ÌâÄ¿ÄÚÈÝ

3£®¹¤Òµ·ÏË®Öг£º¬ÓÐÒ»¶¨Á¿µÄCr2O72-ºÍCrO42-£¬ËüÃÇ»á¶ÔÈËÀ༰Éú̬ϵͳ²úÉúºÜH+¢Ùת»¯ Fe2+¢Ú»¹Ô­ OH-¢Û³Áµí´óµÄÉ˺¦£¬±ØÐë½øÐд¦Àí£®³£ÓõĴ¦Àí·½·¨Óл¹Ô­³Áµí·¨£¬¸Ã·¨µÄ¹¤ÒÕÁ÷³ÌΪ£ºCrO42-$¡ú_{¢Ùת»¯}^{H+}$Cr2O72-$¡ú_{¢Ú»¹Ô­}^{Fe_{2}+}$Cr3+$¡ú_{¢Û³Áµí}^{OH-}$Cr£¨OH£©3ÆäÖеڢٲ½´æÔÚÆ½ºâ£º2CrO42-£¨»ÆÉ«£©+2H+?Cr2O72-£¨³ÈÉ«£©+H2O
£¨1£©ÈôƽºâÌåϵµÄpH=2£¬ÔòÈÜÒºÏÔ³ÈÉ«£®
£¨2£©µÚ¢Ú²½ÖУ¬»¹Ô­1mol Cr2O72-Àë×Ó£¬ÐèÒª6molµÄFeSO4•7H2O£®
£¨3£©Cr3+ÓëAl3+µÄ»¯Ñ§ÐÔÖÊÏàËÆ£¬ÔÚCr2£¨SO4£©3ÈÜÒºÖÐÖðµÎ¼ÓÈëNaOHÈÜÒºÖ±ÖÁ¹ýÁ¿£¬¿É¹Û²ìµ½µÄÏÖÏóÊÇÏÈÓÐÀ¶É«³ÁµíÉú³É£¬ºó³ÁµíÖð½¥Èܽ⣮
µÚ¢Û²½Éú³ÉµÄCr£¨OH£©3ÔÚÈÜÒºÖдæÔÚÒÔϳÁµíÈÜ½âÆ½ºâ£ºCr£¨OH£©3£¨s£©?Cr3+£¨aq£©+3OH-£¨aq£©
£¨4£©³£ÎÂÏ£¬Cr£¨OH£©3µÄÈܶȻýKsp=c£¨Cr3+£©•c3£¨OH-£©=10-32£¬ÒªÊ¹c£¨Cr3+£©½µÖÁ10-5mol/L£¬ÈÜÒºµÄpHÓ¦µ÷ÖÁ5£®
£¨5£©ÒÑÖªAgCl¡¢Ag2CrO4£¨×©ºìÉ«£©µÄKsp·Ö±ðΪ2¡Á10-10ºÍ1.12¡Á10-12£®·ÖÎö»¯Ñ§ÖУ¬²â¶¨º¬ÂȵÄÖÐÐÔÈÜÒºÖÐCl-µÄº¬Á¿£¬ÒÔK2CrO4×÷ָʾ¼Á£¬ÓÃAgNO3ÈÜÒºµÎ¶¨£®µÎ¶¨¹ý³ÌÖÐÊ×ÏÈÎö³ö³ÁµíAgCl£®

·ÖÎö £¨1£©pHµÄÈÜÒº³ÊËáÐÔ£¬Æ½ºâ2CrO42-£¨»ÆÉ«£©+2H+?Cr2O72-£¨³ÈÉ«£©+H2OÕýÏòÒÆ¶¯£»
£¨2£©¸ù¾ÝÑõ»¯»¹Ô­·´Ó¦ÖеÃʧµç×ÓÊØºãÀ´¼ÆË㣻
£¨3£©Cr3+ÓëAl3+µÄ»¯Ñ§ÐÔÖÊÏàËÆ£¬ÔÚCr2£¨SO4£©3ÈÜÒºÖÐÖðµÎ¼ÓÈëNaOHÈÜÒºÖ±ÖÁ¹ýÁ¿£¬ÏÈÉú³É³Áµí£¬ºóÈܽ⣻
£¨4£©¸ù¾ÝÈܶȻý³£ÊýÒÔ¼°Ë®µÄÀë×Ó»ý³£ÊýÀ´½øÐмÆË㣻
£¨5£©¼ÓÈëÏõËáÒø£¬ÒøÀë×ÓŨ¶ÈÏàµÈ£¬¸ù¾ÝÈܶȻý¼ÆËãÉú³É³ÁµíʱÂÈÀë×ÓÒÔ¼°CrO42-µÄŨ¶È´óС£¬Å¨¶ÈСµÄÏÈÉú³É³Áµí£®

