高考重要填空逐空特训(四)和Ksp应用有关的填空特训

高考重要填空逐空特训(四)和Ksp应用有关的填空特训
高考重要填空逐空特训(四)和Ksp应用有关的填空特训

(四)和K sp 应用有关的填空特训

1.某温度下,Ca(OH)2、CaWO 4的相关数据见下表

将钨酸钠溶液加入石灰乳中,发生反应的离子方程式为 WO 2-

4(aq)+Ca(OH)2(s) CaWO 4(s)+2OH -

(aq)

该温度下此反应的平衡常数为________。 答案 100

解析 已知CaWO 4(s) WO 2-

4(aq)+Ca 2+

(aq),

K sp (CaWO 4)=c (WO 2-

4)×c (Ca 2+

)=1×10-

4×1×10-

6=1×10

-10

Ca(OH)2(s) 2OH -

(aq)+Ca 2+

(aq),

K sp [(Ca(OH)2)]=c 2(OH -

)×c (Ca 2+

)=(1×10-

2)2×1×10-

4=1×10-

8;

则WO 2-

4(aq)+Ca(OH)2(s) CaWO 4(s)+2OH -

(aq)该温度下此反应的平衡常数

=K sp [(Ca(OH)2)]÷K sp (CaWO 4)=(1×10-

8)÷(1×10

-10

)=100。

2.已知常温下,H 2S 的电离常数:K 1=1.3×10-7,K 2=7.1×10

-15

,则0.1mol·L -

1H 2S 溶液

的pH 约为________(取近似整数值);向浓度均为0.001mol·L

-1

的AgNO 3和Hg(NO 3)2混合

溶液中通入H 2S 气体,开始生成沉淀的c (S 2-

)为________。(已知:K sp (Ag 2S)=6.3×10-50

K sp (HgS)=1.6×10-52

)

答案 4 1.6×10

-49

mol·L -

1

解析 已知常温下,H 2S 的电离常数:K 1=1.3×10-

7,K 2=7.1×10

-15

,则0.1mol·L -

1H 2S

溶液中氢离子浓度约是 1.3×10-

7×0.1≈10-

4,因此pH 约为4;根据溶度积常数K sp (Ag 2S)=6.3×10

-50

,K sp (HgS)=1.6×10

-52

可知首先产生HgS 沉淀,此时溶液中c (S 2-

)=

1.6×10

-52

0.001

mol·L -

1=1.6×10-49

mol·L -

1。

3.用惰性电极电解浓度较大的CuCl 2溶液,阳极的电极反应式为________________,当电解到一定程度,阴极附近出现蓝色Cu(OH)2絮状物。常温下,经测定阴极附近溶液的pH =m ,此时阴极附近c (Cu 2+

)=______________mol·L -

1 (已知:Cu(OH)2的K sp =2.2×10

-20

)。

答案 2Cl -

-2e -

===Cl 2↑ 2.2×10(8

-2m )

解析 用惰性电极电解浓度较大的CuCl 2溶液,阳极发生氧化反应,其电极反应式为2Cl

-2e -

===Cl 2↑;K sp =c 2

(OH -

)×c (Cu

2+

),c (Cu

2+

)=2.2×10-

20(10m -14)

2mol·L -1=2.2×10(8-

2m)mol·L-1。

4.镁-H2O2酸性燃料电池的反应原理为Mg+H2O2+2H+===Mg2++2H2O, 则正极反应式:________________________________________________________________________

________________________________________________________________________。

常温下,若起始电解质溶液pH=1,则pH=2时,溶液中Mg2+浓度为________。当溶液pH =6时,________(填“有”或“没有”)Mg(OH)2沉淀析出(已知K sp[Mg(OH)2]=5.6×10-12)。答案H2O2+2H++2e-===2H2O0.045mol·L-1没有

解析镁-H2O2酸性燃料电池的反应原理为Mg+H2O2+2H+===Mg2++2H2O,正极上是过氧化氢得到电子生成水的反应,正极反应式H2O2+2H++2e-===2H2O;若起始电解质溶液pH=1,则pH=2时溶液中,氢离子浓度减小0.1mol·L-1-0.01mol·L-1=0.09mol·L-1,依据反应方程式得到Mg2+浓度=0.045mol·L-1;K sp[Mg(OH)2]=5.6×10-12,当溶液pH=6时,c(OH -)=10-8mol·L-1,则Q

c

=c(Mg2+)×c2(OH-)=0.045×10-16=4.5×10-18<K sp[Mg(OH)2],说明无氢氧化镁沉淀生成。

5.(1)食盐精制时所加的四种试剂:纯碱、氯化钡、盐酸和烧碱,是为了除去粗盐水中含有的________等杂质离子,需将粗盐溶解后,按顺序加入药品进行沉淀、过滤等。加入药品的顺序可以是________________。

