南昌大学大学化学第四章
第四章
1.将300mL0.20 mol⋅L -1HAc 溶液稀释到什么体积才能使解离度增加一倍。
-10. 20m ol ⋅L ⨯300m L 解:设稀释到体积为V ,稀释后c =V
2 20. 20 α2=0. 20⨯300⨯(2α) c α由K a =得: 1-αV ⋅(1-2α) 1-
因为K a =1.74⨯10-5 c a =0.2 mol⋅L -1c a K a >20K w c a /K a >500
故由 1-2α=1-α得V =[300⨯4/1]mL =1200mL
此时仍有c a K a >20K w c a /K a >500 。
2.求算 0.20mol ⋅L -1NH 3H 2O 的c (OH-) 及解离度。
解:K b (NH3·H 2O)=1.74⨯10-5 由于c b K b >20K w ,c b /K b >500
θ由c (OH -) =c b ⨯K a
得c (OH -) =0. 20⨯1. 74⨯10-5mol ⋅L -1=1.9⨯10-3 mol⋅L -1
c (OH-) 1. 9⨯10-3α===9. 5⨯10-3=0. 95% b 3.奶油腐败后的分解产物之一为丁酸(C3H 7COOH) ,有恶臭。今有0.40L 含0.20 mol⋅L -1丁酸的溶液, pH 为2.50, 求丁酸的K a 。
解:pH=2.50, c (H+)=10-2.5 mol⋅L -1
α=10-2.5/0.20 = 1.6⨯10-2
-22c α2=0. 20⨯(1. 6⨯10) =5. 2⨯10-5K a = -21-1-1. 6⨯10
-14. What is the pH of a 0.025mol⋅L solution of ammonium acetate at 25℃?p K a of acetic acid
at 25℃ is 4.76,pK a of the ammonium ion at 25℃ is 9.25, pK w is 14.00.
θθ-4. 76解:c (H+)=K a ⨯10-9. 24=10-7. 00 1K a 2=pH= -log c (H+) = 7.00
5.已知下列各种弱酸的K a 值,求它们的共轭碱的K b 值,并比较各种碱的相对强弱。
(1)HCN K a =6.2×10-10; (2)HCOOH K a =1.8×10-4;
(3)C6H 5COOH(苯甲酸) K a =6.2×10-5; (4) C6H 5OH (苯酚) K a =1.1×10-10;
(5)HAsO2 K a =6.0×10-10; (6) H2C 2O 4K a1=5.9⨯10-2;K a2=6.4⨯10-5; 解:(1)HCN K a = 6.2⨯10-10 K b =K w /6.2⨯10-10=1.6⨯10-5
(2)HCOOH K a = 1.8⨯10-4 K b =K w /1.8⨯10-4=5.6⨯10-11
(3)C6H 5COOH K a = 6.2⨯10-5 K b =K w /6.2⨯10-5 =1.61×10-10
(4)C6H 5OH K a =1.1⨯10-10 K b =K w /1.1⨯10-10=9.1⨯10-5
(5)HAsO2 K a =6.0⨯10-10 K b =K w /6.0⨯10-10=1.7⨯10-5
(6)H2C 2O 4 K a1=5.9⨯10-2 K b2=K w /5.9⨯10-2=1.7⨯10-13 K a2=6.4⨯10-5 K b1=K w /6.4⨯10-5=1.5×10-10 碱性强弱:C 6H 5O -> AsO2-> CN-> C6H 5COO ->C2O 42-> HCOO-> HC2O 4-
6.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?
