µ±Ç°Î»ÖãºÊ×Ò³ > ·ÖÎö»¯Ñ§Ìâ¿âÒÔ¼°´ð°¸
HCl ¼ä½Ó·¨ Åðɰ
2£®ºÎνÎü¹â¶ÈºÍ͸¹âÂÊ£¬Á½ÕߵĹØÏµÈçºÎ£¿
3£®Í¨¹ý¼ÆËã˵Ã÷£º0.1mol/lNH4ClÈÜÒºÄÜ·ñ׼ȷµÎ¶¨£¿(NH3µÄKb=1.8¢ª10)
-5
4£®Cu
2£«
¡¢
¡¢Zn
2£«
¡¢
¡¢Cd¡¢NiµÈÀë×Ó¾ùÄÜÓëNHÐγÉÂçºÏÎΪʲô²»ÄÜÒÔ°±Ë®Îª
3
2£«2£«
µÎ¶¨¼ÁÓÃÂçºÏµÎ¶¨·¨À´²â¶¨ÕâЩÀë×Ó£¿
£±£® ÒÑ֪ijËá¼îµÎ¶¨¹ý³ÌµÄͻԾ·¶Î§ÊÇ5.3~8.7(pHÖµ),ӦѡÓÃÄÄÖÖָʾ¼Á_______(A ¼×
»ù³È B¼×»ùºì)£¬ÖÕµãÑÕɫΪ_______(£Á »ÆÉ« £Â ºìÉ«); £²£® ÔÚ2mol/LµÄHSOÈÜÒºÖÐ,Ce(¢ô)/Ce(¢ó)µç¶ÔµÄÌõ¼þµç¼«µçλΪ1.43V, µ±Ce(¢ô)
2
4
ÓëCe(¢ó)µÄŨ¶È±ÈΪ10ʱ,Æäµç¼«µçλΪ_______;
£³£® ÏÂÁÐÎïÖʲ»ÄÜÓÃ×÷»ù×¼ÎïµÄÊÇ______; A ÖØ¸õËá¼Ø B ÁòËáÍ C Åðɰ £´£® H2CO3Ë®ÈÜÒºµÄÖÊ×ÓÌõ¼þʽ___________________________________________; £µ£® H2AÔÚË®ÈÜÒºÖпÉÄÜ´æÔÚµÄÐÍÌå________________________; £¶£® ¼ä½ÓµâÁ¿·¨²âÍʱ,µí·Ûָʾ¼ÁÓ¦ÔÚ___________¼ÓÈë;
A µÎ¶¨Ç° B ¼ÓÈëÁòÇèËá¼Øºó C ½Ó½üÓÚÖÕµãʱ
£·£®Ð´³öÓÐЧÊý×ÖµÄλÊý£º
0.03010
£ß
£ß
£ß
£ß
£ß
,7.02¡ª¡ª¡ª¡ª¡ª¡ª
,5.00§ç10
£²
¡ª¡ª¡ª¡ª¡ª¡ª,pH=10.81¡ª¡ª¡ª¡ª¡ª¡ª,1050¡ª¡ª¡ª¡ª¡ª¡ª;
8£®ÓÃ×î¼òʽ¼ÆËãÏÂÁÐÈÜÒºpHÖµ£º
¢Ù 0.1mol/LµÄNaH2PO4Ë®ÈÜÒº(H3PO4µÄ Ka1=7.6§ç10,Ka2=6.3§ç10,Ka3=4.4§ç
10),pH=¡ª¡ª¡ª£»
¢Ú 0.1mol/LµÄHAcË®ÈÜÒº(HAcµÄKa=1.8¢ª10), pH=¡ª¡ª¡ª£» £¹£®½«ÏÂÁÐÎü¹â¶Èת»»ÎªÍ¸¹âÂÊ0.01£ß£ß£ß£ß£ß£¬0.30£ß£ß
-5
-13
-3
-8
0£®±ê×¼ÈÜÒºµÄÅäÖÆ·½·¨ÓСª¡ª¡ª¡ª¡ª¡ª¡ªºÍ¡ª¡ª¡ª¡ª¡ª¡ª¡ª£®
