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1、第四章 酸碱平衡与酸碱滴定法4-1将 300mL0.20 molL1HAc 溶液稀释到什么体积才能使解离度增加一倍。解：设稀释到体积为 V ，稀释后 VcmL30ol2.01由 12acK得： )(.1.022因为 K a =1.74105 ca = 0.2 molL1 caK a 20K w ca/K a500故由 12 =1 得 V =3004/1mL =1200mL 此时仍有 caK a20K w ca/K a500 。4-2奶油腐败后的分解产物之一为丁酸(C 3H7COOH)，有恶臭。今有 0.40L 含 0.20 molL1 丁酸的溶液， pH 为2.50, 求丁酸的 K a。解：p
2、H=2.50 c(H +) =102.5 molL1 =102.5/0.20 = 1.6102K a=52 10.106.)(c4-3 What is the pH of a 0.025molL1 solution of ammonium acetate at 25? pK a of acetic acid at 25 is 4.76, pK a of the ammonium ion at 25 is 9.25, pKw is 14.00.解： c(H+)= 0.724.976.4210aKpH= logc(H+) = 7.00 4-4已知下列各种弱酸的 K a 值，求它们的共轭碱的 K b
3、 值，并比较各种碱的相对强弱。(1)HCN K a =6.21010； (2)HCOOH K a =1.8104；(3)C6H5COOH(苯甲酸) K a =6.2105； (4) C6H5OH (苯酚 ) K a =1.11010；(5)HAsO2 K a =6.01010； (6) H2C2O4 K a1=5.9102；K a2=6.4105；解： (1)HCN Ka= 6.21010 Kb=Kw/6.21010 =1.6105 (2)HCOOH Ka= 1.8104 Kb=Kw /1.8104 =5.61011 (3)C6H5COOH Ka= 6.2105 Kb=Kw /6.2105 =
4、1.611010(4)C6H5OH Ka=1.11010 Kb=Kw /1.11010 =9.1105 (5)HAsO2 Ka=6.01010 Kb=Kw /6.01010 =1.7105 (6)H2C2O4 Ka1=5.9102 Kb2=Kw /5.9102 =1.71013 Ka2=6.4105 Kb1=Kw /6.4105 =1.5 1010 碱性强弱：C 6H5O AsO2 CN C6H5COOC2O42 HCOO HC2O44-5用质子理论判断下列物质哪些是酸？并写出它的共轭碱。哪些是碱？也写出它的共轭酸。其中哪些既是酸又是碱？H2PO4 ，CO 32 ，NH 3，NO 3 ，H 2
5、O，HSO 4 ，HS ，HCl解： 酸 共轭碱 碱 共轭酸 既是酸又是碱H2PO4 HPO42 H2PO4 H3PO4 H2PO4NH3 NH2 NH3 NH4+ NH3H2O OH H2O H3O+ H2OHSO4 SO42 HSO4 H2SO4 HSO4HS S2 HS H2S HSHCl Cl NO3 HNO3CO32 HCO34-6写出下列化合物水溶液的 PBE：(1) H3PO4 (2) Na2HPO4 (3) Na2S (4)NH4H2PO4(5) Na2C2O4 (6) NH4Ac (7) HCl+HAc (8)NaOH+NH3解：(1) H3PO4： c( H+) = c(H
6、2PO4 ) + 2c( HPO42) + 3c (PO43) + c(OH) (2) Na2HPO4： c(H+) + c(H2PO4 ) + 2c(H3PO4) = c(PO43) + c(OH) (3) Na2S： c(OH) = c(H+) + c(HS) + 2c(H2S )(4)NH4H2PO4： c(H+) + c(H3PO4) = c(NH3) + c(HPO42) + 2c(PO43) + c(OH)(5)Na2C2O4： c(OH) = c(H+) + c(HC2O4) + 2c(H2C2O4)(6)NH4AC： c(HAc) + c(H+) = c(NH3) + c(OH
7、)(7)HCl + HAc： c(H+) = c(Ac) + c(OH) + c(Cl )(8)NaOH + NH3： c(NH4+) + c(H+) = c(OH) c(NaOH)4-7某药厂生产光辉霉素过程中，取含 NaOH 的发酵液 45L (pH=9.0)，欲调节酸度到 pH=3.0，问需加入 6.0 molL1 HCl 溶液多少毫升？解： pH = 9.0 pOH = 14.0 9.0 = 5.0 c(OH) =1.0 105 molL1 n(NaOH)= 45105 mol 设加入 V1 mLHCl 以中和 NaOH V1 = 45105/6.0103mL = 7.5102 mL设
