¯Ñ§·½³ÌʽÊÇPb+PbO₂+2H₂SO₄2PbSO₄+2H₂O£®
£¨2£©¹ý³Ì¢ñÖУ¬Fe²⁺ ´ß»¯¹ý³Ì¿É±íʾΪ£º
i£®2Fe²⁺+PbO₂+4H⁺+SO₄^2-¨T2Fe³⁺+PbSO₄+2H₂O
ii£® ¡­£®
¢Ùд³öiiµÄÀë×Ó·½³Ìʽ£º2Fe³⁺+Pb+SO₄^2-¨TPbSO₄+2Fe²⁺£®
¢ÚÏÂÁÐʵÑé·½°¸¿É֤ʵÉÏÊö´ß»¯¹ý³Ì£®½«ÊµÑé·½°¸²¹³äÍêÕû£®
a£®ÏòËữµÄFeSO₄ÈÜÒºÖмÓÈëKSCNÈÜÒº£¬ÈÜÒº¼¸ºõÎÞÉ«£¬ÔÙ¼ÓÈëÉÙÁ¿PbO₂£¬ÈÜÒº±äºì£®
b£®È¡aÖкìÉ«ÈÜÒºÉÙÁ¿£¬¼ÓÈë×ãÁ¿Pb£¬³ä·Ö·´Ó¦ºó£¬ºìÉ«ÍËÈ¥£®
£¨3£©¹ý³Ì¢òµÄÄ¿µÄÊÇÍÑÁò£®ÈôÂËÒº2ÖÐc£¨SO₄^2-£©=1.6mol•L^-1£¬c£¨CO₃^2-£©=0.1mol•L^-1£¬ÔòPbCO₃Öзñ£¨Ìî¡°ÊÇ¡±»ò¡°·ñ¡±£©»ìÓÐPbSO₄£®
£¨4£©ÄÆÀë×Ó½»»»Ä¤¹ÌÏàµç½â·¨ÊÇ´Óº¬Ç¦·ÏÁÏÖÐÌáȡǦµÄÒ»ÖÖй¤ÒÕ£¬Æä×°ÖÃÈçͼ2Ëùʾ£®½«º¬Ç¦·ÏÁÏͶÈëÒõ¼«ÊÒ£¬º¬Ç¦·ÏÁÏÖеÄPbSO₄ÓëNaOHÈÜÒº·¢Éú·´Ó¦£ºPbSO₄+3OH^-¨THPbO₂^-+SO₄^2-+H₂O£®
¢ÙbÓëÍâ½ÓµçÔ´µÄ¸º¼«ÏàÁ¬£®
¢Úµç½â¹ý³ÌÖУ¬PbO₂¡¢PbO¡¢HPbO₂^-ÔÚÒõ¼«·Åµç£¬ÆäÖÐPbO₂·ÅµçµÄµç¼«·´Ó¦Ê½ÎªPbO₂+4e^-+2H₂O=Pb+4OH^-£®
¢ÛÓ봫ͳÎÞĤ¹ÌÏàµç½â·¨Ïà±È£¬Ê¹ÓÃÄÆÀë×Ó½»»»Ä¤¿ÉÒÔÌá¸ßPbÔªËصÄÀûÓÃÂÊ£¬Ô­ÒòÊÇ×èÖ¹HPbO₂^-½øÈëÑô¼«ÊÒ±»Ñõ»¯£®

Answer 1

·ÖÎö ÒԷϾÉǦËáµç³ØÖеĺ¬Ç¦·ÏÁÏ£¨Pb¡¢PbO¡¢PbO₂¡¢PbSO₄¼°Ì¿ºÚµÈ£©ÎªÔ­ÁÏ£¬ÖƱ¸´ÖǦ£¬º¬Ç¦·ÏÁϼÓÈëÁòËáÑÇÌú¡¢Ï¡ÁòËá¼ÓÈÈ·´Ó¦¹ýÂ˵õ½PbSO₄´ÖÆ·£¬¼ÓÈë̼ËáÄÆÈÜÒº¹ýÂ˵õ½PbCO₃´ÖÆ·£¬ÏòÆäÖмÓÈ뽹̿ÔÚ850¡æÉú³ÉǦ£¬
£¨1£©£¨1£©¸ù¾ÝÌâ¸ø»¯Ñ§¹¤ÒÕÁ÷³ÌÖª£¬¹ý³Ì¢ñÖУ¬ÔÚFe²⁺´ß»¯Ï£¬Pb¡¢PbO₂ºÍH₂SO₄·´Ó¦Éú³ÉPbSO₄ºÍË®£»
