(2) 0.10 mol¡¤L-1HClÈÜÒºÓë0.10mol¡¤L-1Na2CO3ÈÜÒºµÈÌå»ý»ìºÏ£» (3) 0.10 mol¡¤L-1NaOHÈÜÒºÓë0.10 mol¡¤L-1NaHCO3ÈÜÒºµÈÌå»ý»ìºÏ¡£ 12. ͨ¹ý¼ÆËã˵Ã÷£º
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13. ÒÔÏÂÐðÊöÕýÈ·µÄÊÇ
A. ·²ÊǶàÔªÈõËáÈÜÒºÖУ¬ÆäËá¸ùµÄŨ¶È¶¼µÈÓÚÆä×îºóÒ»¼¶µÄËá³£Êý¡£
B. NaAcÈÜÒºÓëHClÈÜÒºÆð·´Ó¦£¬·´Ó¦´ïƽºâʱ£¬Æ½ºâ³£ÊýµÈÓÚ´×ËáµÄËá³£ÊýµÄµ¹Êý¡£ C. H2CO3ÈÜÒºÖмÓÈëNaHCO3ºó²úÉúͬÀë×ÓЧӦ£¬ÍùNaHCO3ÈÜÒºÖмÓÈëNa2CO3²»»á²úÉúͬÀë×ÓЧӦ¡£
D .ÊÒÎÂÏÂ1.0¡Á10-4 mol¡¤L-1°±Ë®ÈÜÒºÖУ¬Ë®µÄÀë×Ó»ý³£ÊýΪ1.0¡Á10-10¡£ 14. ÏÂÁÐÎïÖÊÔÚË®Öа´¼îÐÔÓÉÈõµ½Ç¿ÅÅÁеÄ˳ÐòÊÇ
A. OH-£¼ HPO42-£¼ NH3 £¼ HSO4-£¼H2O B. OH-£¼NH3£¼HPO42-£¼ HSO4-£¼H2O C. HPO42-£¼OH-£¼NH3£¼H2O£¼HSO4- D. HSO4-£¼H2O£¼ HPO42-£¼ NH3£¼ OH- E. HPO42-£¼OH-£¼H2O£¼ NH3£¼HSO4- 15. HAcÔÚÏÂÁÐÄÄÒ»ÖÖÈÜÒºÖнâÀë¶È×î´ó
A£®0.1 mol¡¤L-1NaCl B£®0.1 mol¡¤L-1HCl C£®0.1 mol¡¤L-1NaAc D£®´¿Ë® E£®0.1 mol¡¤L-1H2SO4
16. Tt¡¯s known that the pKa of HA is 4.20, calculate the [H3O+], pH and [A-] of the following solutions:
(1) a solution consisted of 0.10 mol¡¤L-1 of HA and 0.20 mol¡¤L-1 of NaA; (2) a solution consisted of 0.10 mol¡¤L-1 of HA and 0.20 mol¡¤L-1 of HCl.
17. 100 mL of 0.010 mol¡¤L-1 Na2SO4 solution was mixed with 100 mL of 0.20 mol¡¤L-1KNO3 solution, calculate the ionic strength of the mixed solution.
18. 50 mL of 0.100 mol¡¤L-1 HA solution was mixed with 100 mL of 0.100 mol¡¤L-1NaOH solution, the pH of the mixed solution was equal to 5.67. Calculate the proton transfer equilibrium constant of HA.
