分析化学习题答案

cH3BO3 = cH2BO-3 = 0.020mol/L
I
=
1 2
(cNa+
z
2 Na+
+ c z ) 2 H2BO-3 H2BO-3
=
1 2
(0.020 12
+
0.020 (-1)2
)
Fra Baidu bibliotek
=
0.020(mol/L)
lgγm
=
-0.50z
2 i
(
1
+
I
I
- 0.30 I)
lgγ H2BO-3
= -0.5012 ( 0.020 1 + 0.020
查表知氨基乙酸的pKa1=2.30，pKa2=9.60，
+NH3CH2COOH
⇌
+NH3CH2COO-
⇌
NH2CH2COO-
配制pH=2.00的缓冲溶液需 +NH3CH2COOH-
+NH3CH2COO-体系。还需在氨基乙酸溶液中加入
盐酸。加入盐酸的物质的量就等于+NH3CH2COOH
由于C＜20[H+]，[H+]≫[OH-]
=
0.079 100/1000 1.0
=
0.0079(L)
=
7.9(mL)
5.计算0.010mol/L硼砂溶液的pH值(考虑离子强度)。
解：硼砂是一元弱酸，查表知pKa=9.24
Na2B4O7 + 7H2O = 4H3BO3 + 2NaOH
2NaOH + 2H3BO3 = 2NaH2BO + 2H2O 剩余的H3BO3与反应生成的H2BO-3形成缓冲溶液
= 6.210-5mol/L
pH = 4.21
（3）0.10mol/L Na2S溶液
Na2S →2Na+ + S2S2-是二元碱，查表知其解离常数分别为
Kb1 = Kw Ka2 = 1.010-14 1.210-15 = 8.3 Kb2 = Kw Ka1 = 1.010-14 5.7 10-8 = 1.810-7
[H+
]
=
cHA -[H+ ] cA- + [H+ ]
Ka
=
cHA -[H+ ] 0.10 - cHA +[H+
]
Ka
0.010 = cHA - 0.010 4.5 10-3 0.10 - cHA + 0.010
cHA = 0.079(mol/L)
cHA VHA = cHCl VHCl
VHCl
[H+ ] = K K a(HCN) a(NH4+ ) = 6.3 10-10 mol/L
计算结果相同
(2)KHP是两性物质，查表知：
Ka1 = 1.110-3 , Ka2 = 3.9 10-6 cKa2 > 20Kw ,c < 20Ka1，
[H+ ] =
cKa1Ka2 c + Ka1
= 0.0101.110-3 3.910-6 0.010 + 1.110-3
+3[PO43-]+[OH-] PBE [H+]= [NH ]+[HPO 2-]+2[PO 3-] +[OH-]
（2） NH4CN MBE c = [NH3] + [NH4+]
c = [CN-] + [HCN] CEB [H+] + [NH4+] = [OH-] + [CN-] PBE [H+] = [NH3] + [OH-] - [HCN]
2. 计算0.10mol/LNH4CN、0.010mol/L KHP 和 0.10mol/L Na2S溶液的 pH值。
解:(1)0.10mol/LNH4CN溶液，NH4CN是弱
酸弱碱组成的盐 Ka(HCN) = 7.210-10 Ka(NH4+) = Kw/Kb = 1.010-14 /1.810-5 = 5.610-10
解:
[H+ ]2 δH2S = [H+ ]2 + [H+ ]Ka1 + Ka1Ka2
=
(0.10)2
+
0.10
(0.10)2 5.7 10-8 + 5.7
10-8
1.2
10-15
=
1.0
δHS-
= [H+ ]2
[H+ ]Ka1 + [H+ ]Ka1 + Ka1 Ka2
HS-
=
(0.10)2
由于Kb1 ≫Kb2，故可按一元碱处理
c/Kb1＜400，cKb1＞20Kw
[OH- ] = -Kb1 +
K2 b1
+ 4cKb1
2
-8.3 + 8.32 + 4 0.10 8.3 =
2
= 0.099(mol/L)
pOH = 1.00 pH = 13.00
3. 计算pH=1.00时，0.10mol/L H2S溶液中各型体 的浓度。
酸碱滴定法
1. 写出NH4H2PO4和NH4CN的MBE、CEB及 PBE（浓度为c mol/L）。
解（1） NH4H2PO4 MBE c = [NH3]+[NH4+]
c = [H3PO4]+[H2PO4-]+[HPO42]+[PO43-] CEB [H+]+[NH4+]= [H2PO4-]+2[HPO42-]
- 0.300.020)
= -0.059 γH2BO-3 = 0.87
cH3BO3 = cH2BO-3 远远大于溶液中的[H+ ]和[OH- ]
pH
=
pK a
+
lg
α H2BO-3 α H3BO3
= 9.24 + lg 0.87 0.020 0.020
= 9.18
0.10 5.7 10-8 + 0.10 5.7 10-8 + 5.7 10-2 1.2 10-15
= 5.810-7
5.7 10-8 1.2 10-15 δS2- = (0.10)2 + 0.10 5.7 10-8 + 5.7 10-2 1.2 10-15
=6.9×10-21
[H2S] = c H2S H2S = 0.10 1.0 = 0.10(mol/L) [HS- ] = 0.10 5.810-7 = 5.7 10-8(mol/L)
[S2- ] = 0.10 6.910-21 = 6.910-22 (mol/L)
4. 欲配制100mL氨基乙酸缓冲溶液，其总浓度 c=0.10mol/L，pH=2.00，需氨基乙酸多少克？还需 加多少1.0mol/L的酸或碱（M氨基乙酸=75.07g/mol）
解：需氨基乙酸 m =0.10×0.10×75=0.75(g)
[H+ ] =
K (cK a(HCN)
a(NH4+
)
+
Kw
)
c + Ka(HCN)
[H+ ] =
7.210-10 (0.10 5.610-10 + 1.010-14 ) 0.10 + 7.210-10
= 6.310-10(mol/L) pH = 9.20
由于cKa（NH4+）> 20KW；c >> Ka(HCN)