1-7Õ·ÖÎö»¯Ñ§Á·Ï°Ìâ

3¡¢ÔÚpH=10µÄ°±»º³åÈÜÒºÖУ¬µÎ¶¨100.0mLº¬Ca2+¡¢Mg2+µÄË®Ñù£¬ÏûºÄ0.01016mol/LµÄEDTA±ê×¼ÈÜÒº15.28mL£»ÁíÈ¡100.0mLË®Ñù£¬ÓÃNaOH´¦Àí£¬Ê¹Mg2+Éú³ÉMg(OH)2³Áµí£¬µÎ¶¨Ê±ÏûºÄEDTA±ê×¼ÈÜÒº10.43mL£¬¼ÆËãË®ÑùÖÐCaCO3ºÍMgCO3µÄº¬Á¿(ÒÔug/mL±íʾ)¡£ 4¡¢³ÆÈ¡ÂÁÑÎÊÔÑù1.250g£¬Èܽâºó¼ÓÈë0.05000mol/LµÄEDTAÈÜÒº25.00mL£¬ÔÚÊʵ±Ìõ¼þÏ·´Ó¦ºó£¬ÒÔµ÷½ÚÈÜÒºpHΪ5¡«6£¬ÒÔ¶þ¼×·Ó³ÈΪָʾ¼Á£¬ÓÃ0.02000mol/LµÄZn2+±ê×¼ÈÜÒº»ØµÎ¹ýÁ¿µÄEDTA£¬ÏûºÄZn2+ÈÜÒº21.50mL£¬¼ÆËãÂÁÑÎÖÐÂÁµÄÖÊÁ¿·ÖÊý¡£ 5¡¢Ìõ¼þÎȶ¨³£Êý¼ÆËã

(1) pH =10 EDTAÓëMg2+ÐγÉÅäºÏÎïʱ¼ÆËãlgK¡¯MgY (2) pH =8 EDTAÓëCa2+ÐγÉÅäºÏÎïʱ¼ÆËãlgK¡¯CaY (3) pH =5 EDTAÓëZn2+ÐγÉÅäºÏÎïʱ¼ÆËãlgK¡¯ZnY (4) pH =10 NH3=0.1mol/Lʱ,EDTAÓëZn2+ÐγÉÅäºÏÎïʱ¼ÆËãlgK¡¯ZnY

6¡¢·ÖÎöPb¡¢Zn¡¢AlºÏ½ðʱ£¬³ÆÈ¡ºÏ½ð0.4800g£¬Èܽâºó£¬ÓÃÈÝÁ¿Æ¿×¼È·ÅäÖÆ³É100mLÊÔÒº¡£ÎüÈ¡25.00mLÊÔÒº£¬¼ÓKCN½«Zn2+ ÑڱΡ£È»ºóÓÃc(EDTA)=0.02000mol/LµÄEDTAµÎ¶¨Pb2+ ºÍMg2+ £¬ÏûºÄEDTAÈÜÒº46.40mL£»¼ÌÐø¼ÓÈë¶þÛÏ»ù±û´¼(DMP)ÑÚ±ÎPb2+ £¬Ê¹ÆäÖû»³öµÈÁ¿µÄEDTA£¬ÔÙÓÃc(Mg2+)=0.01000mol/LMg2+ ±ê×¼ÈÜÒºµÎ¶¨Öû»³öµÄEDTA£¬ÏûºÄMg2+ Àë×ÓÈÜÒº22.60mL£»×îºó¼ÓÈë¼×È©½â±ÎZn2+£¬ÔÙÓÃÉÏÊöEDTAµÎ¶¨Zn2+ £¬ÓÖÏûºÄEDTAÈÜÒº44.10mL¡£¼ÆËãºÏ½ðÖÐPb¡¢Zn¡¢MgµÄÖÊÁ¿·ÖÊý¡£ ´ð°¸

--WORD¸ñʽ--¿É±à¼­--

1.½â£º(1)µ±pH=5.0ʱ£¬lg¦ÁY(H) = 6.45£»lgK¡ä= 16.31-6.45 9.86= 9.86 ¼´£ºK¡ä= 10

(2)lgcK¡ä= lg0.02 + 9.86 = 8.16 ©ƒ 6 £» ¡àÄÜ׼ȷµÎ¶¨ 2. ½â£º²é±íµÃ lgKBiY = 27.94 £» lgKNiY = 18.62 ¡ßlgK¡ä= lgK - lg¦ÁY(H) ¡Ý 8£¬¼´ lg¦ÁY(H) ¡Ü lgK - 8

