基础化学答案第04
章后习题解答 [TOP]
习题
1. 什么是缓冲溶液? 试以血液中的H 2CO 3--3HCO 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CO 2与
-3
HCO 组成缓冲系。正常人体[-3HCO ]/[CO 2(aq)]为20/1,pH=7.40。若pH<7.35,发生酸中毒,pH>7.45,发生碱中毒。当酸性代谢产物增加时,抗酸成分-3HCO 与H 3O +结合,增加的H 2CO 3可通过加快呼吸以
CO 2的形式呼出;消耗的-3HCO 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
H 3O +生成H 2O ,促使抗碱成分H 2CO 3离解以补充消耗的H 3O +。同理,减少的H 2CO 3及增加的-3HCO 可
通过肺和肾来调控。血液中的H 2CO 3–-3HCO 缓冲系与其他缓冲系共同作用,维持pH 为7.35~7.45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0.10mol·L -1的 HAc-NaAc 和
H 2CO 3--3HCO 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H 2CO 3--3HCO 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 + NH 3·H 2O (2) HCl + Tris (3)HCl + NaOH
(4) Na 2HPO 4 + Na 3PO 4 (5) H 3PO 4 + NaOH (6)NaCl + NaAc
解 可用来配制缓冲溶液的是:(1) HCl + NH 3·H 2O 、(2) HCl + Tris 、(4) Na 2HPO 4 + Na 3PO 4和(5) H 3PO 4
+ NaOH
4. 将0.30 mol·L -1吡啶(C 5H 5N ,p K b =8.77)和0.10 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 C 5H 5N 与HCl 反应生成C 5H 5NH +Cl -(吡啶盐酸盐),混合溶液为0.10 mol·L -1 C 5H 5N 和0.050
mol·L -1 C 5H 5NH +Cl -缓冲溶液,p K a = 14.00 - 8.77 = 5.23
33.505
.010.0lg 23.5)NH H c(C N)
H c(C lg p pH 5555a =+=+=+K
5. 将10.0 gNa 2CO 3和10.0 gNaHCO 3溶于水制备250 mL 缓冲溶液,求溶液的pH 。
解 0.119m ol m ol
84.0g 10.0g )HCO (13=?=--n 0.094m ol m ol
106g 10.0g )(CO 123=?=--n 10.230.119mol
0.094mol lg 3310)(HCO )
(CO lg p pH 323a =+?=+=--n n K 6. 求pH=3.90,总浓度为0.400 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p K a =3.75)
解 设c (HCOONa) = x mol·L -1, 则c (HCOOH) = 0.400 mol·L -1 – x mol·L -1
90.3L
mol )(0.400L mol lg 3.75pH 11
=?-?+=--x x 解得 c (HCOO -) = x mol·L -1 = 0.234 mol·L -1
c (HCOOH)=(0.400 - 0.234) mol·L -1=0.166 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5.60.。已知原缓冲溶液共
轭酸HB 的p K a =5.30,c (HB)=0.25mol·L -1,求原缓冲溶液的pH 。
解 n (NaOH) =1L
1000mL 100mL mol 0.20g/40g 1??-= 0.050 mol·L -1 加入NaOH 后,
60.5L
0.050)mol -(0.25L 0.050mol ][B lg 5.30pH 1--1
=??++=- 解得 [B -] = 0.35 mol·L -1
原溶液 45.5L mol 25.0L mol 35.0lg 30.5pH 1
1=??+=-- 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 K a =3.48) ?
