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22. ÒªÅäÖÆpH = 3.5µÄ»º³åÈÜÒº£¬ÒÑÖªHFµÄpKa = 3.18£¬H2CO3µÄpKa?10.25£¬±ûËáµÄpKa

2= 4.88£¬¿¹»µÑªËáµÄpKa = 4.30£¬»º³å¼ÁӦѡ( )¡£ A. HF-NaF

B. Na2CO3-NaHCO3 C. ±ûËá-±ûËáÄÆ

D. ¿¹»µÑªËá-¿¹»µÑªËáÄÆ

23. Ë®µÄÀë×Ó»ýKwÖµ25?CʱΪ1?10-14£¬60?CʱΪ1?10-13£¬ÓÉ´Ë¿ÉÍÆ¶Ï( )¡£

A. Ë®µÄµçÀëÊÇÎüÈ鵀 C. 60?Cʱ£¬[OH-] = 1?10-7

B. Ë®µÄpH 60?Cʱ´óÓÚ25?Cʱ D. Ë®µÄµçÀëÊÇ·ÅÈȵÄ

24. ÓÉÓÚÓÃÈõËáÈõ¼îÏ໥µÎ¶¨Ê±£¬·´Ó¦½øÐеò»¹»ÍêÈ«£¬ºÜÄÑÓÃָʾ׼ȷָʾ»¯Ñ§¼ÆÁ¿µãµÄ

µ½´ï£¬Òò´ËÔÚËá¼îµÎ¶¨·¨ÖУ¬ÅäÖÆ±ê×¼ÈÜÒº×ÜÊÇÓÃ( )¡£ A. Ç¿Ëá»òÇ¿¼î

B. ÈõËá»òÈõ¼î

C. Ç¿¼îÈõËáÑÎ

D. Ç¿ËáÈõ¼îÑÎ

25. ÏÂÁÐÈõËá»òÈõ¼îÄÜÓÃËá¼îµÎ¶¨·¨Ö±½Ó׼ȷµÎ¶¨µÄÊÇ( )¡£

A. 0.1 mol/L±½·ÓKa = 1.1?10

-10

B. 0.1 mol/L H3BO3 Ka = 7.3?10 D. 0.1 mol/L HF Ka = 3.5?10-4

-10

C. 0.1 mol/L ôǰ·Kb = 1.07?10-8

26. ÓÃÅðɰ±ê¶¨0.1 mol/L HClʱ£¬Ó¦Ñ¡ÓõÄָʾ¼ÁÊÇ( )¡£

A. ÖÐÐÔºì

B. ¼×»ùºì

C. ·Ó̪

D. °ÙÀï·Ó̪

27. ÓÃNaOH²â¶¨Ê³´×µÄ×ÜËáÁ¿£¬×îºÏÊʵÄָʾ¼ÁÊÇ( )¡£

A. ÖÐÐÔºì

B. ¼×»ùºì

C. ·Ó̪

D. ¼×»ù³È

28. ָʾ¼ÁµÄʵ¼Ê±äÉ«·¶Î§ÊǸù¾Ý( )¶øµÃµ½µÄ¡£

A. ÈËÑÛ¹Û²ì

B. ÀíÂÛ±äÉ«µã¼ÆËã C. µÎ¶¨¾­Ñé

D. ±È½ÏµÎ¶¨

29. Na2CO3ºÍNaHCO3»ìºÏÎï¿ÉÓÃHCl±ê×¼ÈÜÒºÀ´²â¶¨£¬²â¶¨¹ý³ÌÖÐÓõ½µÄÁ½ÖÖָʾ¼Á·Ö

±ðÊÇ( )¡£

A. ·Ó̪¡¢°ÙÀï·ÓÀ¶ B. ·Ó̪¡¢°ÙÀï·Ó̪ C. ·Ó̪¡¢ÖÐÐÔºì

D. ·Ó̪¡¢¼×»ù³È