£¨2£©¢Ù´ß»¯¼Áͨ¹ý²Î¼Ó·´Ó¦£¬¸Ä±ä·´Ó¦Àú³Ì£¬½µµÍ·´Ó¦µÄ»î»¯ÄÜ£¬¼Ó¿ì»¯Ñ§·´Ó¦ËÙÂÊ£¬¶ø±¾ÉíµÄÖÊÁ¿ºÍ»¯Ñ§ÐÔÖÊ·´Ó¦Ç°ºó±£³Ö²»±ä£®¸ù¾ÝÌâ¸øÐÅÏ¢Öª·´Ó¦iÖÐFe²⁺±»PbO₂Ñõ»¯ÎªFe³⁺£¬Ôò·´Ó¦iiÖÐFe³⁺±»Pb»¹Ô­ÎªFe²⁺£¬¾Ý´ËÊéдÀë×Ó·½³Ìʽ£»
¢ÚaʵÑéÖ¤Ã÷·¢Éú·´Ó¦i£¬ÔòbʵÑéÐèÖ¤Ã÷·¢Éú·´Ó¦ii£¬ÊµÑé·½°¸ÎªÈ¡aÖкìÉ«ÈÜÒºÉÙÁ¿£¬¼ÓÈë¹ýÁ¿Pb£¬³ä·Ö·´Ó¦ºó£¬ºìÉ«ÍÊÈ¥£»
£¨3£©ÒòΪKsp£¨PbSO₄£©=1.6¡Á10^-5£¬Ksp £¨PbCO₃£©=3.3¡Á10^-14£¬c£¨SO₄^2-£©=1.6mol•L^-1£¬c£¨CO₃^2-£©=0.1mol•L^-1£¬PbCO₃ÖУºc£¨Pb²⁺£©=$\frac{KsP£¨PbC{O}_{3}£©}{c£¨C{{O}_{3}}^{2-}£©}$=$\frac{3.3¡Á1{0}^{-14}}{0.1}$=3.3¡Á10^-13£¬£»PbSO₄ÖУºQ=c£¨Pb²⁺£©¡Ác£¨SO₄^2-£©=3.3¡Á10^-13¡Á1.6=5.28¡Á10^-13£¼Ksp£¨PbSO₄£©£¬ËµÃ÷PbSO₄ÔÚĸҺÖÐδ±¥ºÍ£¬¼´PbCO₃Öв»»ìÓÐPbSO₄£»
£¨4£©¢Ù¸ù¾ÝÄÆÀë×ÓÏòÒõ¼«Òƶ¯Öª£¬aΪÒõ¼«£¬ÓëµçÔ´¸º¼«ÏàÁ¬£»
¢Úµç½â¹ý³ÌÖУ¬PbO₂ÔÚÒõ¼«µÃµç×Ó£¬µç¼«·´Ó¦Ê½ÎªPbO₂+4e^-+2H₂O=Pb+4OH^-£»
¢ÛÄÆÀë×Ó½»»»Ä¤Ö»ÔÊÐíÄÆÀë×Óͨ¹ý£¬×èÖ¹HPbO₂^-½øÈëÑô¼«ÊÒ±»Ñõ»¯£¬´Ó¶øÌá¸ßPbÔªËصÄÀûÓÃÂÊ£®

½â´ð ½â£º£¨1£©¸ù¾ÝÌâ¸ø»¯Ñ§¹¤ÒÕÁ÷³ÌÖª£¬¹ý³Ì¢ñÖУ¬ÔÚ´ß»¯Ï£¬Pb¡¢PbO₂ºÍH₂SO₄·´Ó¦Éú³ÉPbSO₄ºÍË®£¬»¯Ñ§·½³ÌʽΪ£ºPb+PbO₂+2H₂SO₄2PbSO₄+2H₂O£»
¹Ê´ð°¸Îª£ºPb+PbO₂+2H₂SO₄ 2PbSO₄+2H₂O£»
£¨2£©¢Ù´ß»¯¼Áͨ¹ý²Î¼Ó·´Ó¦£¬¸Ä±ä·´Ó¦Àú³Ì£¬½µµÍ·´Ó¦µÄ»î»¯ÄÜ£¬¼Ó¿ì»¯Ñ§·´Ó¦ËÙÂÊ£¬¶ø±¾ÉíµÄÖÊÁ¿ºÍ»¯Ñ§ÐÔÖÊ·´Ó¦Ç°ºó±£³Ö²»±ä£®¸ù¾ÝÌâ¸øÐÅÏ¢Öª·´Ó¦iÖÐFe²⁺±»PbO₂Ñõ»¯ÎªFe³⁺£¬Ôò·´Ó¦iiÖÐFe³⁺±»Pb»¹Ô­ÎªFe²⁺£¬Àë×Ó·½³ÌʽΪ£º2Fe³⁺+Pb+SO₄^2-¨TPbSO₄+2Fe²⁺£¬
