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1、章后习题解答 TOP习题1. 什么是缓冲溶液？ 试以血液中的H2CO3-缓冲系为例，说明缓冲作用的原理及其在医学上的重要意义。答 能抵抗少量外来强酸、强碱而保持其pH基本不变的溶液称为缓冲溶液。血液中溶解的CO2与组成缓冲系。正常人体/CO2(aq)为20/1，pH=7.40。若pH7.45，发生碱中毒。当酸性代谢产物增加时，抗酸成分与H3O+结合，增加的H2CO3可通过加快呼吸以CO2的形式呼出；消耗的则由肾减少对其的排泄而得以补充；当碱性代谢产物增加时，OH-与H3O+生成H2O，促使抗碱成分H2CO3离解以补充消耗的H3O+。同理，减少的H2CO3及增加的可通过肺和肾来调控。血液中的H2
2、CO3缓冲系与其他缓冲系共同作用，维持pH 为7.357.45的正常范围。2. 什么是缓冲容量？影响缓冲溶量的主要因素有哪些?总浓度均为0.10molL-1的 HAc-NaAc和H2CO3-缓冲系的缓冲容量相同吗？解 缓冲容量是衡量缓冲溶液缓冲能力大小的尺度，表示单位体积缓冲溶液pH发生一定变化时，所能抵抗的外加一元强酸或一元强碱的物质的量。影响缓冲容量的主要因素是缓冲系的总浓度和缓冲比：缓冲比一定时，总浓度越大缓冲容量越大；总浓度一定时，缓冲比越接近于1缓冲容量越大。总浓度及缓冲比相同的HAc-NaAc和H2CO3-缓冲系的缓冲容量相同。3. 下列化学组合中，哪些可用来配制缓冲溶液？ (1)
3、 HCl + NH3H2O (2) HCl + Tris (3)HCl + NaOH(4) Na2HPO4 + Na3PO4 (5) H3PO4 + NaOH (6)NaCl + NaAc解 可用来配制缓冲溶液的是：(1) HCl + NH3H2O、(2) HCl + Tris、(4) Na2HPO4 + Na3PO4和(5) H3PO4 + NaOH4. 将0.30 molL-1吡啶(C5H5N，pKb=8.77)和0.10 molL-1HCl溶液等体积混合,混合液是否为缓冲溶液？求此混合溶液的pH。解 C5H5N与HCl反应生成C5H5NH+Cl-(吡啶盐酸盐)，混合溶液为0.10 mol
4、L-1 C5H5N和0.050 molL-1 C5H5NH+Cl-缓冲溶液，pKa = 14.00 - 8.77 = 5.235. 将10.0 gNa2CO3和10.0 gNaHCO3溶于水制备250 mL缓冲溶液，求溶液的pH。 解 6. 求pH=3.90，总浓度为0.400 molL-1的HCOOH (甲酸)HCOONa(甲酸钠)缓冲溶液中，甲酸和甲酸钠的物质的量浓度(HCOOH的pKa=3.75)解 设c(HCOONa) = x molL-1, 则c(HCOOH) = 0.400 molL-1 x molL-1解得 c(HCOO-) = x molL-1 = 0.234 molL-1 c
5、(HCOOH)=(0.400 - 0.234) molL-1=0.166 molL-17. 向100mL某缓冲溶液中加入0.20 g NaOH固体，所得缓冲溶液的pH为5.60.。已知原缓冲溶液共轭酸HB的pKa=5.30，c(HB)=0.25molL-1，求原缓冲溶液的pH。解 n(NaOH) = 0.050 molL-1 加入NaOH后，解得 B- = 0.35 molL-1原溶液 8. 阿司匹林(乙酰水杨酸、以HAsp表示)以游离酸(未解离的)形式从胃中吸收，若病人服用解酸药，调整胃容物的pH为2.95，然后口服阿司匹林0.65 g。假设阿司匹林立即溶解，且胃容物的pH不变，问病人可以从
6、胃中立即吸收的阿司匹林为多少克 (乙酰水杨酸的Mr=180.2、pKa=3.48) ？ 解 依题意 解得 n(HAsp) = 0.0028 mol可吸收阿司匹林的质量 = 0.0028 mol 180.2 gmol-1 = 0.50 g9. 在500 mL 0.20 molL-1 C2H5COOH（丙酸，用HPr表示）溶液中加入NaOH1.8 g，求所得溶液的近似pH和校正后的精确pH。已知C2H5COOH的pKa=4.87，忽略加入NaOH引起的体积变化。解 求近似pHpH = pKa 求精确pH，丙酸钠是强电解质I =cizi2 = (12+12) = 0.09 molL-1 0.1 mo
7、lL-1当Z = 0，I = 0.10时，校正因数 pH = pKa10. 某医学研究中，制作动物组织切片时需pH约为7.0的磷酸盐缓冲液作为固定液。该固定液的配方是：将29 g Na2HPO412H2O和2.6 g NaH2PO42H2O分别溶解后稀释至1 L。若校正因数lg=-0.53，计算该缓冲溶液的精确pH。解 c(Na2HPO4) = 0.081 molL-1c(NaH2PO4) = 0.017 molL-1pH=pKa2 + +lg = 7.21 +(-.0.53) + lg= 7.3611. 将0.10 molL-1HAc溶液和0.10 molL-1NaOH溶液以3:1的体积比混
8、合，求此缓冲溶液的pH及缓冲容量。解 HAc溶液和NaOH溶液的体积分别为3V和V，c(HAc) = (0.103V - 0.10 V) molL-1 / (3V + V) = 0.050 molL-1c(Ac-) = 0.10 molL-1 V / (3V + V ) = 0.025 molL-112. 某生物化学实验中需用巴比妥缓冲溶液，巴比妥(C8H12N2O3)为二元有机酸（用H2Bar表示，pKa17.43)。今称取巴比妥18.4 g，先加蒸馏水配成100 mL溶液，在pH计监控下，加入6.00 molL-1NaOH溶液4.17 mL，并使溶液最后体积为1000 mL。求此缓冲溶液的