30. Na2CO3ºÍNaHCO3»ìºÏÎï¿ÉÓÃHCl±ê×¼ÈÜÒºÀ´²â¶¨£¬²â¶¨¹ý³ÌÖÐÁ½ÖÖָʾ¼ÁµÄµÎ¼Ó˳

ÐòÊÇ( )¡£ A. ·Ó̪¡¢¼×»ù³È

B. ¼×»ù³È¡¢·Ó̪

C. ·Ó̪¡¢°ÙÀï·ÓÀ¶ D. °ÙÀï·ÓÀ¶¡¢·Ó̪

31. ÔÚËá¼îµÎ¶¨ÖУ¬Ò»°ã°ÑµÎ¶¨ÖÁ»¯Ñ§¼ÆÁ¿µã( )µÄÈÜÒºpH±ä»¯·¶Î§³ÆÎªµÎ¶¨Í»Ô¾·¶

Χ¡£

A. ǰºó?0.1%Ïà¶ÔÎó²î C. 0.0%

B. ǰºó?0.2%Ïà¶ÔÎó²î D. ǰºó?1%Ïà¶ÔÎó²î

32. H3PO4ÊÇÈýÔªËᣬÓÃNaOHÈÜÒºµÎ¶¨Ê±£¬ÓÐ( )¸öÃ÷ÏÔµÄpHͻԾ¡£

A. 1

B. 2

-8

C. 3 D. ÎÞ·¨È·¶¨

33. ÓÃÇ¿¼îµÎ¶¨Ò»ÔªÈõËáʱ£¬Ó¦·ûºÏcKa ? 10µÄÌõ¼þ£¬ÕâÊÇÒòΪ( )¡£

A. cKa < 10-8ʱ£¬µÎ¶¨Í»Ô¾·¶Î§Ì«Õ­ C. cKa < 10-8ʱ£¬Ö¸Ê¾¼Á²»·¢ÉúÑÕÉ«±ä»¯

B. cKa < 10-8ʱ£¬ÎÞ·¨È·¶¨»¯Ñ§¼ÆÁ¿¹ØÏµ D. cKa < 10-8ʱ£¬·´Ó¦²»ÄܽøÐÐ

34. ÔÚH2C2O4ÈÜÒºÖУ¬¶ÔÓÚH2C2O4ÐÎÌåµÄ·Ö²¼ÏµÊý?(H2C2O4)¶øÑÔ£¬ÏÂÁÐÕýÈ·µÄÐðÊöÊÇ( )¡£

A. ?(H2C2O4)ËæpHÔö´ó¶ø¼õС C. ?(H2C2O4)ËæËá¶ÈÔö´ó¶ø¼õС

B. ?(H2C2O4)ËæpHÔö´ó¶øÔö´ó D. ?(H2C2O4)ËæËá¶È¼õС¶øÔö´ó

35. ÓÃËá¼îµÎ¶¨·¨ÄÜÖ±½Ó׼ȷ·Ö²½µÎ¶¨¶àÔªËáµÄÌõ¼þÊÇ( )¡£

A. Ka?10?8

i

B. Ka/Kaii?1?10

5

?10

5C. cKa?10?8

iD. cKa?10?8£¬Ka/Kaiii?136. ÏÂÁÐËá¼îµÎ¶¨·´Ó¦ÖУ¬Æä»¯Ñ§¼ÆÁ¿µãpHµÈÓÚ7.00µÄÊÇ( )¡£

A. NaOHµÎ¶¨HAc B. HClµÎ¶¨°±Ë®

C. HClµÎ¶¨Na2CO3 D. NaOHµÎ¶¨HCl

37. ÕôÁ󷨲ⶨNH4+ (Ka = 5.6?10-10)£¬Õô³öµÄNH3ÓùýÁ¿µÄH3BO3 (Ka = 5.8?10-10)ÈÜÒºÎüÊÕ£¬

È»ºóÓñê×¼HClÈÜÒºµÎ¶¨£¬Ëù¼ÓÈëµÄH3BO3ÈÜÒºµÄÁ¿( )¡£ A. ÐèÒÑ֪׼ȷŨ¶È B. ÐèÒÑ֪׼ȷÌå»ý C. ²»Ðè׼ȷÁ¿È¡