----- H 2PO 4,CO 32,NH 3,NO 3,H 2O ,HSO 4,HS ,HCl
解:
酸 共轭碱 碱 共轭酸 既是酸又是碱
---H 3PO 4 H 2PO 4 H 2PO 4 H 2PO 4 HPO 42-
NH 3 NH 3 NH 4+ NH 3 NH 2-
H 2O H 2O H 3O + H 2O OH -
---H 2SO 4 HSO 4 HSO 4 HSO 4 SO 42-
---H 2S HS HS HS S 2-
-HCl HNO 3 NO 3 Cl -
CO 32- HCO 3-
7.写出下列化合物水溶液的PBE :
(1) H3PO 4 (2) Na2HPO 4 (3) Na2S (4)NH4H 2PO 4
(5) Na2C 2O 4 (6) NH4Ac (7) HCl+HAc (8)NaOH+NH3
解:
(1) H3PO 4: c ( H+) = c (H2PO 4- ) + 2c ( HPO42-) + 3c (PO43-) + c (OH-)
(2) Na2HPO 4: c (H+) + c (H2PO 4- ) + 2c (H3PO 4) = c (PO43-) + c (OH-)
(3) Na2S : c (OH-)=c (H+) + c (HS-) + 2c (H2S )
(4)NH4H 2PO 4: c (H+) + c (H3PO 4) = c (NH3) + c (HPO42-)+ 2c (PO43-) + c (OH-)
(5)Na2C 2O 4: c (OH-) = c (H+) + c (HC2O 4-) + 2c (H2C 2O 4)
(6)NH4A C : c (HAc) + c (H+) = c (NH3) + c (OH-)
(7)HCl+HAc: c (H+) = c (Ac-) + c (OH-) +c (Cl- )
(8)NaOH +NH3: c (NH4+) + c (H+) = c (OH-) – c (NaOH)
8.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 mol⋅L -1HCl 溶液多少毫升?
解: pH = 9.0 pOH = 14.0 – 9.0 = 5.0c (OH-) =1.0⨯ 10-5mol ⋅L -1n (NaOH)= 45⨯10-5mol 设加入V 1mLHCl 以中和NaOH V 1= [45⨯10-5/6.0]103mL = 7.5⨯10-2mL
设加入x mLHCl 使溶液pH =3.0 c (H+) =1⨯10-3mol ⋅L -1
6.0⨯x 10-3/(45+7.5⨯10-5 +x 10-3 ) = 1⨯10-3 x = 7.5mL
共需加入HCl :7.5mL + 7.5⨯10-2mL = 7.6mL
9.H 2SO 4第一级可以认为完全电离,第二级K a2=1.2×10-2,,计算0.40 mol⋅L -1 H2SO 4溶液中每种离子的平衡浓度。
解:HSO 4- H + + SO 42-
起始浓度/mol⋅L -10.40 0.40 0
平衡浓度/mol⋅L -1 0.40-x 0.40 +xx
1.2⨯10-2 = x (0.40 +x )/(0.40 -x ) x = 0.011 mol⋅L -1
c (H+) = 0.40 + 0.011 = 0.41 mol⋅L -1pH = -lg0.41 = 0.39
c (HSO4-) =0.40- 0.011 = 0.39mol⋅L -1c (SO42-) =0.011 mol⋅L -1
10.某一元酸与36.12mL 0.100 mol⋅L -1NaOH 溶液中和后,再加入18.06mL 0.100 mol⋅L -1HCl 溶液,测得pH 值为4.92。计算该弱酸的解离常数。
解:36.12mL0.100mol ⋅L -1NaOH 与该酸中和后, 得其共轭碱n b =3.612⨯10-3mol ;
加入18.06mL0.100mol ⋅L -1HCl 后生成该酸n a =1.806⨯10-3mol ;
剩余共轭碱n b =(3.612-1.806) ⨯10-3mol = 1.806⨯10-3mol
pH = pK a -lg c a /c b =pK a = 4.92K a = 10-4.92 = 1.2⨯10-5
11.求1.0×10-6mol ⋅L -1HCN 溶液的pH 值。(提示:此处不能忽略水的解离)
