基础化学答案第04.doc

1、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CO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C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，3HCOOH-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

2、 H2CO3及增加的可通过肺和肾来调控。血液中的 H2CO3 缓冲系与其他缓冲系共同作用，维持 pH 为C37.357.45 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？O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 H

3、2CO3- 缓冲系的缓冲容量相同。CO3. 下列化学组合中，哪些可用来配制缓冲溶液？ (1) 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 + NH 3H2O、(2) HCl + Tris、(4) Na 2HPO4 + Na3PO4和(5) H3PO4 + NaOH4. 将 0.30 molL-1吡啶(C 5H5N，p Kb=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

4、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 molL-1 C5H5N 和 0.050 molL-1 C5H5NH+Cl-缓冲溶液，p Ka = 14.00 - 8.77 = 5.2323.501lg23.5)NHc(ClgpH5a K5. 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 .9molol84.01)O(1n46g2310.23.9ollg31)(HClp23a nK6. 求 pH=3.90，总浓度为 0.400 molL-1的 HCOOH (甲酸)

5、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-190.3Lmol)(0.4lg3.75pH1x解得 c(HCOO-) = x molL-1 = 0.234 molL-1c(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.

6、25molL-1，求原缓冲溶液的 pH。解 n(NaOH) = = 0.050 molL-1 1L0m10ol.2/4加入 NaOH 后， 60.50.)ol-(25Blg.3pH1-解得 B - = 0.35 molL-1原溶液 4.Lmol25.0lg3.p18. 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，问病人可以从胃中立即吸收的阿司匹林为多少克 (乙酰水杨酸的 Mr=180.2、p Ka=3.48) ？ 解 95.2(HA

7、sp)l48.3(HAsp)lgpa nnK295.03依题意 0.36molol18g.65n(HAsp)(s1-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 4.78mol1.8g/0L0.2ol.5l/4lg4.87(

8、Pr)lg 1- n 求精确 pH，丙酸钠是强电解质I = c izi2 = ( 12+ 12) = 0.09 molL-1 0.1 molL -110.5Lmol.l当 Z = 0， I = 0.10 时，校正因数 10)HB(lgpH = pKa 67.4)10(78.4Pr)(lrl n10. 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。)POH(-42解 c(Na2HP

9、O4) = = 0.081 molL-11Lmol358.0g9c(NaH2PO4) = = 0.017 molL-16.pH=pKa2 + +lg = 7.21 +(-.0.53) + lg = 7.36)POH(-42)(42c 1L0.7mol811. 将 0.10 molL-1HAc 溶液和 0.10 molL-1NaOH 溶液以 3:1 的体积比混合，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-) = 0.1

10、0 molL-1 V / (3V + V ) = 0.025 molL-14.5L0.5mol2lg4.7pH1-411-1 Lmol038.L0.25)ol(.ml30.2 12. 某生物化学实验中需用巴比妥缓冲溶液，巴比妥(C 8H12N2O3)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 pH 和缓冲容量。 （已知巴比妥的 Mr=184 gmol-1）解 H 2Bar与NaOH的反应为

11、H2Bar(aq) + NaOH(aq)NaHBar(aq) +H 2O(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 mmol1-mol84.gpHp Ka+lg 7.43+lg 6.95Br)(H275ol2 2.303 0.043 molL -10L5)ol/1(7)l/13. 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

12、的抗酸成分、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、p Ka2=4.76、p Ka3=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- 3.765.76 2:1Na2HCit+HCl H3Cit-H2Cit- H2Cit- H3Cit 2.134.13 2:3Na2HCit+NaOHHCit2-Cit3- Cit3- HCit2- 5.407.40 2:1

13、14. 现有 (1)0.10 molL-1NaOH 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？5解 HCl 与 NaOH 完全反应需 HCl 溶液 50 mL。 HCl(aq) + NH 3H2O(aq) = NH4Cl(aq) + H2O(l)NH4+的 pKa = 14.00-4.75= 9.25， (Cl)L.10mol(Hl).5.l0lg5.90711V解得 V(HCl) =

14、50 mL HCl(aq) + Na 2HPO4(aq) = NaH2PO4(aq) + NaCl(aq)H3PO4的 pKa2=7.21， (HCl)L.10mol(l).5.l0lg1.711V解得 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.