解 95.2(HAsp)
)(Asp lg 48.3(HAsp))(Asp lg p pH a =+=+=--n n n n K 295.0(HAsp)
)(Asp =-n n
依题意 0.0036mol mol 180g 0.65g n(HAsp))n(Asp 1
-=?=
+- 解得 n (HAsp) = 0.0028 mol 可吸收阿司匹林的质量 = 0.0028 mol × 180.2 g·mol -1 = 0.50 g
9. 在500 mL 0.20 mol·L -1 C 2H 5COOH (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C 2H 5COOH 的p K a =4.87,忽略加入NaOH 引起的体积变化。
解 ⑴ 求近似pH
pH = p K a 4.78mol
1.8g/40g L 0.20mol 0.500L mol 1.8g/40g lg 4.87(HPr))(Pr lg 1-1--1-=?-???+=+n n ⑵ 求精确pH ,丙酸钠是强电解质
I =21∑c i z i 2 = 21(0.500L 0.045mol ×12+0.500L
0.045mol ×12) = 0.09 mol·L -1 ≈ 0.1 mol·L -1
当Z = 0,I = 0.10时,校正因数 11.0)
HB ()B (lg -=-γγ pH = p K a 67.4)11.0(78.4Pr)
H ()(Pr lg Pr)H ()(Pr lg =-+=++--γγn n
10. 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 Na 2HPO 4·12H 2O 和2.6 g NaH 2PO 4·2H 2O 分别溶解后稀释至1 L 。若校正因数lg )
PO H ()HPO (-4224γγ-=-0.53,计算该缓冲溶液的精确pH 。 解 c (Na 2HPO 4) =1L
mol 358.0g 29g 1??-= 0.081 mol·L -1 c (NaH 2PO 4) =
1L mol 156g 2.6g 1??-= 0.017 mol·L -1 pH=p K a2 + )PO H ()HPO (-4224γγ-+lg )
PO H ()(HPO 4224--c c = 7.21 +(-.0.53) + lg 11L 0.017m ol L 0.081m ol --??= 7.36 11. 将0.10 mol·L -1HAc 溶液和0.10 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) = (0.10×3V - 0.10 × V ) mol·L -1 / (3V + V ) = 0.050 mol·L -1
c (Ac -) = 0.10 mol·L -1 × V / (3V + V ) = 0.025 mol·L -1
4.45L
0.050mol L 0.025mol lg 4.75pH 1--1
=??+=
11--1
-1L mol 038.0L
0.025)mol (0.050L 0.025mol L 0.050mol 303.2-?=?+????=β 12. 某生物化学实验中需用巴比妥缓冲溶液,巴比妥(C 8H 12N 2O 3)为二元有机酸(用H 2Bar 表示,p K a1
=7.43)。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 mol·L -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 g·mol -1)
解 H 2Bar 与NaOH 的反应为
H 2Bar(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 mol·L -1×4.17 mL=25 mmol ,剩余H 2Bar 的物质的量为
n 余(H 2Bar)=n (H 2Bar) - n (NaOH)=1-mol
1844.18?g g ×1000 - 25 mmol =75 mmol pH =p K a +lg Bar)(H )(HBar 2n n -=7.43+lg 75mmol
mmol 25=6.95 β=2.303×00mL)
25)mmol/10(7500mL)(25mmol/1000mL)(75mmol/10+?=0.043 mol·L -1 13. 分别加NaOH 溶液或HCl 溶液于柠檬酸氢钠(缩写Na 2HCit )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H 3Cit 的p K a1=3.13、p K a2=4.76、p K a3=6.40)`
解.