解:K a (HCN)= 6.2⨯10-10c a ⋅K a
-6-10 c (H +) =c a ⋅K +1. 0⨯10-14=1. 0⨯10-7m ol ⋅L -1 a +K w =. 0⨯10⨯6. 2⨯10
pH= 7.0
12.计算浓度为0.12mol ⋅L -1的下列物质水溶液的pH 值(括号内为p K a 值) :
(1) 苯酚(9.89); (2)丙烯酸(4.25)
(3)氯化丁基胺( C4H 9NH 3Cl) (9.39); (4)吡啶的硝酸盐(C5H 5NHNO 3)(5.25)
解:(1) pK a = 9.89 c ( H+) =
(2) pK a = 4.25c ( H+) =
(3) pK a = 9.39 c ( H+) =
(4)pK a = 5.25 c ( H+)=
-c a ⋅K a = 0. 12⨯10-9. 89=3. 9⨯10-6pH = 5.41 c a ⋅K a =0. 12⨯ 10-4. 25=2. 6⨯10-3pH = 2.59 -9. 39c a ⋅K a =0. 12⨯10=7. 0⨯10-6pH = 5.15 -5. 25a ⋅K a =0. 12⨯ 10=8. 2⨯10-4pH = 3.09 -13.H 2PO 4的K a2 =6.2×10-8,则其共轭碱的K b 是多少?如果在溶液中c (H2PO 4) 和其共轭碱的浓度相等时,溶液的pH 将是多少?
解:K b = K w /K a = 1.0⨯10-14/6.2⨯10-8=1.6⨯10-7
pH = pK a -lg c a /c b = pK a = -lg(6.2⨯10-8)=7.20
14.0.20mol 的NaOH 和0.20molNH 4NO 3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 mol⋅L -1的NH 3·H 2O 溶液K b =1.74⨯10-5
c b K b >20K w c b /K b >500
-1θc (OH-) =c b ⋅K b =0. 20⨯1. 74⨯10-5=1. 87⨯10-3mol ⋅L
pOH = 2.73 pH = 14.00 - 2.73 = 11.27
15.欲配制250mL pH=5.0的缓冲溶液,问在125mL1.0 mol ⋅L -1NaAc 溶液中应加多少6.0 mol ⋅L -1的HAc 和多少水?
解:pH = pK a -lg c a /c b 5.0 = -lg(1.74⨯10-5) -lg c a /c b
c a /c b = 0.575c b =1.0 mol⋅L -1⨯125/250 = 0.50 mol⋅L -1
c a = 0.50 mol⋅L -1⨯0.575 = 0.29 mol⋅L -1
V ⨯6.0mol ⋅L -1 = 250mL ⨯0.29mol ⋅L -1V = 12mL
即要加入12mL 6.0 mol⋅L -1 HAc及 250 mL -125 mL -12 mL =113mL水。
16.今有三种酸(CH3) 2AsO 2H , ClCH 2COOH ,CH 3COOH ,它们的标准解离常数分别为
6.4×10-7, 1.4×10-5 , 1.76×10-5。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1.00mol ⋅L -1?
解:(1)(CH3) 2AsO 2H 的p K a = 6.19;ClCH 2COOH 的p K a = 4.85;CH 3COOH 的p K a = 4.76; 配pH = 6.50的缓冲溶液选(CH3) 2AsO 2H 最好,其p K a 与pH 值最为接近。
(2)pH = pK a - lgc a /c b 6.50 =6.19-lg[c a /(1.00-c a )] c a = 0.329 mol⋅L -1
c b = 1.00-c a = 1.00 mol⋅L -1-0.329 mol⋅L -1= 0.671 mol⋅L -1
应加NaOH :m (NaOH)= 1.00L ⨯0.671 mol⋅L -1⨯40.01g ⋅moL -1=26.8g
需(CH3) 2AsO 2H :m ((CH3) 2AsO 2H) =1.00L ⨯138 g⋅moL -1=138g
17.现有一份HCl 溶液,其浓度为0.20 mol⋅L -1。
(1)欲改变其酸度到pH=4.0应加入HAc 还是NaAc ?为什么?