15、00 - 4.75 = 9.25 (NaOH)Lmol0.1l/53.gl2.901- Vx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 / 53.5gmol-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-1Na

16、OH 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 H 3PO4(aq)+ NaOH(aq)= NaH2PO4(aq) + H2O(l)x mLH3PO4与 x mLNaOH 完全反应，生成 NaH2PO4 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6NaH2PO4(aq) + NaOH(aq) = Na2HP

17、O4(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 mol)(.2lg1.740x=1.55yx依题意又有 2 x + y = 100解得 x = 38.4， y = 23.2即需 H3PO4溶液 38.4 mL，NaOH 溶液(38.4 + 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，

18、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 )HCl(L0.5mol1Lmol05. -lg8.7HC) (Tris.40 1- Vn)l(135.0VV(HCl) = 47.6 mL 设加入 NaCl x g，血浆渗透浓度为 300 mmolL-1=0.018 molL-1m6.1476.47ol05Lol05.)Tris(1c=0.050 molL-1mLHCl （0.018 + 20.050）molL -1 + 0.300

19、 molL-16.147olg5.802xx = 0.79，即需加入 NaCl 0.79 g18. 正常人体血浆中， 24.0 mmolL-1、CO 2(aq)1.20 mmolL-1。若某人因腹泻使3HCO7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3HCOpKa1=6.10。解 pH= p Ka1 7.36L1.2mol094lg6.0(aq)COlg123 pH 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

20、a buffer function? Why are they typically a conjugate acid-base pair? Solution 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 wea

21、k electrolytes present in the same solution at equilibrium.2. When H3O+ is 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

22、 component ratio changes slightly，so the pH changes slightly.3. A certain solution 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

23、 can react with added OH- ions.Likewise the Cl- does not exhibit base behavior 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

24、, pH = pKa, the buffer is most effective. The further the buffer-component 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 c

25、onjugate pairs of suitable for prepare the following beffers（Use Table 4-1 for Ka of acid or Kb of base）：（a）pH4.0 ；（ b）pH7.0；（c ）H 3O+1.010-9 molL-1； Solution （a）HAc and Ac- （b） and （c） and NH342PH244NH6. Choose the factors that determine the capacity of a buffer from among the following and explain

26、 your choices.(a) Conjugate acid-base pair (b) pH of the buffer (c) Buffer ranger8(d) Concentration 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-co

27、mponent ratio. The more concentrated the components of a buffer, the greater the buffer capacity. 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

28、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 HCl 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

29、decreases to a small extent;（c）The pH increases to a small extent;（d）The pH increases to 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

30、 mL of 0.25 molL-1 NaC3H5O3+ 50 mL of 0.25 molL-1 NaOH (d) 100 mL of 0.25 molL-1 C3H5O3H + 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

31、 The pKa of CH3NH2HCl is 10.65. The pKa of is 9.25. The pKa of the former is more close 4NHto 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，

32、 potassium hydrogen tartrate， per liter. What is the pH of the beverage？Ka 1=1.010-3Solution 16.345l.03molg1.508.l0.3)C(Hlp642a1 nK11. 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？K a of benzoic acid=6.310-5。9Solution 13.4850lg2.4L

33、0.3mol28lg.4COH)(lgpH1-56a cKH+=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 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

34、molL-1n(CH3COO-) + n(CH3COOH) = 0.150 molL-1 500 mL 10-3 LmL-1 = 0.0750 mol)COH(lg75.4038.1)(3nn(CH3COOH) = 0.0270 mol, n(CH3COO-) = 0.0480 molMass of sodium acetate = 0.0480 mol 136.1 gmol -1 = 6.53 gmL5.1ol45.170m2)COH(33 V13. 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 = 42POH6.80Solution POHlg80.64742.93PO42