D. Ũ¶È¡¢Ìå»ý¾ùÐè׼ȷ

38. ÓÃ0.1000 mol/L HClµÎ¶¨0.1000 mol/L NaOHʱµÄµÎ¶¨Í»Ô¾·¶Î§ÊÇpH 9.7?4.3£¬ÓÃ0.01000

mol/L HClµÎ¶¨0.01000 mol/L NaOHʱµÄµÎ¶¨Í»Ô¾·¶Î§ÊÇpH( )¡£ A. 9.7?4.3

B. 8.7?4.3

C. 9.7?5.3

D. 8.7?5.3

39. ÓÃ0.1000 mol/L NaOHµÎ¶¨0.1000 mol/L HAc (pKa = 4.7)ʱpHͻԾ·¶Î§Îª7.7?9.7£¬ÓÉ

´Ë¿ÉÍÆ¶ÏÓÃ0.1000 mol/L NaOHµÎ¶¨pKa = 3.7µÄijһԪËáʱ£¬ÆäpHͻԾ·¶Î§Îª( )¡£ A. 6.7?8.7

B. 6.7?9.7

C. 6.7?10.7

D. 7.7?9.7

40. ÓÃͬһNaOHÈÜÒº·Ö±ðµÎ¶¨Ìå»ýÏàͬµÄH2SO4ºÍHAcÈÜÒº£¬ÏûºÄNaOHµÄÌå»ýÏàµÈ£¬

˵Ã÷ÕâÁ½ÖÖÈÜÒºÖÐ( )¡£ A. ÇâÀë×ÓŨ¶ÈÏàµÈ

C. H2SO4Ũ¶ÈΪHAcŨ¶ÈµÄ1/2

B. H2SO4ºÍHAcŨ¶ÈÏàµÈ D. Á½´ÎµÎ¶¨µÄpHͻԾ·¶Î§ÏàµÈ

41. Ëá¼îµÎ¶¨·¨²â¶¨CaCO3º¬Á¿Ê±£¬Ó¦²ÉÓÃ( )¡£

A. Ö±½ÓµÎ¶¨·¨

B. ·µµÎ¶¨·¨

C. Öû»µÎ¶¨·¨

D. ¼ä½ÓµÎ¶¨·¨

42. ÓÃ0.1000 mol/L HClµÎ¶¨0.1000 mol/L NH3?H2O (pKb = 4.75)ʱ£¬ÒÔϼ¸ÖÖָʾ¼ÁÖУ¬×î

¼Ñָʾ¼ÁÊÇ( )¡£ A. °ÙÀï·Ó̪

B. ¼×»ùºì

C. ÖÐÐÔºì

D. ·Ó̪

43. ¿ÕÆøÖÐCO2¶ÔËá¼îµÎ¶¨µÄÓ°Ï죬ÔÚÏÂÁÐÄÄÀàÇé¿öʱ¿ÉºöÂÔ£¿( )¡£

A. ÒÔ¼×»ù³ÈΪָʾ¼Á C. ÒÔ·Ó̪Ϊָʾ¼Á

B. ÒÔÖÐÐÔºìΪָʾ¼Á D. ÒÔ°ÙÀï·Ó̪Ϊָʾ¼Á

44. ±ê¶¨HClºÍNaOHÈÜÒº³£ÓõĻù×¼ÎïÖÊÊÇ( )¡£

A. ÅðɰºÍEDTA

