基本化学沉淀溶解平衡知识题目解析

7 沉淀-溶解平衡习题解答(p180-182)

1. 解答：(1)解：AgI

(2)解：Mg(OH)2

2. 解答：(1) CaF 2 ⇌ Ca 2+ + 2F - s+0.0010 2s

K sp =(s+0.0010)(2s)240.0010s 2

(2) Ag 2CrO 4 ⇌ 2Ag + + CrO 42-

2s+0.010 s

K

sp =(2s+0.010)

2

s 0.0102

s

3. 解答： M 2X = 2M + + X 2- X 2-有酸效应：

4. 解答：(1) CaF 2 ⇌ Ca 2+ + 2F -

(2) BaSO 4 ⇌ Ba 2+ + SO 42-

(3) CuS ⇌ Cu 2+ + S 2-

)

L mol (104.1)L mol ()5.077.234104.1(11612

62

----⋅⨯=⋅⨯⨯==s K sp )

L m ol (102.1)L m ol ()32.581105.8(44)2(11113

33

2

----⋅⨯=⋅⨯⨯⨯==⋅=s s s K sp 1

5111

L mol 102.8L mol 0010.04107.20.0010

4----⋅⨯=⋅⨯⨯=

⨯=

θ

sp

K s 1

82

12

2L mol 100.2010.0100.2010.0---⋅⨯=⨯==θ

sp

K s 19

2

12

2X(H)100.1Ka Ka ][H Ka ][H 1⨯=++=++α)

L (mol 100.14100.1100.44

)2(1103

19

493

X(H)

sp X(H)

sp 'sp 2---⋅⨯=⨯⨯⨯=

⋅=

⋅==⋅ααθθ

θK s K K s s )

L (mol 102.14)10(107.24

)2(1010

8.6101][1133

2

2.1113

2

)

(2

)

(22

.14

2

)

(-----+⋅⨯=⨯⨯=⋅=

⋅=⋅=⨯+=+=H F sp H F sp a H F K s K s s K H αααθθ)

L (mol 104.110101.11010

2.10

.21][1142.210)()

(22

.22

)

(24

24

224----+⋅⨯=⨯⨯=⋅=⋅==⨯+=+

=---

H SO sp H SO sp a H SO

K s K s K H αααθ

θ

5. 解答：(1) AgBr ⇌ Ag + + Br -

3

28.0

Ag(NH )13233-1

1[NH ]+[NH ]=107.110mol L

s αββ-=+==⨯•

(2) BiI 3 ⇌ Bi 3+ + 3I -

-3-4-513.8

Bi(I)3453sp Bi(I)

1

1[I ]+[I ]+[I ]=10s(3s+0.10):s 0.017mol L K θαβββα-=+=⨯=•用逼近法求得

(3) BaSO 4 ⇌ Ba 2+ + SO 42-

由于K BaY 较大且BaSO 4的K sp 较大，所以Ba 2+消耗的EDTA 不能忽略 c Y =[Y]-[BaY]=0.010-s

6. 既考虑配位效应，又考虑酸效应

1510][,5914,9---⋅==-=∴=L mol OH pOH pH -++⋅OH NH O H NH 423

5234θb

101.75O]

H [NH ][OH ][NH --+⨯=⋅⋅=K

52323θb

101.75O]

H [NH ][OH O]}H [NH {2.8--⨯=⋅⋅⋅-=K

1

θ

b

23L 1.02mol ][OH ][OH 2.8O]H [NH ---⋅=+⋅=⋅⇒K )

L (mol 102.21010610][][1189.1936)()

(29

.192

)

(222

122---++⋅⨯=⨯⨯=⋅=⋅==++=---

H S sp H S sp a a a H S

K s K s K K H K H αααθθ

1

475256.556

.310)

(256.556.33

.286.7)(3

.2)

(L mol 1013.60

104104)

1010(101.1101010

010.0101][110010.0-----⋅⨯==⨯-⨯+-⨯⨯=⋅=-=-⨯

+=+=-=

s s s s K s s s

Y K s

c Y Ba sp BaY Y Ba H Y Y

αααθ

3

13

82

9139θa2

θ

a22

θa2

(H)

S

7.4

27.4 3.42

3231)(NH Ag 107.880.977692.31101.3109.5)10(1101.31011][H ]

[H 1101.02101.02101]

[NH ][NH 123

⨯=++=⨯⋅⨯⨯+

⨯⨯+=⋅++

==⋅+⋅+=⨯+⨯+=-----++-

+

K K K αββα

(H)

S 2

)(NH Ag 2θsp (H)

S 2

)

(NH Ag 22(H)

S 22)(NH Ag 2

22θsp 23

2323

S}{Ag ][S ][Ag ][S }]{[Ag '

)'(2]'[S ]'[Ag '-+-+-+

⋅⋅=⋅⋅⋅=⋅⋅⋅=⋅=⋅=-

+-+-+ααK αα

ααs s K

11

3

3

2

7.449

3

(H)

S

2

)(NH Ag 2θsp 106.34

10

7.8)(1010

24

S}{Ag '23

--⨯=⨯⋅⋅⨯=

⋅⋅=

-

+

ααK s

7. 解答： CaCl 2 + 2NaOH = Ca(OH)2 + 2NaCl 1.11/111=0.1 0.12

平衡时：[Ca 2+]=0.1-0.12/2=0.04 mol

L -1

设Ca(OH)2的溶解度为s ，则：(s+0.04)(2s)2=K sp =5.5

10-6

用逼近法求得：s=5.510-3,

[OH -]=2s=0.012mol/L, pH=12.04, [Ca 2+]=0.046 mol L -1

8. 解答： BaSO 4 = Ba 2+ + SO 42- s s+0.01 s(s+0.01)=K

sp =1.1

10-10 s=1.110-8mol L -1 BaSO 4沉淀的损失=1.1

10-8

200

233.4=5.1

10-4mg

9. 解答： BaSO 4的溶解度为：