¡à¶ÔÓÚBi3+Àë×Ó£ºlg¦ÁY(H) ¡Ü 19.94£»¶ÔÓÚNi2+Àë×Ó£ºlg¦ÁY(H) ¡Ü 10.62

²é±í4-3£¬lg¦ÁY(H) ¡Ü 19.94ʱ£¬pH ¡Ý 0.6 lg¦ÁY(H) ¡Ü 10.62ʱ£¬pH ¡Ý 3.2 ¢Ù ×îСµÄÖµpHΪ3.2¡£ ¢Ú ¦¤lg(cK) = 27.94 ¨C 18.62 =9.32£¾5£¬¿ÉÒÔÑ¡ÔñÐԵζ¨3+2+Bi£¬¶øNi²»¸ÉÈÅ£¬µ±pH¡Ý0.6ʱ£¬µÎ¶¨Bi3+£»µ±pH ¡Ý22+ʱ£¬Bi3+½«Ë®½âÎö³ö³Áµí£¬´Ëʱ(pH=0.6¡«2)NiÓëY²»Åä2+3+

λ£»µ±pH ¡Ý 3.2ʱ£¬¿ÉµÎ¶¨Ni£¬¶øBiϽµ¡£ 3. ½â£ºCaCO3% = 0.01016 ¡Á 10.43 ¡Á 10-3 ¡Á 100.09/100.0 = 106.1ug/mL

MgCO3% = 0.01016 ¡Á(15.28 ¨C 10.43) ¡Á 10-3 ¡Á 84.32/100.0 = 41.55ug/mL 4. ½â£º¦Ø(Al)=

[(0.05000¡Á25.00-0.02000¡Á21.50)¡Á10-3¡Á26.98/1.250]¡Á100%=1.77% 5.Ìõ¼þÎȶ¨³£Êý¼ÆËã

(1)½â£º pH=10ʱ lg¦ÁY(H) =0.45 lgKMgY=8.69 lgK¡¯MgY= lgKMgY-lg¦ÁY(H) =8.69-0.45=8.24 (2)½â£º pH=8ʱ lg¦ÁY(H) =2.27 lgKCaY=10.69 lgK¡¯CaY= lgKCaY-lg¦ÁY(H) =10.69-2.27=8.42 (3)½â£ºpH=5ʱ lg¦ÁY(H) =6.45 lgKZnY=16.50 lgK¡¯ZnY= lgKZnY- lg¦ÁY(H) =16.50-6.45=10.05 (4)½â£ºpH =10ʱ lgKZnY=16.50 lgK¡¯MgY= lgKMY -lgKZnY- lg¦ÁY(H)

pH =10 ¦ÁY(H)=100.45

lgK¡¯ZnY= 16.50-5.10£­0.45=10.95 6. ½â:

--WORD¸ñʽ--¿É±à¼­--

6Ñõ»¯»¹Ô­·´Ó¦µÎ¶¨·¨ Ò»Ìî¿ÕÌâ

1¡¢ÅäÖÆI2±ê×¼ÈÜҺʱ£¬±ØÐë¼ÓÈëKI£¬ÆäÄ¿µÄÊÇ

___________________________£»ÒÔAs2O3Ϊ»ù×¼ÎïÖʱ궨I2ÈÜÒºµÄŨ¶Èʱ£¬ÈÜÒºÓ¦¿ØÖÆÔÚpHΪ_________×óÓÒ¡£ 2¡¢ÒÑÖªÔÚ1mol/LHCl½éÖÊÖУ¬ÔòÏÂÁеζ¨·´Ó¦£º2Fe3+Sn2+=2Fe2++Sn4+ƽºâ³£ÊýΪ