8、加入 x mLHCl 使溶液 pH =3.0 c(H+) =1103 molL1 6.0 x103/(45+7.5105 + x103 ) = 1103 x = 7.5mL共需加入 HCl：7.5mL + 7.510 2 mL = 7.6mL4-8H 2SO4 第一级可以认为完全电离，第二级 K a2 =1.2102， ，计算 0.40 molL1 H2SO4 溶液中每种离子的平衡浓度。解： HSO4 H+ + SO42起始浓度/molL 1 0.40 0.40 0平衡浓度/molL 1 0.40x 0.40 +x x 1.2102 = x(0.40 + x)/(0.40 x) x = 0.0
9、11 molL1c(H+) = 0.40 + 0.011 = 0.41 molL1 pH = lg0.41 = 0.39c(HSO4) = 0.40 0.011 = 0.39 molL1 c(SO42) = 0.011 molL14-9求 1.0106 molL1HCN 溶液的 pH 值。 （提示：此处不能忽略水的解离）解：K a(HCN)= 6.21010 caK a108 故可准确进行滴定；(2) Ka(HCN)=6.21010 ca = 0.1 molL1 cKa = 6.21011108 caKa2108 Ka1/Ka2=5.9102/6.4105108 caKa2108 caKa21
10、08 但 Ka1/Ka2 =7.4104/1.7105108 cKa2108 但 Ka1/Ka2 = 9.1104/4.310520Kw ca/K a500c(H+)= 063.5.75.3pH=1.79甲酸钠的 K b= K w/K a = 1014.00/103.751=1010.25cb =1.0molL1 cbK b20K w cb/K b500c(OH)= 625.107. pOH = 5.13 pH = 14.005.2 = 8.874-20Calculate the concentration of sodium acetate needed to produce a pH of
11、 5.0 in a solution of acetic acid (0.10molL1) at 25. pK a for acetic acid is 4.756 at 25.解： pH = pK a lgca/cb lgca/cb = pKa pH = 4.756 5.0 = 0.24 ca/cb = 0.57cb = 0.10/0.57 = 0.18(molL1)4-21Calculate the percent ionization in a 0.20 molL1 solution of hydrofluoric acid, HF (Ka=7.2104)。解： K a= 12c7.21
12、04 = 120.0.202 + 7.2104 7.2104 = 0 = 0.0584-22The concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.10 molL1. Calculated c(H3O+),c(HS),and c(S2) in the solution.解：H 2S 的 Ka1=1.1107 Ka2=1.31013 caKa1=1.3108 20Kw ca/Ka1500c(H+)=471a 10.1.0K(molL1)HS H+ + S2 1.
13、0104x 1.0104+x x 13.10 .)(x =1.31013即 c(S2) = 1.31013molL1c(HS) = 1.0104 1.31013 = 1.0104 molL14-23某一元酸与 36.12mL 0.100 molL1 NaOH 溶液中和后，再加入 18.06mL 0.100 molL1 HCl 溶液，测得 pH 值为 4.92。计算该弱酸的解离常数。解：36.12mL0.100 molL 1NaOH 与该酸中和后, 得其共轭碱 nb=3.612103mol； 加入 18.06mL0.100molL1HCl 后生成该酸 na=1.806103mol； 剩余共轭碱
14、nb=(3.6121.806)103mol = 1.806103molpH = pK a lgca/cb= pK a = 4.92 K a = 104.92 = 1.21054-240.20mol 的 NaOH 和 0.20molNH4NO3 溶于足量水中并使溶液最后体积为 1.0 L，问此时溶液 pH 为多少。解：平衡后为 0.20 molL1 的 NH3H2O 溶液 K b=1.74105cbK b 20K w cb/K b 500 c(OH) = 351087.7.0 molL1 pOH = 2.73 pH = 14.00 2.73 = 11.274-25今有三种酸(CH 3)2AsO2
15、H， ClCH2COOH，CH 3COOH，它们的标准解离常数分别为 6.4107, 1.4105 , 1.76105。试问：(1)欲配制 pH= 6.50 缓冲溶液，用哪种酸最好？(2)需要多少克这种酸和多少克 NaOH 以配制 1.00L 缓冲溶液，其中酸和它的共轭碱的总浓度等于 1.00molL1？解：(1)(CH 3)2AsO2H 的 pKa = 6.19；ClCH 2COOH 的 pKa = 4.85；CH 3COOH 的 pKa = 4.76；配 pH = 6.50 的缓冲溶液选 (CH3)2AsO2H 最好，其 pKa 与 pH 值最为接近。 (2)pH = pKa lgca/cb 6.50 = 6.19 lgca/(1.00
