章后习题解答 [TOP]

习题

1. 什么是缓冲溶液？ 试以血液中的H2CO3-bef42070764bcf36b22037139582acc4.png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bef42070764bcf36b22037139582acc4.png组成缓冲系。正常人体[bef42070764bcf36b22037139582acc4.png]/[CO2(aq)]为20/1，pH=。若pH<，发生酸中毒，pH>，发生碱中毒。当酸性代谢产物增加时，抗酸成分bef42070764bcf36b22037139582acc4.png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bef42070764bcf36b22037139582acc4.png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bef42070764bcf36b22037139582acc4.png可通过肺和肾来调控。血液中的H2CO3–bef42070764bcf36b22037139582acc4.png缓冲系与其他缓冲系共同作用，维持pH 为~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·L-1的 HAc-NaAc和H2CO3-bef42070764bcf36b22037139582acc4.png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-bef42070764bcf36b22037139582acc4.png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) HCl + NH3·H2O (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 + NH3·H2O、(2) HCl + Tris、(4) Na2HPO4 + Na3PO4和(5) H3PO4 + NaOH

4. 将 mol·L-1吡啶(C5H5N，pKb=和 mol·L-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 mol·L-1 C5H5N和 mol·L-1 C5H5NH+Cl-缓冲溶液，pKa = - =

97183584384a7d368ff55c835f2ef89c.png

5. 将 gNa2CO3和 gNaHCO3溶于水制备250 mL缓冲溶液，求溶液的pH。

解 6aa55891a73bea776cd1af824ecbae29.png

0d391afe064a6f67c315789801663ab0.png

bf07e94c5e0e8f583fd1a262e90a2e92.png

6. 求pH=，总浓度为 mol·L-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=

解 设c(HCOONa) = x mol·L-1, 则c(HCOOH) = mol·L-1 – x mol·L-1

d02c498903dbbcc1ebdccc8d0f833ea3.png

解得 c(HCOO-) = x mol·L-1 = mol·L-1

c(HCOOH)= - mol·L-1= mol·L-1

7. 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.。已知原缓冲溶液共轭酸HB的pKa=，c(HB)=·L-1，求原缓冲溶液的pH。

解 n(NaOH) =08a2dfc01ca14cb6467c240f72d741c5.png= mol·L-1

加入NaOH后，

906937714b7615c693b032c1816ef7d7.png

解得 [B-] = mol·L-1

原溶液 c56b22ccf1c7a26637d9ec5cb78e7524.png

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，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=、pKa= ？

解 f6d7c28edec6a25c57a7f1cccb76e11c.png

3eba02f00a237b15bf9454d02e7b91c8.png

依题意 b2db4e90c031069a5eae1f519f69fe45.png

解得 n(HAsp) = mol

可吸收阿司匹林的质量 = mol × 180.2 g·mol-1 = 0.50 g

9. 在500 mL mol·L-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=，忽略加入NaOH引起的体积变化。

解 ⑴ 求近似pH

word/media/image15.gifpH = pKa0aa81bb1c91416d1677b141146814034.png

⑵ 求精确pH，丙酸钠是强电解质

I =93b05c90d14a117ba52da1d743a43ab1.png∑cizi2 = 93b05c90d14a117ba52da1d743a43ab1.png(ead7fad0f08376123c9c94b2380ad302.png×12+ead7fad0f08376123c9c94b2380ad302.png×12) = mol·L-1 ≈ mol·L-1

当Z = 0，I = 时，校正因数 ec4de0ed54fd5b87eb489216656b29b2.png

pH = pKa8fa27074926b35134aeb7bf3a1302e8d.png

10. 某医学研究中，制作动物组织切片时需pH约为的磷酸盐缓冲液作为固定液。该固定液的配方是：将29 g Na2HPO4·12H2O和2.6 g NaH2PO4·2H2O分别溶解后稀释至1 L。若校正因数lgad9a9fd2c5d14e2c0d44c5a5fc13a2c7.png=，计算该缓冲溶液的精确pH。

