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4 版1高二化学选修4·第10~17期版 详细答案 详细答案
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2019
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9~10
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10~17
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3
01234567
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10
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3
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1.A 2.B 3.B 4.A 5.D 6.D
7.D pH
!
5
"#$%&'()*!
:c(H+)=
1×10-5mol/L,
+ , - ' ( " ) * !
:c(SO2-4 )=
1
2c(H
+)=5×10-6mol/L;
#$./
1000
01
,
&'(
)*234567
1×10-7mol/L,
8+,-'()*
!
:c(SO2-4 )=5×10
-9 mol/L,
9:./1#$%
c(SO2-4 );c(H
+)
"<=7>!
:5×10-9mol/L∶1×
10-7mol/L=1∶20,
?@A
D.
8.D 25℃
B
,
C#$%DEF'G"
c(OH-)=1
×10-13mol/L,
!,HI#$
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NH+4、Fe
3+,
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A
JA
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3;,OPT3;IOP
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RA
D.
9.B
10.D X
WXN
25℃
B
c(H+)=10-7mol/L,pH=
7,
8
Z
WXY*Z[
25℃,
\
pH<7,D
]^
.
?@A
D.
11.(1)①D2 幑幐O D
++OD- ②7.4 ③12.8
(2)①H2O 幑幐2 H
+ +HO-2 ②H2O2+Ba(OH)
2
BaO2+2H2O ③H2O2+H2O 2 H3O
+
2 +HO
-
2
,-
:(1)①_<E"F'`ab,cdeE"F'
`ab!
D2 幑幐O D
+ +OD-;②CY*feE(D2O)
"'(ghi!
1.6×10-15,
j
c(D+)=c(OD-)=4×
10-8mol/L,pD=8-lg4=7.4;③0.01mol/L" NaOD#
$%
,c(OD-)=0.01mol/L,c(D+)=1.6×10
-15
0.01 =1.6
×10-13mol/L,
\
pD=13.0-lg1.6=12.8.
(2)①klH2O2mnopqr,,jNE%stF
'
,
uUtF'"`ab!
H2O 幑幐2 H
+ +HO-2;②
H2O2vr,w,mnopqr,,xyIz{|n}~,
j
H2O2; Ba(OH)2z{|n}~"��`ab!
H2O2+Ba(OH)幑幐2 BaO2+2H2O;③E"��F'!
2H2 幑幐O H3O
++OH-,
j
H2O2"��F'`ab!:
H2O2+H2O 2 H3O
+
2 +HO
-
2.
12.(1)E=A=D<C<B
(2)bc
(3)10∶1
(4)9∶11
13.(1)
�Z
(2)
�Z
(3)
�Z
(4)
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(5)
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(6)
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(7)
�Z
(8)
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14.Ⅰ.(1)1×10-4mol/L 1×10-7
(2)106
(3)①B ②<
Ⅱ.0.011
,-
:Ⅰ.(1)�d�YB,0.1mol/LCUq, HA
NE%�
0.1%
��F'
,
j��F'"
HA
o
0.1mol/L
×0.1% =1×10-4mol/L,
,oUq,
,
9:�#$%
c(H+)=1×10-4 mol/L.HA
"F'��hi
Ka =
c(H+)c(A-)
c(HA) =
10-4×10-4
0.1-10-4
≈1×10-7.
(2)
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,
)*!
10-10mol/L,
��D
HA
F'G"
c(H+)
�!EF'G
"
c(H+)
"
10-4
10-10
=106
0
.
(3)①HAorF��,NOP�a% HAF'G&
'(
,
9:OP�a%
HA
"
c(H+)
K[
HCl,
j
HA
%
pH
�Ka*�[
HCl,
9:��
HA
#$%
pH
"��
WX!
B;②���OP1,HA�~,% pH� ,j¡
¢OP"
n(HA)>n(HCl).
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j¡¢OP
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Zn
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m1<m2.
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10mL,
OP1#$o
100mL,
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90mL.
OP1&'("¨�"Lo
0.01mol/L
×0.1L=0.001mol,
��#$%«'("¨�"Lo
0.
001mol,
j©
BaCl2#$% Cl
-
")*!
0.001mol÷0.
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