Question

£¨9·Ö£©ÓÐA¡¢B¡¢C¡¢D¡¢EÎåÖÖ³£¼û»¯ºÏÎ¶¼ÊÇÓÉϱíÖеÄÀë×ÓÐγɵģº

ÑôÀë×Ó	K+ Na+ Cu2+ Al3+
ÒõÀë×Ó	SO42£­ HCO3£­ NO3£­ OH£­

ΪÁ˼ø±ðÉÏÊö»¯ºÏÎ·Ö±ðÍê³ÉÒÔÏÂʵÑ飬Æä½á¹ûÊÇ£º

¢Ù½«ËüÃÇÈÜÓÚË®ºó£¬DΪÀ¶É«ÈÜÒº£¬ÆäËû¾ùΪÎÞÉ«ÈÜÒº£»

¢Ú½«EÈÜÒºµÎÈ˵½CÈÜÒºÖгöÏÖ°×É«³Áµí£¬¼ÌÐøµÎ¼Ó£¬³ÁµíÈܽ⣻

¢Û½øÐÐÑæÉ«·´Ó¦£¬½öÓÐB¡¢CΪ×ÏÉ«£¨Í¸¹ýÀ¶É«îܲ£Á§£©£»

¢ÜÔÚ¸÷ÈÜÒºÖмÓÈëÏõËá±µÈÜÒº£¬ÔÙ¼Ó¹ýÁ¿Ï¡ÏõËᣬֻÓÐAÖзųöÎÞÉ«ÆøÌ壬ֻÓÐC¡¢DÖвúÉú°×É«³Áµí£»

¢Ý½«B¡¢DÁ½ÈÜÒº»ìºÏ£¬Î´¼û³Áµí»òÆøÌåÉú³É¡£¸ù¾ÝÉÏÊöʵÑéÌî¿Õ£º

£¨1£©Ð´³öB¡¢DµÄ»¯Ñ§Ê½£ºB £¬D ¡£

£¨2£©½«º¬l mol AµÄÈÜÒºÓ뺬l mol EµÄÈÜÒº·´Ó¦ºóÕô¸É£¬½öµÃµ½Ò»ÖÖ»¯ºÏÎ¸Ã»¯ºÏÎﻯѧʽΪ________________¡£

£¨3£©C³£ÓÃ×÷¾»Ë®¼Á£¬ÓÃÀë×Ó·½³Ìʽ±íʾÆ侻ˮԭÀí ¡£

£¨4£©ÔÚAÈÜÒºÖмÓÈë¹ýÁ¿³ÎÇåʯ»ÒË®£¬ÆäÀë×Ó·½³ÌʽΪ .

£¨5£©Ïò20mL 2mol/L CÈÜÒºÖмÓÈë60mL EÈÜÒº£¬³ä·Ö·´Ó¦ºóµÃµ½0.78g³Áµí£¬ÔòE

ÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶È¿ÉÄÜÊÇ moI/L¡£

 