B. ²ÝËáºÍK2Cr2O7

12C. CaCO3ºÍ²ÝËá

3D. ÅðɰºÍÁÚ±½¶þ¼×ËáÇâ¼Ø

?3?8?1345. ÒÔNaOHµÎ¶¨H3PO4 (Ka?7.6?10, Ka?6.2?10, Ka?4.4?10)ÖÁÉú³ÉNaH2PO4,

ÈÜÒºµÄpHΪ( )¡£

A. 2.3 ËÄ. ¼ÆËãÌâ

B. 4.7 C. 5.8 D. 9.8

1. ÓÐÒ»ÈõËáHA£¬Æä½âÀë³£ÊýKa = 6.4?10-7£¬Çóc(HA) = 0.30 mol/LʱÈÜÒºµÄ[H+]¡£ 2. Ũ¶ÈΪ0.010 mol/L HAcÈÜÒºµÄ? = 0.042£¬ÇóHAcµÄKa¼°ÈÜÒºÖеÄ[H]¡£ÔÚ500 mLÉÏ

ÊöÈÜÒºÖмÓÈë2.05¿Ë¹Ì̬NaAcºó£¬Èô²»¿¼ÂÇÌå»ý±ä»¯£¬Çó´ËʱÈÜÒºµÄ[H+]ºÍpH£¬½«Á½Õ߱Ƚϣ¬²¢¸ø³öÏàÓ¦µÄ½áÂÛ¡£

3. ÓûÅäpH = 10.0µÄ»º³åÈÜÒº£¬Ó¦ÔÚ300 mL 0.5 mol/L NH3?H2OÈÜÒºÖмÓÈëNH4Cl¶àÉÙ¿Ë£¿ 4. ¼ÆËãŨ¶ÈΪ0.02 mol/LµÄH2CO3ÈÜÒºÖÐ[H]¡¢[HCO3]¡¢[CO3]¸÷ÊǶàÉÙ£¿ÒÑÖª£º

Ka1?4.2?10?7+

+-2-

,Ka2?5.61?10?11

5. ÔÚ1Éý0.2 mol/L HAcÈÜÒºÖмÓÈë¶àÉÙ¿ËNaOH£¬²ÅÄÜʹÈÜÒºÖеÄ[H+]´ïµ½6.5?10£­5

mol/L£¿

6. ³Æ´¿CaCO3 0.5000 g£¬ÈÜÓÚ50.00 mL¹ýÁ¿µÄHClÖУ¬¶àÓàËáÓÃNaOH»ØµÎ£¬ÓÃÈ¥6.20 mL¡£

Èô1.000 mL NaOHÏ൱ÓÚ1.010 mL HClÈÜÒº£¬ÇóÕâÁ½ÖÖÈÜÒºµÄŨ¶È¡£

7. ³ÆÈ¡»ìºÏ¼îÑù0.3400 gÖÆ³ÉÈÜÒººó£¬ÓÃ0.1000 mol/L HClµÎÖÁ·Ó̪Öյ㣬ÏûºÄHCl 15.00

mL£¬¼ÌÐøÓÃÏàͬµÄÑÎËá±ê×¼ÈÜÒºµÎÖÁ¼×»ù³ÈÖյ㣬ÓÖÓÃÈ¥HCl 36.00 mL£¬´ËÑùÆ·Öк¬ÓÐÄÄЩ¼îÐÔÎïÖÊ£¿º¬Á¿¸÷ÊǶàÉÙ£¿