解 c(Na2HPO4) =68e7115aa26f08cdd267aca9f357904d.png= mol·L-1

c(NaH2PO4) =834739cb39e465c7e916fc99bf15dc15.png= mol·L-1

pH=pKa2 + ad9a9fd2c5d14e2c0d44c5a5fc13a2c7.png+lg186153e764a7179f3e2882737e0f03a6.png = + + lg5f60e9016434ba3e133756c3e5bf730c.png=

11. 将 mol·L-1HAc溶液和 mol·L-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) = ×3V - × V) mol·L-1 / (3V + V) = mol·L-1

c(Ac-) = mol·L-1 × V / (3V + V ) = mol·L-1

ff5c96015ce8eff9b32efdba7f34fdf3.png

6a059dc14974a1a69c1106b79cb92a64.png

12. 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，pKa1＝。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 mol·L-1NaOH溶液 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 g·mol-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)＝ mol·L-1× mL＝25 mmol，剩余H2Bar的物质的量为

n余(H2Bar)＝n(H2Bar) - n(NaOH)＝0ed0e7d72b6ab3e4d2ffb326c9e28899.png×1000 - 25 mmol＝75 mmol

pH＝pKa+lg012fbff6e83d9101e80449e87bcffe14.png＝+lgb64f5c2a1e6dc60d15ddc4b76ef1266a.png＝

β＝×ca8c5f951e76e046ca979d9354521f64.png＝ mol·L-1

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=、pKa2=、pKa3=）`

解.

14. 现有(1) mol·L-1NaOH溶液，(2) mol·L-1NH3溶液，(3) mol·L-1Na2HPO4 溶液各50 mL，欲配制pH=的溶液，问需分别加入 mol·L-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 ⑴ HCl与NaOH完全反应需HCl溶液50 mL。

⑵ HCl(aq) + NH3·H2O(aq) = NH4Cl(aq) + H2O(l)

NH4+的pKa = ，

8bf685572ac9f9ef4d26a1991f2ef59f.png

解得 V(HCl) = 50 mL

⑶ HCl(aq) + Na2HPO4(aq) = NaH2PO4(aq) + NaCl(aq)

H3PO4的pKa2=，

ecd13b3020bd453b8b45fe78efe26241.png

解得 V (HCl) = 31 mL

第一种混合溶液无缓冲作用；第二种pHKa -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 mol·L-1，pH=的缓冲溶液，问需NH4Cl多少克？求需 mol·L-1的NaOH溶液的体积（mL）。

解 设需NH4Cl 的质量为 xg

pKa(NH4+) = - =

71c71430e83c34b2363d325eb2e3d7db.png

得 mol·L-1 × V(NaOH) = [x/53.5 g·mol-1 - mol·L-1 × V(NaOH)]

又 [ mol·L-1 × VNaOH) + (x / 53.5g·mol-1 - mol·L-1 × V(NaOH))] / 1L

= mol·L-1

解得 x = , V(NaOH) = 0.045 L

即：需NH4Cl 6.69 g，NaOH溶液0.045 L。

16. 用 mol·L-1H3PO4溶液和 mol·L-1NaOH溶液配制100 mL pH=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

⑴ H3PO4(aq)+ NaOH(aq)= NaH2PO4(aq) + H2O(l)

x mLH3PO4与x mLNaOH完全反应，生成NaH2PO4 mol·L-1 × x mL = x mmol

⑵ 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 + +

变化量mmol +

word/media/image37.gif 平衡量mmol (x-y) 0 +

a12b159bd007064d0fb2aab03a61b360.png

0659fce19a10fd2b37cff378d104afea.png=

依题意又有 2x + y = 100

解得 x = ，y =

即需H3PO4溶液 mL，NaOH溶液 + mL = mL。

17. 今欲配制37℃时，近似pH为的生理缓冲溶液，计算在Tris和Tris·HCl浓度均为 mol·L-1的溶液l00 mL中，需加入 mol·L-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Tris·HCl在37℃时的pKa＝，忽略离子强度的影响。）