(2)如果向这个溶液中加入等体积的2.0 mol⋅L -1NaAc 溶液,溶液的pH 是多少?
(3)如果向这个溶液中加入等体积的2.0 mol⋅L -1HAc 溶液,溶液的pH 是多少?
(4)如果向这个溶液中加入等体积的2.0 mol⋅L -1NaOH 溶液,溶液的pH 是多少? 解:(1) 0.20 mol⋅L -1HCl 溶液的pH=0.70,要使pH = 4.0,应加入碱NaAc ;
(2)加入等体积的2.0 mol⋅L -1NaAc 后,生成0.10mol ⋅L -1HAc ;
余(2.0-0.20)/2 = 0.90mol⋅L -1NaAc ;
pH = pK a -lg c a /c b pH = -lg(1.74⨯10-5) -lg(0.10/0.90) = 5.71
(3)加入2.0 mol⋅L -1的HAc 后, c (HAc) =1.0 mol⋅L -1
HAc H + + Ac -
1.0-x 0.10 +xx
1.74⨯10-5 = (0.10 +x ) x /(1.0-x ) x =1.74⨯10-4 mol⋅L -1
c (H+) = 0.10 mol⋅L -1+1.74⨯10-4 mol⋅L -1 = 0.10 mol⋅L -1pH = 0.10
(4)反应剩余NaOH 浓度为0.9 mol⋅L -1
pOH = -lg0.9 = 0.05 pH = 14.00-0.05 = 13.95
18.0.5000mol ⋅L -1 HNO3溶液滴定0.5000mol ⋅L -1 NH3⋅H 2O 溶液。试计算滴定分数为0.50及1.00时溶液的pH 值。应选用何种指示剂?
解:滴定分数为0.50时,NH 3⋅H 2O 溶液被中和一半,为NH 3⋅H 2O 和NH 4+的混合溶液; pOH = pK b -lg c b /c a 其中c a = c b
pOH = -lg(1.74⨯10-5) = 4.76 pH = 14.00 -4.75 = 9.24
滴定分数为1.00时,NH 3⋅H 2O 刚好完全被中和,溶液为0.2500mol ⋅L -1 NH4+; K a (NH4+)= K w /K b = 1.00⨯10-14/1.74⨯10-5 = 5.75⨯10-10 c a K a >20K w c a /K a >500
θc (H +) =cK a =0. 2500⨯5. 75⨯10-10=1. 20⨯10-5pH= 4.92
可选指示剂:甲基红较好(4.4~6.2);溴甲酚绿(3.8~5.4)。
-19.人体中的CO 2在血液中以H 2CO 3和HCO 3存在,若血液的pH 为7.4,求血液中 H 2CO 3
-与HCO 3的摩尔分数x (H2CO 3) 、x (HCO3-) ?
解:H 2CO 3的K a1 = 4.2⨯10-3 (pK a1 = 6.38);K a2 = 5.6⨯10-11 (pK a2 = 10.25)
pH = pK a -lg c a /c b 7.4 = 6.38 -lg c (H2CO 3)/c (HCO3-)
c (H 2CO 3) n (H 2CO 3) -==0. 095n (HCO ) = 0.095n (HCO) 233--c (HCO 3) n (HCO 3)
n (H 2CO 3) 0. 095n (HCO 3) x (H2CO 3) ===0. 087---n (H 2CO 3) +n (HCO 3) 0. 095n (HCO 3) +n (HCO 3)
x (HCO 3) =--n (HCO 3) n (HCO 3) ==0. 91---n (H 2CO 3) +n (HCO 3) 0. 095n (HCO 3) +n (HCO 3) --
或x (HCO3-) = 1-x (H2CO 3) = 1- 0.087 = 0.913