½â´ð ½â£º£¨1£©c£¨H+£©Ôö´ó£¬Æ½ºâ2CrO42-£¨»ÆÉ«£©+2H+?Cr2O72-£¨³ÈÉ«£©+H2OÓÒÒÆ£¬ÈÜÒº³Ê³ÈÉ«£¬¹Ê´ð°¸Îª£º³ÈÉ«£»
£¨2£©¸ù¾Ýµç×ÓµÃÊ§ÊØºã£ºn£¨Cr2O72-£©¡Á6=n£¨FeSO4•7H2O£©¡Á1£¬n£¨FeSO4•7H2O£©=$\frac{1mol¡Á6}{1}$=6mol£¬¹Ê´ð°¸Îª£º6mol£»
£¨3£©Cr3+ÓëAl3+µÄ»¯Ñ§ÐÔÖÊÏàËÆ£¬ÔÚCr2£¨SO4£©3ÈÜÒºÖÐÖðµÎ¼ÓÈëNaOHÈÜÒºÖ±ÖÁ¹ýÁ¿£¬ÏÈÉú³ÉÇâÑõ»¯¸õÀ¶É«³Áµí£¬ºóÈܽ⣬
¹Ê´ð°¸Îª£ºÏÈÓÐÀ¶É«³ÁµíÉú³É£¬ºó³ÁµíÖð½¥Èܽ⣻
£¨4£©c£¨Cr3+£©=10-5mol/Lʱ£¬ÈÜÒºµÄc£¨OH-£©=$\root{3}{\frac{10{\;}^{-32}}{10{\;}^{-5}}}$=10-9 mol/L£¬c£¨H+£©¨T$\frac{10{\;}^{-14}}{10{\;}^{-9}}$=10-5mol/L£¬pH=5£¬¼´ÒªÊ¹c£¨Cr3+£©½µÖÁ10-5mol/L£¬ÈÜÒºµÄpHÓ¦µ÷ÖÁ5£¬¹Ê´ð°¸Îª£º5£»
£¨5£©µÎ¶¨ÖÕµãʱ£¬ÈÜÒºÖÐc£¨Cl-£©¡Ü1.0¡Á10-5mol•L-1]£¬ÓÉKsp£¨AgCl£©=1.0¡Á10-10£¬c£¨Ag+£©¡Ý1.0¡Á10-5 mol•L-1£¬ÓÖKsp£¨Ag2CrO4£©=1.12¡Á10-12£¬
Ôòc£¨CrO42-£©=$\frac{{K}_{SP}}{{C}^{2}£¨A{g}^{+}£©}$=$\frac{1.12¡Á1{0}^{-12}}{£¨1¡Á1{0}^{-5}£©^{2}}$=0.0112mol•L-1£¬¿ÉÖªÓ¦ÏÈÉú³ÉAgCl³Áµí£¬
¹Ê´ð°¸Îª£ºAgCl£®

µãÆÀ ±¾Ì⿼²éÁËÎïÖʺ¬Á¿µÄ²â¶¨£¬Îª¸ßƵ¿¼µã£¬²â×Ü¿¼²éѧÉúµÄ·ÖÎö¡¢¼ÆËãÄÜÁ¦£¬ÌâÄ¿Éæ¼°ÄÑÈÜÎïµÄÈÜ½âÆ½ºâ¡¢Ñõ»¯»¹Ô­·´Ó¦µÈ֪ʶµã£¬Ã÷ȷʵÑéÔ­ÀíÊǽⱾÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø