ÌâÄ¿ÄÚÈÝ

Âȼµç½â±¥ºÍʳÑÎË®ÖÆÈ¡NaOHµÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º

ÒÀ¾ÝÉÏͼ£¬Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©ÓëµçÔ´¸º¼«ÏàÁ¬µÄµç¼«¸½½ü£¬ÈÜÒºpHÖµ£¨Ñ¡Ì²»±ä¡¢Éý¸ß»òϽµ£©£¬ÓëµçÔ´Õý¼«ÏàÁ¬µÄµç¼«½Ð    ¼«£¬¼ìÑé¸Ã¼«ÉϲúÎïµÄ·½·¨ÊÇ                 ¡£
£¨2£©Ð´³öµç½â±¥ºÍʳÑÎË®µÄ»¯Ñ§·½³Ìʽ                       ¡£
£¨3£©Èç¹û´ÖÑÎÖÐSOº¬Á¿½Ï¸ß£¬±ØÐëÌí¼Ó±µÊ½¼Á³ýÈ¥SO£¬¸Ã±µÊÔ¼Á¿ÉÒÔÊÇ£¨Ñ¡ÌîA.B.c£¬¶àÑ¡¿Û·Ö£©  
A£®Ba(OH)2      B£®Ba(NO3)2      C£®BaCl2 
£¨4£©ÅжϱµÊÔ¼ÁÒѾ­¹ýÁ¿µÄ·½·¨ÊÇ                                 ¡£
£¨5£©ÎªÓÐЧ³ýÈ¥Ca2+¡¢Mg2+¡¢SO£¬¼ÓÈëÊÔ¼ÁµÄºÏÀí˳ÐòΪ£¨Ñ¡Ìîa£¬b£¬c¶àÑ¡¿Û·Ö£©
A£®ÏȼÓNaOHÈÜÒº£¬ºó¼ÓNa2CO3ÈÜÒº£¬ÔÙ¼Ó±µÊÔ¼Á 
B£®ÏȼÓNaOHÈÜÒº£¬ºó¼Ó±µÊÔ¼Á£¬ÔÙ¼ÓNa2CO3ÈÜÒº   
C£®ÏȼӱµÊÔ¼Á£¬ºó¼ÓNaOHÈÜÒº£¬ÔÙ¼ÓNa2CO3ÈÜÒº
£¨6£©Îª¼ìÑ龫Ñδ¿¶È£¬ÐèÅäÖÆ150 mL0.2 mol/LNaCl£¨¾«ÑΣ©ÈÜÒº£¬ÏÂͼÊǸÃͬѧתÒÆÈÜÒºµÄʾÒâͼ£¬Í¼ÖеĴíÎóÊÇ                                ¡£

£¨1£©Éý¸ß  Ñô¼«  ½«µí·Ûµâ»¯¼ØÊÔÖ½ÓÃË®ÈóʪºóÕ³ÔÚ²£Á§°ôÒ»¶Ë,¿¿½ü×°Óдý²âÆøÌåµÄ¼¯ÆøÆ¿ Èç¹û»ÆÂÌÉ«ÆøÌåÄÜʹÊÔÖ½±äÀ¶É«,Ö¤Ã÷º¬ÓÐCl2¡£
£¨2£©µç½â±¥ºÍʳÑÎË®µÄ·´Ó¦£º2NaCl+2H2O  2NaOH+Cl2¡ü+H2¡ü
£¨3£©C
£¨4£©¾²Ö¹£¬ÏòÉϲãÇåÒºÖмÌÐøµÎ¼ÓBaCl2£¬Èç²»²úÉú³Áµí£¬ÒѹýÁ¿¡£
£¨5£©B.C¡£
£¨6£©Î´Óò£Á§°ôÒýÁ÷       100 mLµÄÈÝÁ¿Æ¿²»ÄÜÅäÖÆ150 mL ÈÜÒº

½âÎöÊÔÌâ·ÖÎö£º£¨1£©ÓëµçÔ´¸º¼«ÏàÁ¬ÎªÒõ¼«£¬Éú³ÉÇâÆøºÍNaOH£¬Éú³É¼î£¬ËùÒÔpHÉý¸ß£»ÓëÕý¼«ÏàÁ¬ÎªÑô¼«£¬Éú³ÉÂÈÆø£»½«µí·Ûµâ»¯¼ØÊÔÖ½ÓÃË®ÈóʪºóÕ³ÔÚ²£Á§°ôÒ»¶Ë,¿¿½ü×°Óдý²âÆøÌåµÄ¼¯ÆøÆ¿ Èç¹û»ÆÂÌÉ«ÆøÌåÄÜʹÊÔÖ½±äÀ¶É«,Ö¤Ã÷º¬ÓÐCl2¡£
£¨2£©µç½â±¥ºÍʳÑÎË®µÄ·´Ó¦£º2NaCl+2H2O  2NaOH+Cl2¡ü+H2¡ü
£¨3£©Ìí¼Ó±µÊÔ¼Á³ýÈ¥SO42-£¬×¢Òâ²»ÄÜÒýÈëеÄÔÓÖÊ£¬Ñ¡Ba£¨NO3£©2»áÒýÈëÔÓÖÊÏõËá¸ùÀë×Ó£¬ËùÒԸñµÊÔ¼Á²»ÄÜÑ¡Óá£Í¬ÀíBa(OH)2ÔòÒýÈëÁËOH-Àë×Ó¡£
£¨4£©¾²Ö¹£¬ÏòÉϲãÇåÒºÖмÌÐøµÎ¼ÓBaCl2£¬Èç²»²úÉú³Áµí£¬ÒѹýÁ¿¡£
SO42-¡¢Ca2+¡¢Mg2+µÈ·Ö±ðÓëBaCl2ÈÜÒº¡¢Na2CO3ÈÜÒº¡¢NaOHÈÜÒº·´Ó¦Éú³É³Áµí£¬¿ÉÔÙͨ¹ý¹ýÂ˳ýÈ¥£¬Na2CO3ÈÜÒºÄܳýÈ¥¹ýÁ¿µÄBaCl2ÈÜÒº£¬ÑÎËáÄܳýÈ¥¹ýÁ¿µÄNa2CO3ÈÜÒººÍNaOHÈÜÒº£¬ËùÒÔÓ¦ÏȼÓBaCl2ÈÜÒºÔÙ¼ÓNa2CO3ÈÜÒº£¬×îºó¼ÓÈëÑÎËá¡£¹Ê´ð°¸Îª:B.C¡£
£¨6£©Î´Óò£Á§°ôÒýÁ÷£»100 mLµÄÈÝÁ¿Æ¿²»ÄÜÅäÖÆ150 mL ÈÜÒº¡£
¿¼µã£ºµç½â³Ø¡¢ÊµÑéÊÒ³ýÔÓÖÊ¡¢Àë×Ó·½³ÌʽÊéд¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ijÖÖ̼ËáÃÌ¿óµÄÖ÷Òª³É·ÖÓÐMnCO3¡¢MnO2¡¢FeCO3¡¢MgO¡¢SiO2¡¢Al2O3µÈ¡£ÒÑ֪̼ËáÃÌÄÑÈÜÓÚË®¡£Ò»ÖÖÔËÓÃÒõÀë×ÓĤµç½â·¨µÄм¼Êõ¿ÉÓÃÓÚ´Ó̼ËáÃÌ¿óÖÐÌáÈ¡½ðÊôÃÌ£¬Á÷³ÌÈçÏ£º

ÒõÀë×ÓĤ·¨µç½â×°ÖÃÈçͼËùʾ£º

£¨1£©Ð´³öÓÃÏ¡ÁòËáÈܽâ̼ËáÃÌ·´Ó¦µÄÀë×Ó·½³Ìʽ                               ¡£
£¨2£©ÔÚ½þ³öÒºÀïÃÌÔªËØÖ»ÒÔMn2+µÄÐÎʽ´æÔÚ£¬ÇÒÂËÔüÖÐÒ²ÎÞMnO2£¬Çë½âÊÍÔ­Òò                                         .
£¨3£©£¨5·Ö£©ÒÑÖª²»Í¬½ðÊôÀë×ÓÉú³ÉÇâÑõ»¯Îï³ÁµíËùÐèµÄpHÈçÏÂ±í£º

¼Ó°±Ë®µ÷½ÚÈÜÒºµÄpHµÈÓÚ6£¬ÔòÂËÔüµÄ³É·ÖÊÇ                             £¬ÂËÒºÖк¬ÓеÄÑôÀë×ÓÓÐH+ºÍ                       ¡£
£¨4£©µç½â×°ÖÃÖмýÍ·±íʾÈÜÒºÖÐÒõÀë×ÓÒƶ¯µÄ·½Ïò£¬ÔòAµç¼«ÊÇ    ¼«¡£Êµ¼ÊÉú²úÖУ¬Ñô¼«ÒÔÏ¡ÁòËáΪµç½âÒº£¬Ñô¼«µÄµç¼«·´Ó¦Ê½Îª                     ¡£
£¨5£©¸Ã¹¤ÒÕÖ®ËùÒÔ²ÉÓÃÒõÀë×Ó½»»»Ä¤£¬ÊÇΪÁË·ÀÖ¹Mn2+½øÈëÑô¼«Çø·¢Éú¸±·´Ó¦Éú³ÉMnO2Ôì³É×ÊÀË·Ñ£¬Ð´³ö¸Ã¸±·´Ó¦µÄµç¼«·´Ó¦Ê½                                  ¡£