Answer 1

£¨1£©KNO3 £¬£¨1·Ö£© CuSO4£¨1·Ö£© £¨2£©Na2CO3£¨1·Ö£©

£¨3£©Al3+ + 3H2O Al£¨OH£©3£¨½ºÌ壩+3H+ £¨2·Ö£©

£¨4£©Ca2+ + HCO3¡ª + OH¡ª £½ CaCO3¡ý+ H2O £¨2·Ö£©£¨5£©0.5 ºÍ2.5

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º¢Ù½«ËüÃÇÈÜÓÚË®ºó£¬DΪÀ¶É«ÈÜÒº£¬ÆäËû¾ùΪÎÞÉ«ÈÜÒº£¬ÔòÔÚDÖк¬ÓÐCu2+£»¢Ú½«EÈÜÒºµÎÈëµ½CÈÜÒºÖгöÏÖ°×É«³Áµí£¬¼ÌÐøµÎ¼Ó£¬³ÁµíÈܽ⣻ÔòEÖк¬ÓÐOH-£¬CÖк¬ÓÐAl3+£»¢Û½øÐÐÑæÉ«·´Ó¦£¬½öÓÐB¡¢CΪ×ÏÉ«£¬ÔòB¡¢CÖк¬ÓÐK+£»¢ÜÔÚ¸÷ÈÜÒºÖмÓÈëÏõËá±µÈÜÒº£¬ÔÙ¼Ó¹ýÁ¿Ï¡ÏõËᣬֻÓÐAÖзųöÎÞÉ«ÆøÌ壬ÔòAÖк¬ÓÐHCO3¡ª£»Ö»ÓÐC¡¢DÖвúÉú°×É«³Áµí£»ÔòC¡¢DÖк¬ÓÐSO42-£»¢Ý½«B¡¢DÁ½ÈÜÒº»ìºÏ£¬Î´¼û³Áµí»òÆøÌåÉú³É¡£ÔòBÊÇKNO3 £¬DÊÇCuSO4£»¸ù¾ÝÀë×ӵĹ²´æµÄ֪ʶ¼°ÒÑÖªÌõ¼þ¿ÉÖªAÊÇNaHCO3£»BÊÇKNO3 £¬CÊÇKAl(SO4)2¡¤12H2O£»DÊÇCuSO4£»EµÄNaOH¡££¨1£©BµÄ»¯Ñ§Ê½ÊÇKNO3 £»D »¯Ñ§Ê½ÊÇCuSO4£»£¨2£©½«º¬l mol AµÄÈÜÒºÓ뺬l mol EµÄÈÜÒº£¬»á·¢Éú·´Ó¦£ºNaOH + NaHCO3= Na2CO3+ H2O£»·´Ó¦ºó½«ÈÜÒºÕô¸ÉµÃµ½Ò»ÖÖ»¯ºÏÎ¸Ã»¯ºÏÎﻯѧʽΪNa2CO3£»£¨3£©Ã÷·¯³£ÓÃ×÷¾»Ë®¼Á£¬ÊÇÒòΪÂÁÀë×ÓË®½â²úÉú±íÃæ»ýºÜ´óµÄÇâÑõ»¯ÂÁ½ºÌåÓÐÎü¸½×÷Óá£ÓÃÀë×Ó·½³Ìʽ±íʾÆ侻ˮԭÀíÊÇAl3+ + 3H2O Al£¨OH£©3£¨½ºÌ壩+3H+£»£¨4£©ÔÚAÈÜÒºÖмÓÈë¹ýÁ¿³ÎÇåʯ»ÒË®£¬·´Ó¦µÄÀë×Ó·½³ÌʽӦ¸ÃÒÔ²»×ãÁ¿µÄÎïÖÊNaHCO3Ϊ±ê×¼£»ÆäÀë×Ó·½³ÌʽΪCa2+ + HCO3¡ª + OH¡ª £½ CaCO3¡ý+ H2O£¨5£©20mL 2mol/L Ã÷·¯Ë®ÈÜÒºÖк¬ÓÐÂÈÀë×ÓµÄÎïÖʵÄÁ¿ÊÇ£ºn(Al3+)=0.02L¡Á2mol/L=0.04mol;n(Al(OH) 3)= 0.78g¡Â78g/mol=0.01mol£¬n(Al3+)> n(Al(OH) 3).¿ÉÄÜÊǼ×ãÁ¿£¬¸ù¾ÝÀë×Ó·½³Ìʽ¿ÉÖª£ºAl3£«£«3OH-£½Al(OH)3¡ý¿ÉÖª£ºn(OH-)=3 n(Al(OH) 3)=0.03mol£¬c(OH-)=0.03mol¡Â0.06L=0.5mol/L£»Ò²¿ÉÄÜÊǼî¹ýÁ¿£¬Ê×ÏÈ·¢Éú·´Ó¦Al3£«£«3OH-£½Al(OH)3¡ý£¬Ê¹ÂÈÀë×ÓÍêÈ«³Áµí£¬·´Ó¦ÏûºÄn(OH-)1=3 n(Al3+)= 3¡Á0.04mol= 0.12mol£»n(Al(OH) 3)=0.04mol£¬È»ºó·¢Éú·´Ó¦£º Al(OH)3+ OH-= AlO2-+ 2H2O£»ÓÖÈܽâ0.03molµÄAl(OH)3£¬·´Ó¦ÏûºÄn(OH-)2= n(Al(OH) 3)=0.03mol£¬n(OH-)×Ü=0.12mol+0.03mol=0.15mol£»c(OH-)=0.15mol ¡Â0.06L =2.5mol/L.

¿¼µã£º¿¼²éÀë×ӵļø¶¨¡¢ÎïÖÊ»¯Ñ§Ê½µÄÈ·¶¨¡¢ÑεÄË®½âµÄÓ¦Óá¢Àë×Ó·½³ÌʽµÄÊéд¡¢ÂÁÀë×ÓÓë¼î·´Ó¦Ê±ÓйØŨ¶ÈµÄ¼ÆËãµÄ֪ʶ¡£