£±£® Ó°ÏìÑõ»¯»¹ÔµÎ¶¨Í»Ô¾·¶Î§µÄÒòËØ
_____________________________________
£» 2£®ÊÊÓÃÓڵζ¨
·ÖÎö·¨µÄ»¯Ñ§·´Ó¦±ØÐë¾ß±¸µÄÌõ¼þÊÇ£º(!) £¬(2) £¬(3) £»·²ÄÜÂú×ãÉÏÊöÒªÇóµÄ·´Ó¦¶¼¿ÉÓ¦ÓÃÓÚ µÎ¶¨·¨¡£
3£®ÂÁµÄ²â¶¨Ò»°ã²»²ÉÓÃEDTAÖ±½ÓµÎ¶¨·½Ê½£¬ÓÃÅäλµÎ¶¨·¨²â¶¨ÂÁºÏ½ðÖÐÂÁµÄº¬Á¿¿É²ÉÓà µÎ¶¨·½Ê½¡£
4£®´Ó²úÉúµÄÔÒòÉÏ£¬Îó²î·ÖΪ Îó²î, Îó²î£» £µ£® д³öÓÐЧÊý×ÖµÄλÊý£º
0.02110_______,6.02_______,6.100x10______,pH=10.2______,
1040______£»
6£®ÒÑ֪ijËá¼îµÎ¶¨¹ý³ÌµÄͻԾ·¶Î§ÊÇ8.7~5.3(pHÖµ),¼ÆÁ¿µãΪ7£¬Ó¦Ñ¡ÓÃÄÄÖÖָʾ¼Á_______(A ¼×»ù³È B¼×»ùºì)£¬ÖÕµãÑÕɫΪ_______(£Á »ÆÉ« £Â ºìÉ« C ³ÈÉ«)£» 7£®Ä³¼îÒº¿ÉÄܺ¬ÓÐNaOH£¬NaCO»òNaHCO»òËüÃǵĻìºÏÎ½ñÓÃÑÎËá±êÒºµÎ¶¨£¬ÓÃ
2
3
3
2
˫ָʾ¼Á·¨£¨ÏÈÓ÷Ó̪ºóÓü׻ù³È£©Ö¸Ê¾Öյ㣬¸ù¾ÝVºÍVÖµµÄ´óС£¬ÅжϼîÒº×é
1
2
³É¡£
a.ÈôV>V£¬¸Ã¼îÒº×é³ÉΪ £»
1
2
b.ÈôV 1 2 c.ÈôV=V£¬¸Ã¼îÒº×é³ÉΪ £» 1 2 d.ÈôV=0,V>0£¬¸Ã¼îÒº×é³ÉΪ £» 1 2 e.ÈôV>0,V=0£¬¸Ã¼îÒº×é³ÉΪ ¡£ 1 2 8£®Ð´³öÖÊ×ÓÌõ¼þʽ£º H3PO4Ë®ÈÜÒº_______________________________________£» 9£®³£Óõĵζ¨·½Ê½ÓУºÖ±½ÓµÎ¶¨·¨¡¢_______________¡¢_________________¡¢ _________________£» 10£®½«ÏÂÁÐÎü¹â¶Èת»»ÎªÍ¸¹âÂÊA=0.02 T=______£ß£¬A=0.40 T=_________¡£ 11£®µâÁ¿·¨µÄÁ½¸öÖ÷ÒªÎó²îÀ´Ô´Îª ºÍ ¡£ 12£®ÒÔ²£Á§µç¼«ºÍ±¥ºÍ¸Ê¹¯µç¼«ÓëÊÔÒº×é³É¹¤×÷µç³Ø²â¶¨pHÖµ£¬ Ϊ²Î±Èµç¼«£¬ Ϊָʾµç¼«¡£ £±£® Á½µç¶ÔµÄµçλ²î 2·Ö2£® (!)