ÈçͼΪµç½â×°Öã¬X¡¢YΪµç¼«²ÄÁÏ£¬aΪµç½âÖÊÈÜÒº¡£

£¨1£©ÈôaΪº¬ÓзÓ̪µÄKClÈÜÒº£¬XΪFe£¬YΪʯī£¬µç½âÒ»¶Îʱ¼äºó:
Xµç¼«¸½½ü¿É¹Û²ìµ½µÄʵÑéÏÖÏóÊÇ                                               £»
д³öYµç¼«µÄµç¼«·´Ó¦Ê½                                                           ¡£
£¨2£©ÈôҪʵÏÖCu  +H2SO4£½CuSO4+H2¡ü£¬
ÔòYµç¼«²ÄÁÏÊÇ                                               £»
д³öXµç¼«µÄµç¼«·´Ó¦Ê½                                        ¡£
£¨3£©ÈôÒªÀûÓøÃ×°ÖÃÔÚÌúÖÆÆ·±íÃæ¶ÆÉÏÒ»²ãÒø£¬ÔòaΪ                       £¬·´Ó¦Ç°Á½µç¼«µÄÖÊÁ¿ÏàµÈ£¬·´Ó¦ºóµç¼«ÖÊÁ¿Ïà²î2.16g£¬Ôò¸Ã¹ý³ÌÀíÂÛÉÏͨ¹ýµçÁ÷±íµÄµç×ÓÊýΪ                           ¡£
£¨4£©ÈôX¡¢Y¾ùΪ¶èÐԵ缫£¬aΪNaOHÈÜÒº£¬µç½âÒ»¶Îʱ¼äºó£¬ÈÜÒºµÄpH           £¨Ìî¡°Ôö´ó¡±¡°²»±ä¡±¡°¼õС¡±£©£¬ÈôҪʹÈÜÒº»Ö¸´Ô­À´µÄ״̬£¬¿ÉÍùÈÜÒºÖмÓÈë                                        ¡£

¼×¡¢ÒÒÁ½³ØµÄµç¼«²ÄÁÏÈçͼËùʾ£¬Çë°´ÒªÇó»Ø´ðÏÂÁÐÎÊÌ⣺

(1)ÈôÁ½³ØÖоùΪCu(NO3)2ÈÜÒº£¬·´Ó¦Ò»¶Îʱ¼äºó£º
¢ÙÓкìÉ«ÎïÖÊÎö³öµÄÊǼ׳ØÖеĠ   £¨Ìî¡°Ìú¡±»ò¡°Ì¼¡±£©°ô£»ÒÒ³ØÖеĠ   £¨Ìî
¡°Òõ¡±»ò¡°Ñô¡±£©¼«¡£
¢ÚÒÒ³ØÖÐÑô¼«ÉÏ·¢ÉúµÄµç¼«·´Ó¦·½³ÌʽÊÇ                      ¡£
(2)ÈôÁ½³ØÖоùΪ±¥ºÍNaClÈÜÒº£º
¢Ùд³öÒÒ³ØÖÐ×Ü·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º                                 ¡£
¢Ú¼×³ØÖÐ̼¼«Éϵ缫·´Ó¦·½³ÌʽÊÇ                            £¬ÒÒ³Ø̼¼«ÉÏ
µç¼«·´Ó¦ÊôÓÚ          £¨Ìî¡°Ñõ»¯·´Ó¦¡±»ò¡°»¹Ô­·´Ó¦¡±£©¡£
¢Û½«ÊªÈóµÄµí·ÛKIÊÔÖ½·ÅÔÚÒÒ³Ø̼¼«¸½½ü£¬·¢ÏÖÊÔÖ½±äÀ¶£¬´ýÒ»¶Îʱ¼äºóÓÖ·¢ÏÖ
À¶É«ÍÊÈ¥¡£ÕâÊÇÒòΪ¹ýÁ¿µÄCl2½«Éú³ÉµÄI2ÓÖÑõ»¯¡£Èô·´Ó¦µÄC12ºÍI2ÎïÖʵÄÁ¿
Ö®±ÈΪ5£º1£¬ÇÒÉú³ÉHClºÍÁíÒ»ÖÖÇ¿Ëᣬ¸ÃÇ¿ËáµÄ»¯Ñ§Ê½Îª            ¡£
¢ÜÈôÒÒ³ØÖÐתÒÆ0£®1 mol e-ºóֹͣʵÑ飬³ØÖÐÈÜÒºÌå»ýÊÇ1L£¬ÔòÈÜÒº»ìÔȺóµÄpH=   £¨²»¿¼ÂÇËùÉú³ÉµÄÆøÌåÈܽâÔÚÈÜÒºÖеÄÇé¿ö£©

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

¾«Ó¢¼Ò½ÌÍø