8. Ôڱ궨KMnO4ÈÜÒºµÄŨ¶Èʱ£¬³ÆÈ¡²ÝËáÄÆ(Na2C2O4) 0.2262 g£¬¾­Ë®Èܽâºó£¬ÔÚÊʵ±Ìõ

¼þÏ£¬ÓÃKMnO4ÈÜÒºµÎ¶¨£¬ÏûºÄ41.50 mL£¬¼ÆËã´ËKMnO4ÈÜÒºµÄŨ¶È¼°¶ÔFeµÄµÎ¶¨¶È¡£

9. ²â¶¨Ò»¹¤ÒµH2SO4º¬Á¿£¬È¡ÑùΪ1.125 g£¬¶¨Á¿Ï¡ÊͳÉ250.00 mL£¬Ò¡ÔȺóÒÆÈ¡25.00 mL£¬

ÏûºÄ0.1340 mol/L NaOHÈÜÒº15.40 mL£¬Çó¸Ã¹¤ÒµÁòËáÖÐH2SO4µÄÖÊÁ¿·ÖÊý¡£

10. ÓÐһʯ»ÒʯÊÔÑù£¬¼ÙÉèÆä½öº¬CaCO3¼°ÆäËü²»ÓëËá×÷ÓõÄÎïÖÊ£¬ÏÖ³ÆÈ¡¸ÃÊÔÑù0.3000 g£¬

¼ÓÈë25.00 mL 0.2500 mol/LµÄHClÈÜÒº£¬Öó·ÐÈÃËüÃÇ·´Ó¦ÍêÈ«£¬ÓÃ0.2000 mol/L NaOHÈÜÒº·µµÎ¶¨£¬ÓÃÈ¥5.84 mL£¬ÇóÕâÖÖʯ»ÒʯÖÐCaCO3¼°CaOµÄÖÊÁ¿·ÖÊý¡£

²Î¿¼´ð°¸ Ò». Ìî¿ÕÌ⣺

1. ¼õС£¬¼õС£¬Í¬Àë×ÓЧӦ£»2. 1.0?10-3 mol/L£»3. H3O+£¬OH-£»4. KaKb = Kw£»5.

Ka1Ka2£»6. ¼õС£¬Ï½µ£»7. CH3CH2CH2COOH£¬1.30?10£­5£»8. <5£¬5?7£¬>7£»9. »ù±¾²»±ä£¬¸üÇ¿£»10. NaCN > NaAc > NaNO2£»11. [Fe(H2O)6]£¬[Fe(H2O)4(OH)2]£»12. ±äÉ«·¶Î§£¬ËáÉ«ºÍ¼îÉ«µÄ»ìºÏÉ«£»13. ¼õС£¬c?Ka/c£»14. Ka2£¬Ka2£»15. 0.1 mol/L£¬0.1£»16. Ëá¼î£¬NH3 +

3+

+

H2O = NH4+ + OH-£»17. 4.4?8.0£»18. 4.8£»19. 9.0£»20. Ëá¼îÐÔ£¬ËáʽÐÎÌ壬¼îʽÐÎÌ壬0.1£¬10£¬2£»21. 1010£»22. 1.8?10-4£»23. µÎ¶¨Í»Ô¾·¶Î§£¬ÑÕÉ«±ä»¯ÊÇ·ñÒ×Óڹ۲죻24. ¸ß£¬¶à£»25. Õ­£»26. 3.4£¬2.4?4.4£»27. ÀíÂÛ±äÉ«µã£»28. ¶èÐÔȾÁÏ£¬Ö¸Ê¾¼Á£»29. ˫ָʾ¼Á£»30. »¯Ñ§¼ÆÁ¿µã£»31. ÈÜÒºµÄpH£¬Ìå»ý£¬µÎ¶¨ÇúÏߣ»32. Ç¿Ëá»òÇ¿¼î£¬Å¨¶È£¬»¯Ñ§¼ÆÁ¿£»33. cKa?10?8£¬

2Ka1/Ka2?10£»34. ÄÜÌṩÖÊ×ÓµÄÎïÖÊ£¬ÄܽÓÊÜÖÊ×ÓµÄÎïÖÊ£¬Ka?Kb = Kw£»35. Æ«¸ß£¬²»±ä¡£

5¶þ. ÕýÎóÅжÏÌâ

1. ?£»2. ?£»3. ?£»4. ?£»5. ?£»6. ?£»7. ?£»8. ?£»9. ?£»10. ?£»11. ?£»12. ?£»13. ?£»14. ?£»15. ?£»16. ?£»17. ?£»18. ?£»19. ?£»20. ?£»21. ?£»22. ?£»23. ?£»24. ?£»25. ?£»26. ?£»27. ?£»28. ?¡£ Èý. µ¥Ñ¡Ìâ