·´Ó¦±ØÐ붨Á¿½øÐУ¬(2)·´Ó¦±ØÐëѸËÙÍê³É£¬(3)±ØÐëÓÐÊʵ±µÄ·½·¨È·¶¨Öյ㣻ֱ½Ó ÿ¿Õ1·Ö 3£®Öû»µÎ¶¨·½Ê½ 2·Ö 4£®ÏµÍ³ żȻ ÿ¿Õ1·Ö £µ£® 4£¬3£¬4£¬1£¬º¬ºý ÿ¿Õ1·Ö 6£®B£¬C ÿ¿Õ2·Ö 7£® a. NaOH NaCOb. NaCONaHCOc. NaCO 2 3 2 3 3 2 3 d. NaHCOe. NaOH ÿ¿Õ1·Ö 3 9£®¼ä½ÓµÎ¶¨·¨¡¢·µµÎ¶¨·¨¡¢Öû»µÎ¶¨·¨ ÿ¿Õ1·Ö 10£® 95.5%£¬39.8% ÿ¿Õ1·Ö 11£®µâµÄ»Ó·¢ µâÀë×Ó±»¿ÕÆøÑõ»¯ ÿ¿Õ1·Ö 12£®±¥ºÍ¸Ê¹¯µç¼«²£Á§µç¼« ÿ¿Õ1·Ö 1£® ʲôÊǾ«Ãܶȣ¿Ê²Ã´ÊÇ׼ȷ¶È£¿¶þÕß¹ØÏµÈçºÎ£¿ £¨8·Ö£© £® ¾«ÃܶÈÊÇÖ¸¼¸´ÎƽÐвⶨ½á¹ûÖ®¼äÏ໥½Ó½üµÄ³Ì¶È¡£ 3·Ö ׼ȷ¶ÈÊÇÖ¸²â¶¨½á¹ûÓëÕæÊµÖµÖ®¼äµÄ½Ó½ü³Ì¶È¡£ 3·Ö ¹²Í¬ºâÁ¿·ÖÎö½á¹ûµÄºÃ»µ£¬¸ßµÄ¾«ÃܶÈÊǸߵÄ׼ȷ¶ÈµÄǰÌᣬµ«¸ßµÄ¾«ÃܶȲ»Ò»¶¨Äܱ£Ö¤¸ßµÄ׼ȷ¶È¡£ 2·Ö 2£®ÈÜÒºÖÐMgºÍCaŨ¶ÈΪ2.0¢ª10mol/L,ÓÃÏàͬŨ¶ÈµÄEDTAµÎ¶¨,ͨ¹ý¼ÆËã˵Ã÷ÄÜ·ñͨ¹ý¿ØÖÆËá¶È·Ö²½µÎ¶¨?£¨£Ë£Ã£á£Ù",=10.69 KMgY",=8.69£© £¨5·Ö£© 2+ 2+ -2 3£®Ê²Ã´ÊÇ»¯Ñ§¼ÆÁ¿µã£¿ £¨3·Ö£© 4£®Ê²Ã´ÊÇָʾ¼ÁµÄ·â±ÕÏÖÏó£¿ÈçºÎ´¦Àí£¿£¨5·Ö£© 5£®·Ö¹â¶È¼ÆµÄÖ÷Òª×é³É²¿¼þÓÐÄÄЩ£¿£¨4·Ö T=-LGA £±£® Ò»ÔªÈõËáHAcÔÚË®ÈÜÒºÖÐÓÐ_HAc_ºÍ_Ac-_Á½ÖÖÐÍÌå´æÔÚ£¬µ±pH=pKaʱ£¬¦Ä£½0.5 £¬ 0 ¦Ä1£½¡ª0.5£» £²£® д³öÓÐЧÊý×ÖµÄλÊý£º 0.0301¡ª3, 7.020_4, 5.10§ç10_3__, pH=10.8_1__¡ª, 1060_º¬ºý_£» £³£® ÒÒ¶þ°·ËÄÒÒËá¼ò³Æ¡ª__EDTA___£» 4.ÏÂÁÐÎïÖÊÄÜÓÃ×÷»ù×¼ÎïµÄÊÇ__A__¡£ A ÁÚ±½¶þ¼×ËáÇâ¼Ø B ÇâÑõ»¯ÄÆ C Áò´úÁòËáÄÆ D EDTA 5£®ÒÑÖªH3PO4µÄKa1=7.6§ç10,Ka2=6.3§ç10,Ka3=4.4§ç10,ÇóNa3PO4 µÄKb1=_2.27x10-2_£¬ -3 -8 -13 2
¹²·ÖÏí92ƪÏà¹ØÎĵµ