_________£»»¯Ñ§¼ÆÁ¿µãµçλΪ_________£»·´Ó¦½øÐеÄÍêÈ«³Ì¶Èc(Fe2+)£¯c(Fe3+)Ϊ________¡£

3¡¢ÒÑÖªÔÚ1mol/LHCl½éÖÊÖУ»£¬ÔòÒÔFe3+µÎ¶¨Sn2+»¯Ñ§¼ÆÁ¿µãµçλΪ_________£»¼ÆÁ¿µãǰºóµçλ¸Ä±ä²»¶Ô³ÆÊÇÓÉÓÚ______________________¡£ 4¡¢Ñõ»¯»¹Ô­µÎ¶¨£¬ÔÚ»¯Ñ§¼ÆÁ¿µã¸½½üµÄµçλͻԼµÄ³¤¶ÌÓë________Á½µç¶ÔµÄ________Óйء£ËüÃÇÏà²îÔ½________£¬µçλͻԾԽ________¡£

--WORD¸ñʽ--¿É±à¼­--

5¡¢Ñõ»¯»¹Ô­µÎ¶¨µÄָʾ¼Á´ó¶¼Êǽṹ¸´ÔÓµÄ________£¬ËüÃǾßÓÐ________ÐÔÖÊ£¬ËüÃǵÄ________ºÍ________Ð;ßÓв»Í¬µÄ________¡£

6¡¢Ö±½ÓµâÁ¿·¨ÊÇÀûÓÃ________×÷±ê×¼µÎ¶¨ÈÜÒºÀ´Ö±½ÓµÎ¶¨Ò»Ð©________ÎïÖʵķ½·¨£¬·´Ó¦Ö»ÄÜÔÚ________ÐÔ»ò________ÐÔÈÜÒºÖнøÐС£

7¡¢¼ä½ÓµâÁ¿·¨·ÖÎö¹ý³ÌÖмÓÈëKIºÍÉÙÁ¿HClµÄÄ¿µÄÊÇ________¡¢________¡¢________¡£

8¡¢KMnO4ÊÇÒ»ÖÖÇ¿________¼Á£¬ÔÚËáÐÔÈÜÒºÖÐKMnO4Ó뻹ԭ¼Á×÷ÓÃʱ¿ÉÒÔ»ñµÃ________¸öµç×Ó¶ø±»________³É________¡£·´Ó¦Öг£ÓÃ________À´Ëữ¡£

9¡¢ÀûÓõ缫µçλ¿ÉÒÔÅжÏÑõ»¯»¹Ô­·´Ó¦½øÐеÄ________¡¢________ºÍ________¡£

10¡¢±ê¶¨Áò´úÁòËáÄÆÒ»°ã¿ÉÒÔÑ¡ÔñΪ________»ù×¼Îï,±ê¶¨¸ßÃÌËá¼ØÒ»°ã¿ÉÒÔÑ¡ÔñΪ________»ù×¼Îï¡£

11¡¢Ñõ»¯»¹Ô­µÎ¶¨ÖÐ,²ÉÓõÄָʾ¼ÁÓÐ________¡¢________ºÍ________¡£

12¡¢Ñõ»¯»¹Ô­·´Ó¦ÊÇ»ùÓÚ________×ªÒÆµÄ·´Ó¦,±È½Ï¸´ÔÓ,·´Ó¦³£ÊÇ·Ö²½½øÐÐ,ÐèÒªÒ»¶¨Ê±¼ä²ÅÄÜÍê³É,Òò´Ë,Ñõ»¯»¹Ô­µÎ¶¨ÊÇ,Ðè×¢Òâ________ËÙ¶ÈÓÚ________ËÙ¶ÈÏàÊÊÓ¦¡£ 13¡¢Ó°ÏìÑõ»¯»¹Ô­·´Ó¦ËٶȵÄÒòËØÓÐ________¡¢________¡¢________¡¢________¡£

14¡¢¸ßÃÌËá¼ØÔÚÇ¿Ëá½éÖÊÖб»»¹Ô­Îª________,ÔÚ΢Ëá,ÖÐÐÔ»òÈõ¼îÐÔ½éÖÊÖл¹Ô­Îª________,Ç¿¼îÐÔ½éÖÊÖл¹Ô­Îª________¡£ ´ð°¸

--WORD¸ñʽ--¿É±à¼­--

ÁªÏµ¿Í·þ£º779662525#qq.com(#Ìæ»»Îª@)