19. In a solution of NH4Ac, if c(NH4Ac)= 0.1 mol¡¤L-1 , the concentration of proton would be£º A£®0.1KHAcD£®
B£®
E£®
0.1KNH?4 C£®
KHAc?KNH3
KHAc?KwKNH3KNH3?KwKHAc
20. At the temperature of the human body, 370C, the value of Kw is 2.4¡Á10-14. Which the follow is right?
A.[H+] = [OH-] =10-7mol¡¤L-1 B. [H+] = [OH-] = 1.55¡Á10-7mol¡¤L-1 C. pH + pOH = 14 D. [H+]£¼[OH-] =10-7mol¡¤L-1 E. pH + pOH£¾14
21. All of the following are acid-base conjugate pairs EXCEPT
A. HONO, NO2- B. H3O+, OH- C. CH3NH3+, CH3NH2 D. HS-, S2- E. C6H5COOH, C6H5COO-
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1. ʲôÊÇ»º³åÈÜÒº?ʲôÊÇ»º³åÈÝÁ¿?¾ö¶¨»º³åÈÜÒºpHºÍ»º³åÈÝÁ¿µÄÖ÷ÒªÒòËظ÷ÓÐÄÄЩ? 2. ÊÔÒÔKH2PO4-Na2HPO4»º³åÈÜҺΪÀý£¬ËµÃ÷ΪºÎ¼ÓÉÙÁ¿µÄÇ¿Ëá»òÇ¿¼îʱÆäÈÜÒºµÄpH»ù±¾±£³Ö²»±ä¡£
3. ÒÑÖªÏÂÁÐÈõËáµÄpKa£¬ÊÔÇóÓëNaOHÅäÖƵĻº³åÈÜÒºµÄ»º³å·¶Î§£® ¢Å ÅðËá (H3BO3)µÄpKa£½9.27
¢Æ ±ûËá(CH3CH2COOH)µÄpKa£½4.86
¢Ç Á×Ëá¶þÇâÄÆ(NaH2PO4) µÄpKa(H2PO4-)£½7.21 ¢È ¼×Ëá(HCOOH) µÄpKa£½3.75
4. 0.20 mol?L-1 NH3 ºÍ0.10 mol?L-1 NH4Cl×é³ÉµÄ»º³åÈÜÒºµÄpHΪ¶àÉÙ? pKb= 4.75 5. ¼ÆËãÏÂÁÐNH3 ºÍNH4Cl×é³ÉµÄ²»Í¬pHµÄ»º³åÈÜÒºµÄ[NH3]/[NH4+]»º³å±È¡£ ¢Å pH = 9.00 ¢Æ pH = 8.80 ¢Ç pH = 10.00 ¢È pH = 9.60
6. Ò»¸öÓÉ 0.25 mol?L-1 KHCO3 Óë0.32 mol?L-1 K2CO3×é³ÉµÄ»º³åÈÜÒº,H2CO3ÊǶþÔªË᣺ Ka1 = 4.5?10-7 ;Ka2 = 4.7?10-11 (25¡æ)¡£
¢Å ÄÄÒ»¸öKa¶ÔÓÚÕâ¸ö»º³åÈÜÒºÊǸüÖØÒª? ¢Æ Õâ¸ö»º³åÈÜÒºµÄpHΪ¶àÉÙ?
7. ÓÃ0.055 mol¡¤L-1KH2PO4ºÍ0.055 mol¡¤L-1Na2HPO4Á½ÖÖÈÜÒºÅä³ÉpH½üËÆֵΪ7.40µÄ»º³åÈÜÒº1 000 mL£¬ÎÊÐèÈ¡ÉÏÊöÈÜÒºÌå»ý¸÷Ϊ¶àÉÙ? ÒÑÖªpKa (H2PO4-) = 7.21 (25¡æ)¡£
8. °¢Ë¾Æ¥ÁÖ(ÒÒõ£Ë®ÑîËá¡¢ÒÔHasp±íʾ)ÒÔÓÎÀëËᣨδ½âÀëµÄ£©ÐÎʽ´ÓθÖÐÎüÊÕ£¬Èô²¡ÈË·þÓýâËáÒ©£¬µ÷ÕûθÈÝÎïµÄpHΪ2.95£¬È»ºó¿Ú·þ°¢Ë¾Æ¥ÁÖ0.65g¡£¼ÙÉ谢˾ƥÁÖÁ¢¼´Èܽ⣬ÇÒθÈÝÎïµÄpH²»±ä£¬Îʲ¡ÈË¿ÉÒÔ´ÓθÖÐÁ¢¼´ÎüÊյݢ˾ƥÁÖΪ¶àÉÙ¿Ë£¨ÒÒõ£Ë®ÑîËáµÄMr=180.2¡¢pKa=3.50£©£¿
9. ÅäÖÆpH £½ 7.40µÄ»º³åÈÜÒº1 500 mL¡£
¢Å ½ñÓлº³åϵHAc-NaAc¡¢KH2PO4-Na2HPO4¡¢NH4Cl-NH3£¬ÎÊÑ¡ÓúÎÖÖ»º³åϵ×îºÃ? ¢Æ ÈçÑ¡ÓõĻº³åϵµÄ×ÜŨ¶ÈΪ0.200 mol¡¤L-1£¬ÐèÒª¹ÌÌå¹²éîËáºÍ¹ÌÌå¹²éî¼îÎïÖʵÄÁ¿
Ϊ¶àÉÙ (¼ÙÉè²»¿¼ÂÇÌå»ýµÄ±ä»¯)?
