第五章 习题参考答案
9 解：（1）因H 2C 2O 4·2H 2O 部分风化（失水），H 2C 2O 4有效质量增加，使实际消耗的NaOH 体积增加，故NaOH 浓度偏低。

用此NaOH 溶液测定某有机酸的摩尔
质量时，实际消耗的NaOH 体积减小，会使结果偏高。

（2）因Na 2CO 3潮解，Na 2CO 3有效质量减小，使实际消耗的HCl 体积减小，故HCl 浓度偏高。

用此HCl 溶液测定某有机碱 的摩尔质量时，实际消耗的HCl
体积增大，会使结果偏低。

14 解：（1）32110100.50.203
.17%
2989.0-⨯⨯⨯=⨯⨯V
V 1=66mL
（2） V 2=57mL
（3）
32310100.50.209
.98%
9684.1-⨯⨯⨯=⨯⨯V
V 3=56mL
17 解：CaO + 2HCl = CaCl 2 + H 2O 依
CaO HCl A B CaO HCl M C M C b a T ⨯⨯⨯=⨯⨯⨯=
--33/102
1
10 所以
0.005000=08.56102
13⨯⨯⨯-HCl C
C HCl =0.1783mol·L -1 1.000×0.2000=0.1783(1.000+V)
V=0.1217L=121.7mL
18 解：Na 2CO 3 + 2HCl = 2NaCl + H 2O+CO 2↑
2:1:32=H Cl CO Na n n
故
32)(2
1
CO Na HCl M CV m ⋅=
V=20mL 时，g m 11.099.105102010.02
13=⨯⨯⨯⨯=- 称量误差E r =
%2.011
.00002.0=±
V=25mL 时，g m 13.099.105102510.02
13=⨯⨯⨯⨯=- 称量误差E r =%2.013
.00002.0=±
32210100.50.205
.60%
10005.1-⨯⨯⨯=⨯⨯V
19 解：C=105545.01000
.099
.105/5877.0/-⋅==
L mol V
M
m
Na 2CO 3 + 2HCl = 2NaCl + H 2O + CO 2↑
依
H C l n ：1:232=CO Na n
C HCl ×0.02196:(0.02000×0.05545)=2:1
解得：C HCl =0.1010mol•L -1
21解：MgCO 3 + 2HCl = MgCl 2 + H 2O + CO 2↑ NaOH + HCl = NaCl + H 2O 30.33NaOH C =36.40C HCl NaOH C =1.2C HCl 依
H C l n ：1:23
=MgCO n 所以 (48.48×C HCl -3.83×1.2C HCl )×10-3:32
.84850.1=2:1
C HCl =1.000 mol•L -1
NaOH C =1.2C HCl =1.2×1.000=1.200 mol•L -1 22解：(1)1000.05000
.018.294/709.14/===V
M m C mol•L -1
（2） 6Fe 2+ + Cr 2O 72- + 14H + = 6Fe 3+ + 2Cr 3+ + 7H 2O
依 T B/A = a/b 3
10
-⨯A B M C
Fe O Cr K T
/722 = 67
22O C K r C ×e F M ×10-3 =6×0.1000×55.845×10-3
=0.03351g·mL -1
因 Cr 2O 72- ≌ 6Fe 2+ ≌ 3Fe 2O 3
3
2722/O Fe O Cr K T
= 3722O C K r C ×32
O F e M ×10-3 =3×0.1000×159.69×10-3
=0.04791g·mL -1
或：3
27
22
/O Fe O Cr K T =Fe O Cr K T
/722·e
e F O F M M 23
2
=0.03351×845.55269.159⨯=0.04791g·mL -1 23解：1200.010
00.146
.36/00.1004374.03
=⨯⨯-HCl
C mol•L -1
(1) NaOH + HCl = NaCl + H 2O 依 T B/A = a/b 310-⨯A B M C
NaOH HCl NaOH HCl M C T ⨯⨯⨯=-3/101=1×
0.1200×10-3×40.00=0.004800 g·mL -1 (2) CaO + 2HCl = CaCl 2 + H 2O
CaO HCl CaO HCl M C T ⨯⨯⨯=
-3/1021=2
1
×0.1200×10-3×56.08=0.003365 g·mL -1 24解：
%
47.3%10000
.10055.105.6002017.03024.0%10010)(%1003=⨯⨯⨯⨯=⨯⨯⨯=⨯=-S
HAc
NaOH S HAc HAc
m M CV m m ω
25解：CaCO 3 + 2HCl = CaCl 2 + H 2O + CO 2↑ HCl n ：1:23
=CaCO n
(0.5100×50.00-0.4900×25.00)×10-3:
1:209
.100=m
m=0.6631g ω=m/m s =0.6631/1.000=66.31%
27解：依题意 ω=V%
2NaOH + H 2C 2O 4 = Na 2C 2O 4 + 2H 2O 解法一：1:2:4
22=O C Na NaOH n n
0.1018×V×10-3:
1:204
.9010
2
=⨯⨯-V m s
m S =0.4583g
解法二：ω=
%1021
/3B s
A B B s
A
B B s A V m M V
C m M V bC a m m =⨯==-
故g V M V C m B A B B s 4583.001
.01004.901018.021%102
13
3=⨯⨯⨯=⨯=-- 解法三：004583.01004.901018.02
13422/=⨯⨯⨯=-O C H NaOH T g·mL -1
%4224
22/V m V
T m m s
O C H NaOH s
O C H =⋅=
=ω
=⋅=
%
422/V V
T m O C H NaOH S 100422/⨯O C H NaO H T =0.004583×100=0.4583g
解法四：依题意01.0%1%
%
/422===
=
V
V V
T
O C H NaOH ω
01.01004.901018.021
3/%/4
22422=⨯⨯⨯==
-S
S
O C H NaOH O C H NaOH m m T T
m S =0.4583g。