9、pH和缓冲容量。（已知巴比妥的Mr=184 gmol-1）解 H2Bar与NaOH的反应为H2Bar(aq) + NaOH(aq)NaHBar(aq) +H2O(l)反应生成的NaHBar的物质的量n(NaHBar) c(NaOH)V(NaOH)6.0 molL-14.17 mL25 mmol，剩余H2Bar的物质的量为n余(H2Bar)n(H2Bar) - n(NaOH)1000 - 25 mmol75 mmolpHpKa+lg7.43+lg6.95 2.3030.043 molL-113. 分别加NaOH溶液或HCl溶液于柠檬酸氢钠（缩写Na2HCit）溶液中。写出可能配制的缓冲溶液的抗酸
10、成分、抗碱成分和各缓冲系的理论有效缓冲范围。如果上述三种溶液的物质的量浓度相同，它们以何种体积比混合，才能使所配制的缓冲溶液有最大缓冲容量?（已知H3Cit的pKa1=3.13、pKa2=4.76、pKa3=6.40）解.溶液组成缓 冲 系抗酸成分抗碱成分有效缓冲范围最大时体积比Na2HCit+HClH2Cit-HCit2-HCit2-H2Cit-3.765.762:1Na2HCit+HClH3Cit-H2Cit-H2Cit-H3Cit2.134.132:3Na2HCit+NaOHHCit2-Cit3-Cit3-HCit2-5.407.402:114. 现有(1)0.10 molL-1NaOH
11、溶液，(2)0.10 molL-1NH3溶液，(3)0.10 molL-1Na2HPO4 溶液各50 mL，欲配制pH=7.00的溶液，问需分别加入0.10 molL-1 HCl溶液多少mL？配成的三种溶液有无缓冲作用？哪一种缓冲能力最好？解 HCl与NaOH完全反应需HCl溶液50 mL。 HCl(aq) + NH3H2O(aq) = NH4Cl(aq) + H2O(l)NH4+的pKa = 14.00-4.75= 9.25， 解得 V(HCl) = 50 mL HCl(aq) + Na2HPO4(aq) = NaH2PO4(aq) + NaCl(aq)H3PO4的pKa2=7.21， 解得
12、 V (HCl) = 31 mL 第一种混合溶液无缓冲作用；第二种pHpKa -1，无缓冲能力；第三种缓冲作用较强。15. 用固体NH4Cl和NaOH溶液来配制1 L总浓度为0.125 molL-1，pH=9.00的缓冲溶液，问需NH4Cl多少克？求需1.00 molL-1的NaOH溶液的体积（mL）。解 设需NH4Cl 的质量为 xgpKa(NH4+) = 14.00 - 4.75 = 9.25得 1.00 molL-1 V(NaOH) = 0.562x/53.5 gmol-1 - 1.00 molL-1 V(NaOH)又 1.00 molL-1 VNaOH) + 0.562 (x / 53
13、.5gmol-1 - 1.00 molL-1 V(NaOH) / 1L=0.125 molL-1解得 x = 6.69, V(NaOH) = 0.045 L即：需NH4Cl 6.69 g，NaOH溶液0.045 L。16. 用0.020 molL-1H3PO4溶液和0.020 molL-1NaOH溶液配制100 mL pH=7.40的生理缓冲溶液，求需H3PO4溶液和NaOH溶液的体积（mL）。解 设第一步反应需H3PO4和NaOH溶液体积各为x mL H3PO4(aq)+ NaOH(aq)= NaH2PO4(aq) + H2O(l)x mLH3PO4与x mLNaOH完全反应，生成NaH2P
14、O4 0.020 molL-1 x mL = 0.020 x mmol 第二步反应：设生成的NaH2PO4再部分与NaOH y mL反应，生成Na2HPO4，其与剩余NaH2PO4组成缓冲溶液 NaH2PO4(aq) + NaOH(aq) = Na2HPO4(aq) + H2O(l)起始量mmol +0.020x +0.020y 变化量mmol -0.020y -0.020y +0.020y 平衡量mmol 0.020(x-y) 0 +0.020y =1.55依题意又有 2x + y = 100解得 x = 38.4，y = 23.2即需H3PO4溶液38.4 mL，NaOH溶液(38.4 +
15、 23.2) mL = 61.6 mL。17. 今欲配制37时，近似pH为7.40的生理缓冲溶液，计算在Tris和TrisHCl浓度均为0.050 molL-1的溶液l00 mL中，需加入0.050 molL-1HCl溶液的体积（mL）。在此溶液中需加入固体NaCl多少克，才能配成与血浆等渗的溶液？（已知TrisHCl在37时的pKa7.85，忽略离子强度的影响。） 解 V(HCl) = 47.6 mL 设加入NaCl x g，血浆渗透浓度为300 mmolL-1=0.018 molL-1=0.050 molL-1（0.018 + 20.050）molL-1 + 0.300 molL-1x =
16、 0.79，即需加入NaCl 0.79 g18. 正常人体血浆中，24.0 mmolL-1、CO2(aq)1.20 mmolL-1。若某人因腹泻使血浆中减少到为原来的90%，试求此人血浆的pH，并判断是否会引起酸中毒。已知H2CO3的pKa1=6.10。解 pH= pKa1pH虽接近7.35，但由于血液中还有其他缓冲系的协同作用，不会引起酸中毒。Exercises1. How do the acid and base components of a buffer function? Why are they typically a conjugate acid-base pair? Solut
17、ion A buffer solution consists of a conjugate acid-base pair. The conjugate base can consume the added strong acid, and the conjugate acid can consume the added strong base, to maintain pH。The conjugate acid-base pairs of weak electrolytes present in the same solution at equilibrium.2. When H3O+ is
18、added to a buffer，does the pH remain constant or does it change slightly？Explain. Solution The pH of a buffer depends on the pKa of the conjugate acid and the buffer component ratio. When H3O+ is added to a buffer， the buffer component ratio changes slightly，so the pH changes slightly.3. A certain s
19、olution contains dissolved HCl and NaCl. Why cant this solution act as a buffer?Solution This solution cant act as a buffer. HCl is not present in solution in molecular form. Therefore, there is no reservoir of molecules that can react with added OH- ions.Likewise the Cl- does not exhibit base behav
20、ior in water, so it cannot react with any H3O+ added to the solution.4. What is the relationship between buffer range and buffer-component ratio? Solution The pH of a buffer depends on the buffer component ratio. When B-/HB=1, pH = pKa, the buffer is most effective. The further the buffer-component
21、ratio is from 1，the less effective the buffering action is. Practically, if the B-/HB ratio is greater than 10 or less than 0.1, the buffer is poor. The buffer has a effective range of pH = pKa1.5. Choose specific acid-base conjugate pairs of suitable for prepare the following beffers（Use Table 4-1
22、for Ka of acid or Kb of base）：（a）pH4.0；（b）pH7.0；（c）H3O+1.010-9 molL-1； Solution （a）HAc and Ac- （b） and （c） and NH36. Choose the factors that determine the capacity of a buffer from among the following and explain your choices.(a) Conjugate acid-base pair (b) pH of the buffer (c) Buffer ranger(d) Con
23、centration of buffer-component reservoirs(e) Buffer-component ratio (f) pKa of the acid componentSolution Choose (d) and (e). Buffer capacity depends on both the concentration of the reservoirs and the buffer-component ratio. The more concentrated the components of a buffer, the greater the buffer c