10. ÈܽâÓÐNH4ClºÍ NH3 µÄÈÜÒºÖÐNH3 µÄŨ¶ÈΪ0.500 mol?L-1 ²¢ÇÒÈÜÒºµÄpH ÊÇ 8.95. pKa(NH4+ )= 9.25
¢Å ¼ÆËãNH4+µÄƽºâŨ¶È
¢Æ µ±4.00 g µÄ NaOH(s)±»¼ÓÈëµ½1.00 LµÄ´ËÈÜÒººó, pHΪ¶àÉÙ?(ºöÂÔÌå»ýµÄ±ä»¯) 11. µ¥´¿ÐÔËá¼îʧºâÖ÷Òª¿¿ÑªÆø·ÖÎöÕï¶Ï£¬¸ù¾ÝpHµÄ±ä»¯¿ÉÅжÏËáÖж¾»¹ÊǼîÖж¾¡£ÁÙ´²¼ìÑé²âµÃÈýÈËѪ½¬ÖÐHCO3-ºÍÈܽâµÄCO2µÄŨ¶ÈÈçÏ£º
¢Å [HCO3-]£½24.0 mmol¡¤L-1¡¢[CO2(aq)]£½1.20 mmol¡¤L-1 ¢Æ [HCO3-]£½21.6 mmol¡¤L-1¡¢[CO2(aq)]£½1.34 mmol¡¤L-1 ¢Ç [HCO3-]£½56.0 mmol¡¤L-1¡¢[CO2(aq)]£½1.40 mmol¡¤L-1
ÊÔ¼ÆËãÈýÈËѪ½¬µÄpH£¬²¢ÅжϺÎÈËÊôÕý³££¬ºÎÈËÊôËáÖж¾(pH? 7.35), ºÎÈËÊô¼îÖж¾ (pH ? 7.45)¡£ÒÑÖªpKa1[CO2(aq)] = 6.10£¨37¡æ£©
12. ÓÉ184 mL 0.442 mol?L-1 HCl ºÍ 0.500 L 0.400 mol?L-1 NaAc»ìºÏÐγɵĻº³åÈÜÒº¡£pKa(HAc) = 4.76 (25¡æ)
¢Å ´Ë»º³åÈÜÒºµÄpHΪ¶àÉÙ?
¢Æ ¶àÉÙÖÊÁ¿µÄ KOH¹ÌÌå¼Óµ½ 0.500 L´Ë»º³åÈÜÒº¿ÉÒÔʹpH ¸Ä±ä0.15 (?pH = 0.15)? 13. ÄûÃÊËᣨËõдH3Cit£©¼°ÆäÑÎΪһÖÖ¶àÔªËỺ³åϵ£¬³£ÓÃÓÚÅäÖƹ©ÅàÑøϸ¾úµÄ»º³åÈÜÒº¡£ÈçÓÃ1 000 mLµÄ0.200 mol¡¤L-1ÄûÃÊËᣬÐë¼ÓÈë¶àÉÙÖÊÁ¿NaOH¹ÌÌ壬²ÅÄÜÅä³ÉpHΪ5.00µÄ»º³åÈÜÒº? (ÒÑÖªÄûÃÊËáµÄpKa1£½3.14£¬pKa2£½4.77£¬pKa3£½6.39)
14. ½ñÓûÅäÖÆ37¡æʱpH½üËÆΪ7.40µÄ»º³åÈÜÒº£¬ÊÔÇóÔÚTrisºÍTris?HClŨ¶È¾ùΪ0.050 mol¡¤L-1£¬×ÜÌå»ýΪ1.0 LµÄÈÜÒºÖУ¬Ðè¼ÓÈë0.10 mol¡¤L-1HCl¶àÉÙÌå»ý¡£ÔÚ´ËÈÜÒºÖÐÐè¼ÓÈë¶àÉÙÖÊÁ¿NaCl(¹Ì)²ÅÄÜÅä³ÉÓëÉø͸Ũ¶ÈΪ300 mmol?L-1µÈÉøµÄÈÜÒº? (ÒÑÖªTris?HClÔÚ37¡æʱµÄpKa£½7.85£»ºöÂÔÀë×ÓÇ¿¶ÈµÄÓ°Ïì)
15. Choose the factor that determine the capacity of a buffer from among the following and explain your choices:
¢Å Conjugate acid-base pair ¢Æ pH of the buffer
¢Ç Concentration of buffer-component reservoirs ¢È Buffer range
¢É Buffer-component ratio ¢Ê pKa of the acid component
16. How much are the [H3O+] and the pH of a propanoate £¨±ûËáÑΣ©buffer that consists of 0.25 mol?L-1 CH3CH2COONa and 0.15 mol?L-1 CH3CH2COOH ? Ka(CH3CH2COOH) = 1.3?10-5
£¬
17. How much is the pH of a buffer that consists of 0.55 mol?L-1 HCOOH and 0.63 mol?L-1 HCOONa? pKa=3.75 (25¡æ).
18. A buffer is prepared by mixing 50.0 mL of 0.050 mol?L-1 sodium bicarbonate and 10.0 mL of 0.10 mol?L-1 NaOH. pKa1(H2CO3) = 6.35; pKa2(H2CO3) =10.32 (25¡æ)¡£
¢Å How much is the pH?