解 ⑴

89f3caa3d5b54ac1b7522bdb3147ef48.png 20cc839086c71425cbc47cbfd1abbdb0.png

V(HCl) = mL

⑵ 设加入NaCl x g，血浆渗透浓度为300 mmol·L-1

65889db6e29095fc8dddd0f81afe7a65.png= mol·L-1

353d29a0a34e83ea66afde1675cbdced.png= mol·L-1

（ + 2×）mol·L-1 + 2d030272924bec12b3261cc7a0585749.png mol·L-1

x = ，即需加入NaCl 0.79 g

18. 正常人体血浆中，[bef42070764bcf36b22037139582acc4.png]＝ mmol·L-1、[CO2(aq)]＝ mmol·L-1。若某人因腹泻使血浆中[bef42070764bcf36b22037139582acc4.png]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pKa1ˊ=。

解 pH= pKa1′f1b48d34f2efce410fc37e6f9a1f8c8f.png

pH虽接近，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。

Exercises

1. How do the acid and base components of 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 weak 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 component ratio changes slightly，so the pH changes slightly.

3. A certain solution contains dissolved HCl and NaCl. Why can’t this solution act as a buffer?

Solution This solution can’t 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- 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, 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 , the buffer is poor. The buffer has a effective range of pH = pKa±1.

5. Choose specific acid-base conjugate pairs of suitable for prepare the following beffers（Use Table 4-1 for Ka of acid or Kb of base）：

（a）pH≈；（b）pH≈；（c）[H3O+]≈×10-9 mol·L-1；

Solution （a）HAc and Ac- （b）4e28ab46ad355f43f75399911a47afc3.png and 9d613954814ba7710fd61c7c4bd2ae5b.png （c）018b0a796be2edcc470273ba8f839771.png and NH3

6. 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) Concentration of buffer-component reservoirs

(e) Buffer-component ratio (f) pKa of the acid component

Solution 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 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 of mol·L-1 NaOH to 100 mL of mol·L-1 acetate buffer

(b) Add 5 drops of mol·L-1 HCl to 100 mL of mol·L-1 acetate buffer

(c) Add 5 drops of mol·L-1 NaOH to 100 mL of mol·L-1 HCl

(d) Add 5 drops of mol·L-1 NaOH to distilled water

Solution（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 a larger extent.

8. Which of the following solutions will show buffer properties?

(a) 100 mL of mol·L-1 NaC3H5O3 + 150 mL of mol·L-1 HCl

(b) 100 mL of mol·L-1 NaC3H5O3 + 50 mL of mol·L-1 HCl

(c) 100 mL of mol·L-1 NaC3H5O3+ 50 mL of mol·L-1 NaOH

(d) 100 mL of mol·L-1 C3H5O3H + 50 mL of mol·L-1 NaOH

Solution (b) and (d)

9. A chemist needs a pH 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 CH3NH2·HCl is . The pKa of 018b0a796be2edcc470273ba8f839771.png is . The pKa of the former is more close to . 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=×10-3

Solution 4f6d293b017a1c13312f1dfd0de07502.png

11. What are the [H3O+] and the pH of a benzoate buffer that consists of mol·L-1 C6H5COOH and mol·L-1 C6H5COONa？Ka of benzoic acid=×10-5。

Solution 43c48728d1734e45dd2d369fd5e2c5c1.png

[H+]=×10-5 mol·L-1

12. What mass of sodium acetate (NaC2H3O2·3H2O,Mr=136.1g·mol-1) and what volume of concentrated acetic acid ·L-1) should be used to prepare 500 mL of a buffer solution at pH= that is mol·L-1 overall?

Solution c(CH3COO-) + c(CH3COOH)= mol·L-1

n(CH3COO-) + n(CH3COOH) = mol·L-1 × 500 mL × 10-3 L·mL-1 = mol

fa5aebe355db5e04da9021499822eaae.png

2d748cc4651d4ffdbdbb2a3c57c59c3b.png

n(CH3COOH) = mol, n(CH3COO-) = mol

Mass of sodium acetate = mol × 136.1 g·mol -1 = 6.53 g

1540db454ec7a5ce63ea66b642c900b5.png

13. Normal arterial blood has an average pH of . 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 pKa2ˊof 4e28ab46ad355f43f75399911a47afc3.png=

Solution 84366e573601361ff011eb6bd68c6e78.png

8e0c2b9ae1fcd64da403ffd1cbcea5c1.png

基础化学答案第