24、apacity. When the component ratio is close to one, a buffer is most effective.7. Would the pH increase or decrease, and would it do so to a larger or small extent, in each of the following cases:(a) Add 5 drops of 0.1 molL-1 NaOH to 100 mL of 0.5 molL-1 acetate buffer(b) Add 5 drops of 0.1 molL-1 HC
25、l to 100 mL of 0.5 molL-1 acetate buffer(c) Add 5 drops of 0.1 molL-1 NaOH to 100 mL of 0.5 molL-1 HCl(d) Add 5 drops of 0.1 molL-1 NaOH to distilled waterSolution（a）The pH increases to a small extent;（b）The pH decreases to a small extent;（c）The pH increases to a small extent;（d）The pH increases to
26、a larger extent.8. Which of the following solutions will show buffer properties?(a) 100 mL of 0.25 molL-1 NaC3H5O3 + 150 mL of 0.25 molL-1 HCl(b) 100 mL of 0.25 molL-1 NaC3H5O3 + 50 mL of 0.25 molL-1 HCl (c) 100 mL of 0.25 molL-1 NaC3H5O3+ 50 mL of 0.25 molL-1 NaOH (d) 100 mL of 0.25 molL-1 C3H5O3H
27、+ 50 mL of 0.25 molL-1 NaOH Solution (b) and (d)9. A chemist needs a pH 10.5 buffer. Should she use CH3NH2 and HCl or NH3 and HCl to prepare it? Why? What is the disadvantage of choosing the other base? Solution The pKa of CH3NH2HCl is 10.65. The pKa of is 9.25. The pKa of the former is more close t
28、o 10.5. A buffer is more effective when the pH is close to pKa. She should choose CH3NH2. The other is not a good choice. 10. An artificial fruit contains 11.0 g of tartaric acid H2C4H4O6，and 20.0 g its salt， potassium hydrogen tartrate，per liter. What is the pH of the beverage？Ka1=1.010-3Solution 1
29、1. What are the H3O+ and the pH of a benzoate buffer that consists of 0.33 molL-1 C6H5COOH and 0.28 molL-1 C6H5COONa？Ka of benzoic acid=6.310-5。Solution H+=7.4110-5 molL-112. What mass of sodium acetate (NaC2H3O23H2O,Mr=136.1gmol-1) and what volume of concentrated acetic acid (17.45molL-1) should be
30、 used to prepare 500 mL of a buffer solution at pH=5.00 that is 0.150 molL-1 overall?Solution c(CH3COO-) + c(CH3COOH)=0.150 molL-1n(CH3COO-) + n(CH3COOH) = 0.150 molL-1 500 mL 10-3 LmL-1 = 0.0750 moln(CH3COOH) = 0.0270 mol, n(CH3COO-) = 0.0480 molMass of sodium acetate = 0.0480 mol 136.1 gmol -1 = 6.53 g 13. Normal arterial blood has an average pH of 7.40. Phosphate ions form one of the key buffering systems in the blood. Find the buffer-component ratio of a KH2PO4/Na2HPO4 solution with this pH pKa2of = 6.80Solution 9