溶液组成
缓 冲 系 抗酸成分 抗碱成分 有效缓冲范围 β最大时体积比 Na 2HCit+HCl
H 2Cit --HCit 2- HCit 2- H 2Cit - 3.76~5.76 2:1 Na 2HCit+HCl H 3Cit-H 2Cit -
H 2Cit - H 3Cit 2.13~4.13 2:3 Na 2HCit+NaOH HCit 2--Cit 3-
Cit 3- HCit 2- 5.40~7.40 2:1 14. 现有(1)0.10 mol·L -1NaOH 溶液,(2)0.10 mol·L -1NH 3溶液,(3)0.10 mol·L -1Na 2HPO 4 溶液各50
mL ,欲配制pH=7.00的溶液,问需分别加入0.10 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) + NH 3·H 2O(aq) = NH 4Cl(aq) + H 2O(l)
NH 4+的p K a = 14.00-4.75= 9.25,
(HCl)
L .10mol 0(HCl)L .10mol 00mL 5L .10mol 0lg 25.900.7111V V ????-??+=--- 解得 V (HCl) = 50 mL
⑶ HCl(aq) + Na 2HPO 4(aq) = NaH 2PO 4(aq) + NaCl(aq)
H 3PO 4的p K a2=7.21,
(HCl)
L .10mol 0(HCl)L .10mol 00mL 5L .10mol 0lg 21.700.7111V V ????-??+=--- 解得 V (HCl) = 31 mL
第一种混合溶液无缓冲作用;第二种pH
15. 用固体NH 4Cl 和NaOH 溶液来配制1 L 总浓度为0.125 mol·L -1,pH=9.00的缓冲溶液,问需NH 4Cl
多少克?求需1.00 mol·L -1的NaOH 溶液的体积(mL )。
解 设需NH 4Cl 的质量为 x g
p K a (NH 4+) = 14.00 - 4.75 = 9.25
(NaOH)
L mol 00.1mol /53.5g (NaOH)L mol 00.1lg 25.900.911-1V x V ??-???+=-- 得 1.00 mol·L -1 × V (NaOH) = 0.562[x /53.5 g·mol -1 - 1.00 mol·L -1 × V (NaOH)]
又 [1.00 mol·L -1 × V NaOH) + 0.562 (x / 53.5g·mol -1 - 1.00 mol·L -1 × V (NaOH))] / 1L
=0.125 mol·L -1
解得 x = 6.69, V (NaOH) = 0.045 L
即:需NH 4Cl 6.69 g ,NaOH 溶液0.045 L 。
16. 用0.020 mol·L -1H 3PO 4溶液和0.020 mol·L -1NaOH 溶液配制100 mL pH=7.40的生理缓冲溶液,
求需H 3PO 4溶液和NaOH 溶液的体积(mL )。
解 设第一步反应需H 3PO 4和NaOH 溶液体积各为x mL
⑴ H 3PO 4(aq)+ NaOH(aq)= NaH 2PO 4(aq) + H 2O(l)
x mLH 3PO 4与x mLNaOH 完全反应,生成NaH 2PO 4 0.020 mol·L -1 × x mL = 0.020 x mmol
⑵ 第二步反应:设生成的NaH 2PO 4再部分与NaOH y mL 反应,生成Na 2HPO 4,其与剩余NaH 2PO 4
组成缓冲溶液
NaH 2PO 4(aq) + NaOH(aq) = Na 2HPO 4(aq) + H 2O(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
mmol
)(.0200mmol .0200lg 21.740.7y x y -+= y
x y -=1.55 依题意又有 2x + y = 100
解得 x = 38.4,y = 23.2
即需H 3PO 4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Tris·HCl 浓度均为0.050
mol·L -1的溶液l00 mL 中,需加入0.050 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p K a =7.85,忽略离子强度的影响。)
解 ⑴
)HCl (L 0.0500mol 100mL L mol 0500.0)HCl (L 0.0500mol -100mL L mol 0500.0lg
85.7HCl) (Tris (Tris)lg
7.857.401-1-1-1-V V n n ??+??????+=?+= )
HCl (100)HCl (100355.0V V +-= V (HCl) = 47.6 mL
⑵ 设加入NaCl x g ,血浆渗透浓度为300 mmol ·L -1
m L
6.147m L 6.47L m ol 050.0m L 100L m ol 050.0)Tris (11??-??=--c =0.018 mol·L -1 m L
6.147m L 6.47L m ol 050.0m L 100L m ol 050.0)HCl Tris (11??+??=?--c =0.050 mol·L -1 (0.018 + 2×0.050)mol·L -1 + =????--mL 6.147mol g 5.58L mL 1000g 211x 0.300 mol·L -1 x = 0.79,即需加入NaCl 0.79 g
18. 正常人体血浆中,[-3HCO ]=24.0 mmol·L -1、[CO 2(aq)]=1.20 mmol·L -1。若某人因腹泻使血浆中
[-3HCO ]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H 2CO 3的p K a1ˊ=6.10。
解 pH= p K a1′7.36L
1.2mmol 0.90L 24mmol lg 6.10(aq)][CO ][HCO lg 1123=???+=+--- 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。
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 H 3O + 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 p K a of the conjugate acid and the buffer component ratio. When
H 3O + 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 ca n’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 - ions.Likewise the Cl - does not exhibit base behavior in water, so it cannot react with any H 3O + 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 = p K a , 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 = p K a ±1.
