ÎÞ»ú»¯Ñ§Ï°Ìâ´ð°¸

Ôò[HAc]¡Ý0.2 mol¡¤dm3

£­

5.40 ¼ÓÈëÈÜÒººó[Mn2+]=

2?10?150.100.20£­

=0.10 mol¡¤dm3 2[OH]¡Ü

£­

?1.4?10?7 mol¡¤dm3 pH¡Ü7.15

£­

c(NH4?)n(NH4?)pH=pKa?lg=7.15 ?9.26?lgc(NH3)n(NH3)n(NH4?)ÇóµÃ=128.8

n(NH3)n(NH4+)=128.8¡Á100¡Á10-3¡Á0.10=1.29(mol) m(NH4Cl)=1.29¡Á53.5=69.0(g)

µÚÁùÕ Ñõ»¯»¹Ô­·´Ó¦

Ï° Ìâ

6.1 ʲôÊÇÑõ»¯Êý£¿ËüÓ뻯ºÏ¼ÛÓкÎÒìͬµã£¿Ñõ»¯ÊýµÄʵÑéÒÀ¾ÝÊÇʲô£¿ 6.2 ¾ÙÀý˵Ã÷ʲôÊÇÆ绯·´Ó¦£¿ 6.3 Ö¸³öÏÂÁл¯ºÏÎïÖи÷ÔªËصÄÑõ»¯Êý£º

Fe3O4 PbO2 Na2O2 Na2S2O3 NCl3 NaH KO2 KO3 N2O4 6.4 ¾ÙÀý˵Ã÷³£¼ûµç¼«µÄÀàÐͺͷûºÅ¡£

6.5 д³ö5ÖÖÓɲ»Í¬ÀàÐ͵缫×é³ÉµÄÔ­µç³ØµÄ·ûºÅºÍ¶ÔÓ¦µÄÑõ»¯»¹Ô­·´Ó¦·½³Ìʽ¡£ 6.6 ÅäƽÏÂÁз´Ó¦·½³Ìʽ

?? Zn(NO3)2 £« NH4NO3 £« H2O (1) Zn £« HNO3(¼«Ï¡) ??? HIO3 £« NO2 £« H2O (2) I2 £« HNO3 ?

45

??? Cu(NO3)2 £« NO £« H2O (3) Cu £« HNO3(Ï¡)???H3PO4 £« NO (4) P4 £« HNO3 £« H2O ??? Mg(NO3)2 £« N2O £« H2O (5) Mg £« HNO3(Ï¡) ??? CuSO4 £« NO2 £« H2O (6) CuS £« HNO3(Ũ) ??? H3AsO4 £« H2SO4 (7) As2S3 £« HNO3(Ũ) £« H2O ??? NaH2PO2 £« PH3 (8) P4 £« NaOH £« H2O ??? Cr2(SO4)3 £« K2SO4 £« I2 £« H2O (9) K2Cr2O7 £« KI £« H2SO4 ???MnSO4£«K2SO4£«Na2SO4£«CO2£«H2O (10) Na2C2O4£«KMnO4£«H2SO4??? MnSO4 £« K2SO4 £« O2 £« H2O (11) H2O2 £« KMnO4 £« H2SO4 ??? K2CrO4 £« K2SO4 £« H2O (12) H2O2 £« Cr2(S O4)3 £« KOH ??? Na2S4O6 £« NaI (13) Na2S2O3 £« I2 ??? NaCl £« Na2SO4 £« H2O (14) Na2S2O3 £« Cl2 £« NaOH ?? H2SO4 £« KMnO4 (15) K2S2O8 £« MnSO4 £« H2O ???6.7 ÅäƽÏÂÁÐÀë×Ó·´Ó¦Ê½(ËáÐÔ½éÖÊ)£º

Ag????? I2 (1) IO3 £« I ?£­

£­

?? MnO4 £« Bi3 (2) Mn2 £« NaBiO3 ?£«

£­

£«

?? CrO72 £« Pb2 (3) Cr3 £« PbO2 ?£«

£­

£«

?? C3H6O2 £« Mn2 (4) C3H8O £« MnO4 ?£­

£«

?? Cl £« H3PO4 (5) HClO £« P4 ?£­

46

6.8 ÅäƽÏÂÁÐÀë×Ó·´Ó¦Ê½(¼îÐÔ½éÖÊ)£º

?? CrO2 £« HSnO3 (1) CrO42 £« HSnO2 ?£­

£­

£­

£­

?? CrO42 (2) H2O2 £« CrO2 ?£­

£­

?? AsO43 £« I (3) I2 £« H2AsO3 ?£­

£­

£­

?? SiO32 £« H2 (4) Si £« OH ?£­

£­

?? BrO3 £« Br (5) Br2 £« OH ?£­

£­

£­

6.9 ¸ù¾Ýµç¼«µçÊÆÅжÏÔÚË®ÈÜÒºÖÐÏÂÁи÷·´Ó¦µÄ²úÎ²¢Åäƽ·´Ó¦·½³Ìʽ¡£

?? (1) Fe £« Cl2 ??? (2) Fe £« Br2 ??? (3) Fe £« I2 ??? (4) Fe £« HCl ??? (5) FeCl3 £« Cu ??? (6) FeCl3 £« KI ?6.10 ÒÑÖªµç¼«µçÊƵľø¶ÔÖµÊÇÎÞ·¨²âÁ¿µÄ£¬ÈËÃÇÖ»ÄÜͨ¹ý¶¨ÒåijЩ²Î±Èµç¼«µÄµç¼«µçÊÆ