¹Ê´ð°¸Îª£º2Fe³⁺+Pb+SO₄^2-¨TPbSO₄+2Fe²⁺£»
¢ÚaʵÑéÖ¤Ã÷·¢Éú·´Ó¦i£¬ÔòbʵÑéÐèÖ¤Ã÷·¢Éú·´Ó¦ii£¬ÊµÑé·½°¸Îª£º
a£®ÏòËữµÄFeSO₄ÈÜÒºÖмÓÈëKSCNÈÜÒº£¬ÈÜÒº¼¸ºõÎÞÉ«£¬ÔÙ¼ÓÈëÉÙÁ¿PbO₂£¬ÈÜÒº±äºì£¬ÑÇÌúÀë×Ó±»Ñõ»¯ÎªÌúÀë×Ó£¬
b£®È¡aÖкìÉ«ÈÜÒºÉÙÁ¿£¬ÈÜÒºÖдæÔÚƽºâ£¬Fe³⁺+3SCN^-=Fe£¨SCN£©₃£¬¼ÓÈë¹ýÁ¿Pb£¬ºÍƽºâ״̬ÏÂÌúÀë×Ó·´Ó¦Éú³ÉÑÇÌúÀë×Ó£¬Æ½ºâÄæÏò½øÐгä·Ö·´Ó¦ºó£¬ºìÉ«ÍÊÈ¥£¬
¹Ê´ð°¸Îª£ºÈ¡aÖкìÉ«ÈÜÒºÉÙÁ¿£¬¼ÓÈë×ãÁ¿Pb£¬³ä·Ö·´Ó¦ºó£¬ºìÉ«ÍËÈ¥£»
£¨3£©ÒòΪKsp£¨PbSO₄£©=1.6¡Á10^-5£¬Ksp £¨PbCO₃£©=3.3¡Á10^-14£¬c£¨SO₄^2-£©=1.6mol•L^-1£¬c£¨CO₃^2-£©=0.1mol•L^-1£¬PbCO₃ÖУºc£¨Pb²⁺£©=$\frac{KsP£¨PbC{O}_{3}£©}{c£¨C{{O}_{3}}^{2-}£©}$=$\frac{3.3¡Á1{0}^{-14}}{0.1}$=3.3¡Á10^-13£¬£»PbSO₄ÖУºQ=c£¨Pb²⁺£©¡Ác£¨SO₄^2-£©=3.3¡Á10^-13¡Á1.6=5.28¡Á10^-13£¼Ksp£¨PbSO₄£©£¬ËµÃ÷PbSO₄ÔÚĸҺÖÐδ±¥ºÍ£¬¼´PbCO₃Öв»»ìÓÐPbSO₄£»
¹Ê´ð°¸Îª£º·ñ£»
£¨4£©¢Ù¸ù¾ÝÄÆÀë×ÓÏòÒõ¼«Òƶ¯Öª£¬aΪÒõ¼«£¬ÓëµçÔ´¸º¼«ÏàÁ¬£»
¹Ê´ð°¸Îª£º¸º£»
¢Úµç½â¹ý³ÌÖУ¬PbO₂ÔÚÒõ¼«µÃµç×Ó£¬µç¼«·´Ó¦Ê½ÎªPbO₂+4e^-+2H₂O=Pb+4OH^-£»
¹Ê´ð°¸Îª£ºPbO₂+4e^-+2H₂O=Pb+4OH^-£»
¢ÛÄÆÀë×Ó½»»»Ä¤Ö»ÔÊÐíÄÆÀë×Óͨ¹ý£¬×èÖ¹HPbO₂^-½øÈëÑô¼«ÊÒ±»Ñõ»¯£¬´Ó¶øÌá¸ßPbÔªËصÄÀûÓÃÂÊ£»
¹Ê´ð°¸Îª£º×èÖ¹HPbO₂^-½øÈëÑô¼«ÊÒ±»Ñõ»¯£®

µãÆÀ ±¾Ì⿼²éδ֪·´Ó¦Êéд¡¢ÈܶȻý¡¢µç»¯Ñ§¼°»ù´¡ÊµÑéÎÊÌ⣬ÊǸ߿¼³£¼ûÌâÐÍ£¬Ïò¿¼ÉúչʾÔËÓû¯Ñ§ÖªÊ¶½â¾öÉú²úÉú»îÖеÄʵ¼ÊÎÊÌ⣬ÌâÄ¿ÄѶȽϴó£®