¢Æ How much the mass of HCl must be added to 25.0 mL of the buffer to change the pH Value by 0.07?
19. Normal arterial blood has an average pH of 7.40. Phosphate ions is one of the key buffering systems in the blood. Find out the buffer-component ratio of a Na2HPO4/KH2PO4¡£pKa(H2PO4-)=7.21
20. As an FDA (Food and Drug AdministrationʳƷ¼°Ò©Îï¹ÜÀí¾Ö) physiologist, you need 0.600 L of formate buffer with pH of 3.75. (pKa(HCOOH) =3.75)
¢Å What is the required buffer-component ratio?
¢Æ How would you prepare this solution from stock solutions of 1.0 mol?L-1 HCOOH and 1.0 mol?L-1 NaOH? [Hint: Mix x mL of HCOOH with (600-x) mL of NaOH.]
¢Ç How much is the final concentration of HCOOH in this solution?
21. A buffer solution is made by adding 75g of sodium acetate to 500.0 mL of a 0.64 mol?L-1 solution of acetic acid. What is the pH of the final solution? (Assume no volume change.). pKa(acetic acid)= 4.76
22. The pH of human blood is always close to 7.40.Calculate the ratio of [HCO3-]/ [CO2] in blood at this pH. pKa1[H2CO3] =6.10 (37¡æ)
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A. S = (Ksp)1/2 B. S = (1/4Ksp)1/2 C. S = (Ksp)1/3 D. S = (1/27Ksp)1/4 10.ÓûʹMg(OH)2µÄÈܽâ¶È½µµÍ£¬×îºÃ¼ÓÈë
A. NaOH B. Fe(OH)3 C. H2O D. HCl
11. When NaCl is added to the saturated solution of AgCl, the solubility of AgCl would A. increase slightly B. increase greatly C. decrease slightly D. decrease greatly 12. ÒÑ֪ijÄÑÈÜÇ¿µç½âÖÊA2B(Mr = 80 g¡¤mol-1)£¬³£ÎÂÏÂÔÚË®ÖÐÈܽâ¶ÈΪ2.4¡Á10-3g¡¤L-1£¬ÔòA2BµÄÈܶȻýKspΪ
A. 1.1¡Á10-13 B. 2.7¡Á10-14 C. 1.8¡Á10-9 D. 9.0¡Á10-10
E. 1.8¡Á10-11
13. Fe2S3µÄÈܶȻýKsp±í´ïʽÊÇ
A. Ksp.= [Fe3+][S2-] B. Ksp = [Fe23+][S32-] C. Ksp = 2 [Fe3+]¡Á3[S2-] D. Ksp = [Fe3+]2[S2-]3. E. Ksp = [1/2Fe3+]2[1/3S2-]3
14. ½«1.0 mol¡¤L-1CaCl2ÈÜҺͨÈëCO2ÆøÌåÖÁ±¥ºÍ£¬ÔòÈÜÒºÖÐÓÐÎÞCaCO3³Áµí£¿[ÒÑÖªCaCO3µÄKsp= 2.9¡Á10-9£¬H2CO3µÄKa1= 4.2¡Á10-7£¬Ka2 = 5.6¡Á10-11]
A. ¸ÕºÃ±¥ºÍ£¬ÎÞ³Áµí B. δ´ïµ½±¥ºÍ£¬ÎÞ³Áµí C. ÓгÁµí D. ÓëÈÜÒºpHÎ޹أ¬ÎÞ³Áµí E. ÓëÈÜÒºpHÎ޹أ¬ÓгÁµí
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£12
£39
, Ksp[Mg(OH)2]
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17. ÔÚ0.20 mol¡¤L-1 MnCl2 100 mLÈÜÒºÖУ¬¼ÓÈ뺬ÓÐNH4ClµÄ0.10 mol¡¤L-1 NH3¡¤H2OÈÜÒº100.0 mL£¬ÎªÁ˲»Ê¹Mn(OH)2³ÁµíÐγɣ¬Ð躬NH4Cl¶àÉÙ¿Ë£¨?ÒÑÖªMn(OH)2µÄKsp =2.06