1. D£»2. A£»3. B£»4. D£»5. C£»6. A£»7. A£»8. D£»9. A£»10. A£»11. A£»12. B£»13. C£»14. D£»15. A£»16. C£»17. D£»18. A£»19. B£»20. D£»21. B£»22. A£»23. A£»24. A£»25. D£»26. B£»27. C£»28. A£»29. D£»30. A£»31. A£»32. B£»33. A£»34. A£»35. D£»36. D£»37. C£»38. D£»39. B£»40. C£»41. B£»42. B£»43. A£»44. D£»45. B¡£ ËÄ. ¼ÆËãÌâ 1. [H?]?0.30?6.4?10?7?4.4?10[H]0.010??4mol/L

2. ¸ù¾Ý?µÄ¶¨Ò壺0.042?K??£¬[H+] = 4.2?10-4 [mol/L]

?5(4.2?10?4)2?40.010?4.2?10?5?1.8?10

?4[H]?1.8?10cNaAc??0.010?4.2?10mol/L

?0.0500mol/L

2.05?(22.99?12.01?2?1.008?3?16.00?2)0.500[H]?Ka?cacb?1.8?10?50.010.05?3.6?10?6mol/L

pH = 5.44

3. 1?10?10?1?10?14?5cNH4Cl0.51.77?10

cNH4Cl?0.088mol/L

0.088?0.3?(14.01+1.008?4+35.45) = 1.4 g 4. [H?]?0.02?4.2?10?7?9.2?10?5

?5

[HCO]??30.02?9.2?10(9.2?10)?9.2?10?52?5?4.2?10?7?7?5?4.2?10?9.2?10?5.6?10?11= 9.1?10 mol/L

-5

[CO2?3]?0.02?4.2?10(9.2?10)?9.2?10?52?5?7?5.6?10?7?11?5?4.2?10?9.2?10?5.6?10?11= 5.5?10 mol/L

-11

5. É裺[H+] = 6.5?10-5ʱ£¬NaAcµÄŨ¶ÈΪx mol/L

6.5?10?5?1.76?10?50.2?xx

6.5x = 0.352 ¨C 1.76x, x = 0.043 mol/L

0.043?(22.99+16.00+1.008)?1.0 = 1.7 g 6. 0.5000?(40.08+12.01+16.00?3) = 5.005?10-3 mol

É裺NaOHÈÜÒºµÄŨ¶ÈΪx mol/L

5.005?10-3?2 = x?1.010?0.05000 ¨C 0.00620x x = 0.2312 mol/L

HClÈÜÒºµÄŨ¶È= 0.2289 mol/L

7. Na2CO3% = 0.1000?0.01500?(22.99?2 + 12.01 + 16.00?3)?0.3400?100% = 46.76%

NaHCO3% = 0.1000?(0.03600 ¨C 0.01500)?(22.99 + 1.008 + 12.01 + 16.00?3)?0.3400?100%

= 51.89%

0.2262134.08. 2MnO4- + 5C2O42- +16H+ = 2Mn2+ + 10CO2? + 8H2O

cKMnO4??5?2?0.04150?0.01627mol/L

MnO4- + 5Fe2- + 8H+ = Mn2+ + 5Fe3+ + 4H2O

0.01627?10?3?5?56.85?4.625?10?3g/mL

9. H2SO4 + 2NaOH = Na2SO4 + 2H2O

mH2SO4%?0.1340?0.01540?2?250.0025.00?98.03?1.125?100%?89.91%

10. CaCO3 + 2HCl = CaCl2 + H2CO3£¬ ¼ÓÈÈÖó·Ð£ºH2CO3 = CO2? + H2O

HCl + NaOH = NaCl + H2O

mCaCO3%?(0.2500?0.02500?0.2000?0.00584)?2?100.1?0.3000?100%?84.78%

mCaO%?(0.2500?0.02500?0.2000?0.00584)?2?56.08?0.3000?100%?47.50%