À´²âÁ¿±»²âµç¼«µÄÏà¶Ôµç¼«µçÊÆ¡£Èô¼ÙÉèHg2Cl2 £« 2e£½2Hg £« 2Clµç¼«·´Ó¦µÄ

£­

£­

±ê×¼µç¼«µçÊÆΪ0£¬ÔòECu2£«/Cu¡¢EZn2£«/Zn±äΪ¶àÉÙ£¿

??6.11 ÒÑÖªNO3 £« 3H £« 2e===HNO2 £« H2O·´Ó¦µÄ±ê×¼µç¼«µçÊÆΪ0.94V£¬Ë®µÄÀë

£­

£«

£­

×Ó»ýΪKw£½10

£­14

£¬HNO2µÄµçÀë³£ÊýΪKa?£½5.1¡Á104¡£ÊÔÇóÏÂÁз´Ó¦ÔÚ298KʱµÄ

£­

±ê×¼µç¼«µçÊÆ¡£

NO3 £« H2O£« 2e===NO2 £« 2OH

£­

£­

£­

£­

47

6.12 ÒÑÖªÑÎËá¡¢ÇâäåËá¡¢ÇâµâËᶼÊÇÇ¿Ëᣬͨ¹ý¼ÆËã˵Ã÷£¬ÔÚ298K±ê׼״̬ÏÂAgÄÜ´Ó

ÄÄÖÖËáÖÐÖû»³öÇâÆø£¿

?ÒÑÖªEAg£½0.799 V£¬Ksp?[AgCl]£½1.8¡Á10£«/Ag£­10

£¬Ksp?[AgBr]£½5.0¡Á10

£­13

£¬

Ksp?[AgI]£½8.9¡Á10

£­17

¡£

¨D

¨D

¨D

6.13 ijËáÐÔÈÜÒºº¬ÓÐCl¡¢Br¡¢IÀë×Ó£¬ÓûÑ¡ÔñÒ»ÖÖÑõ»¯¼ÁÄܽ«ÆäÖеÄIÀë×ÓÑõ»¯¶ø²»

Ñõ»¯ClÀë×ÓºÍBrÀë×Ó¡£ÊÔ¸ù¾Ý±ê×¼µç¼«µçÊÆÅжÏӦѡÔñH2O2¡¢Cr2O72¡¢Fe3ÖÐ

¨D

¨D

£­

£«

¨D

µÄÄÄÒ»ÖÖ£¿

6.14 ÒÑÖªECu2£«/Cu£½0.34V£¬ECu2£«/Cu£«£½0.16V£¬Ksp?[CuCl]£½2.0¡Á106¡£Í¨¹ý¼ÆËãÅжÏ

??£­

·´Ó¦Cu2 £« Cu £« 2Cl === 2CuClÔÚ298K¡¢±ê׼״̬ÏÂÄÜ·ñ×Ô·¢½øÐУ¬²¢¼ÆËã

£«

£­

·´Ó¦µÄƽºâ³£ÊýK?ºÍ±ê×¼×ÔÓÉÄܱ仯?rGm?¡£

6.15 ͨ¹ý¼ÆËã˵Ã÷£¬ÄÜ·ñÓÃÒÑ֪Ũ¶ÈµÄ²ÝËá(H2C2O4)±ê¶¨ËáÐÔÈÜÒºÖÐKMnO4µÄŨ¶È£¿ 6.16 ΪÁ˲ⶨCuSµÄÈܶȻý³£Êý£¬Éè¼ÆÔ­µç³ØÈçÏ£ºÕý¼«ÎªÍ­Æ¬½þÅÝÔÚ0.1 mol¡¤dm

£«

£­3

Cu2µÄÈÜÒºÖУ¬ÔÙͨÈëH2SÆøÌåʹ֮´ï±¥ºÍ£»¸º¼«Îª±ê׼пµç¼«¡£²âµÃµç³Øµç¶¯ÊÆΪ0.67V¡£

ÒÑÖªECu2£«/Cu£½0.34V£¬EZn2£«/Zn£½£­0.76V£¬H2SµÄµçÀë³£ÊýΪKa1?£½1.3¡Á107£¬

??£­

Ka2?£½7.1¡Á10

£­15

¡£ÇóCuSµÄÈܶȻý³£Êý¡£

£­3

6.17 298Kʱ£¬Ïò1 mol¡¤dm

£«

µÄAgÈÜÒºÖеμӹýÁ¿µÄҺ̬¹¯£¬³ä·Ö·´Ó¦ºó²âµÃÈÜÒº

£­3

£«

ÖÐHg22Ũ¶ÈΪ0.311 mol¡¤dm

?£¬·´Ó¦Ê½Îª2Ag£«2Hg===2Ag£«Hg22

£«

£«

(1) ÒÑÖªEAg£«/Ag£½0.799 V£¬ÇóEHg2£«/Hg£»

?2(2) ½«·´Ó¦Ê£ÓàµÄAgºÍÉú³ÉµÄAgÈ«²¿³ýÈ¥£¬ÔÙÏòÈÜÒºÖмÓÈëKCl¹ÌÌåʹHg22Éú

£«

£«

³ÉHg2Cl2³Áµí£¬²¢Ê¹ÈÜÒºÖÐClŨ¶È´ïµ½1 mol¡¤dm3¡£½«´ËÈÜÒº£¨Õý¼«£©Óë±ê

£­

£­

×¼Çâµç¼«£¨¸º¼«£©×é³ÉÔ­µç³Ø£¬²âµÃµç¶¯ÊÆΪ0.280V£¬ÊÔÇóHg2Cl2µÄÈܶȻý³£Êý²¢Ð´³ö¸Ãµç³ØµÄ·ûºÅ£»

48

ÁªÏµ¿Í·þ£º779662525#qq.com(#Ì滻Ϊ@)