5. Choose specific acid-base conjugate pairs of suitable for prepare the following beffers (Use Table 4-1 for K a
of acid or K b of base ):
(a )pH≈4.0;(b )pH≈7.0;(c )[H 3O +]≈1.0×10-9 mol ·L -1;
Solution (a )HAc and Ac - (b )-42PO H and -24HPO (c )+4
NH and NH 3 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) p K a 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 0.1 mol·L -1 NaOH to 100 mL of 0.5 mol·L -1 acetate buffer
(b) Add 5 drops of 0.1 mol·L -1 HCl to 100 mL of 0.5 mol·L -1 acetate buffer
(c) Add 5 drops of 0.1 mol·L -1 NaOH to 100 mL of 0.5 mol·L -1 HCl
(d) Add 5 drops of 0.1 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 0.25 mol ·L -1 NaC 3H 5O 3 + 150 mL of 0.25 mol ·L -1 HCl
(b) 100 mL of 0.25 mol ·L -1 NaC 3H 5O 3 + 50 mL of 0.25 mol ·L -1 HCl
(c) 100 mL of 0.25 mol ·L -1 NaC 3H 5O 3+ 50 mL of 0.25 mol ·L -1 NaOH
(d) 100 mL of 0.25 mol ·L -1 C 3H 5O 3H + 50 mL of 0.25 mol ·L -1 NaOH
Solution (b) and (d)
9. A chemist needs a pH 10.5 buffer. Should she use CH 3NH 2 and HCl or NH 3 and HCl to prepare it? Why?
What is the disadvantage of choosing the other base?
Solution The p K a of CH 3NH 2·HCl is 10.65. The p K a of +4
NH is 9.25. The p K a of the former is more close to 10.5. A buffer is more effective when the pH is close to p K a . She should choose CH 3NH 2. The other is not a good choice.
10. An artificial fruit contains 11.0 g of tartaric acid H 2C 4H 4O 6,and 20.0 g its salt , potassium hydrogen
tartrate ,per liter. What is the pH of the beverage ?K a 1=1.0×10-3
Solution 16.345.1lg .00.3mol g 1.15011.0g mol g 2.18820.0g lg 00.3)O H C (H )O H (KHC lg p pH 1
16442644a 1=+=??+=+=--n n K 11. What are the [H 3O +] and the pH of a benzoate buffer that consists of 0.33 mol ·L -1 C 6H 5COOH and 0.28
mol ·L -1 C 6H 5COONa ?K a of benzoic acid=6.3×10-5。
Solution 13.485.0lg 20.4L
0.33mol L 0.28mol lg 20.4COOH)H (C )COO H (C lg p pH 1--15656a =+=??+=+=-c c K [H +]=7.41×10-5 mol ·L -1
12. What mass of sodium acetate (NaC 2H 3O 2·3H 2O,Mr=136.1g ·mol -1) and what volume of concentrated
acetic acid (17.45mol ·L -1) should be used to prepare 500 mL of a buffer solution at pH=5.00 that is 0.150 mol ·L -1 overall?
Solution c (CH 3COO -) + c (CH 3COOH)=0.150 mol ·L -1
n (CH 3COO -) + n (CH 3COOH) = 0.150 mol ·L -1 × 500 mL
× 10-3 L ·mL -1 = 0.0750 mol
)
COOH CH ()COO CH (lg 75.400.533n n -+= 78.1)
COOH CH ()COO CH (33=-n n n (CH 3COOH) = 0.0270 mol, n (CH 3COO -) = 0.0480 mol
Mass of sodium acetate = 0.0480 mol × 136.1 g ·mol -1 = 6.53 g
m L 55.1L
m ol 45.17L m l 10m ol 0270.0)COOH CH (11
33=???=--V 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 KH 2PO 4/Na 2HPO 4 solution with this pH p K a2ˊof -4
2PO H = 6.80
Solution ]
PO H []HPO [lg 80.640.74224--
+= 00.498.3]PO [H ][HPO 422
4≈=--