(2)加入足量纯碱时,会有部分BaSO4转化成BaCO3。若 1.0LBaSO4的悬浊液中含0.010molBaSO4固体,需加入________molNa2CO3才能将BaSO4完全转化为BaCO3。(已知K sp(BaSO4)=1.1×10-10,K sp(BaCO3)=5.5×10-9)

答案(1)Ca2+、Mg2+、SO2-4烧碱、氯化钡、碳酸钠、盐酸(可写化学式,顺序合理即可) (2)0.51

解析(1)NaOH是为了除去Mg2+,氯化钡是为了除去SO2-4,纯碱是为了除去Ca2+和过量的Ba2+,稀盐酸是为了除去OH-和CO2-3,食盐精制时所加四种试剂是为了除去粗盐水中含有的Ca2+、Mg2+、SO2-4等杂质离子;除去粗盐中的Mg2+,使用的除杂试剂是过量的NaOH 溶液;除去SO2-4使用的是过量的BaCl2;除去Ca2+和过量的Ba2+使用的是Na2CO3,由于Na2CO3的目的是除去Ca2+和过量的Ba2+,故Na2CO3的加入必须在BaCl2之后,待沉淀完全后,过滤除去生成的沉淀,此时溶液中的杂质离子是OH-和CO2-3,故加入过量的盐酸即可除去,加入药品的顺序可以是烧碱、氯化钡、碳酸钠、盐酸或氯化钡、烧碱、碳酸钠、盐酸或氯化钡、碳酸钠、烧碱、盐酸。(2)溶液中SO2-4的浓度为0.01mol·L-1,根据CO2-3+

BaSO4===BaCO3+SO2-4当开始有BaCO3沉淀时,K=c(SO2-4)

c(CO2-3)

K sp(BaSO4)

K sp(BaCO3)

1.1×10-10

5.5×10-9

0.02,则c(CO2-3)=0.5mol·L-1,此时溶液中Na2CO3的物质的量为0.5mol·L-1×1L=0.5mol,转化的BaSO4为0.01mol,需要0.5mol+0.01mol=0.51molNa2CO3才能将BaSO4完全转化为BaCO3。

6.利用喷雾干燥法脱硫工艺是除去SO 2的常见方法,先将含SO 2的废气溶于水,再用饱和石灰浆吸收,具体步骤如下:

SO 2(g)+H 2O(l) H 2SO 3(l) H +

(aq)+HSO -

3 (aq) Ⅰ HSO -

3(aq) H +

(aq)+SO 2-

3(aq) Ⅱ

Ca(OH)2(s) Ca 2+

(aq)+2OH -

(aq) Ⅲ

Ca 2+

(aq)+SO 2-

3(aq) CaSO 3(s) Ⅳ

该温度下,测定吸收后液体中c (Ca 2+

)一直保持为0.70mol·L -

1,

已知K sp (CaSO 3)=1.4×10-

7,则吸收后溶液中的SO 2-

3的浓度为________。

答案 2×10-

7mol·L -

1

解析 根据硫酸钙的溶度积常数表达式K sp (CaSO 3)=c (Ca 2+

)·c (SO 2-

3)可知,c (SO 2-

3)=

K sp (CaSO 3)c (Ca 2+

)

=1.4×10-7

0.7mol·L -1=2.0×10-7mol·L -

1。 7.将氢氧化钙加入钨酸钠碱性溶液中可得到钨酸钙(CaWO 4),已知温度为T 时K sp (CaWO 4) =1×10

-10

,K sp [Ca(OH)2]=4×10-

7,当溶液中WO 2-

4恰好沉淀完全(离子浓度等于10-

5

mol·

L -1

)时,溶液中c (WO 2-

4)

c (OH -)

等于________________。

答案 5×10-

5

解析 当溶液中WO 2-

4恰好沉淀完全,

其离子浓度等于10-

5mol·L -

1,根据K sp (CaWO 4)=c (Ca 2+

)×c (WO 2-

4)=1×10

-10

,则溶液中c (Ca 2+

)=1×10-

5mol·L -

1,再根据K sp [Ca(OH)2]=c (Ca 2

)×c 2

(OH -)=4×10-7

,可知c (OH -)=0.2 mol·L

-1

,则溶液中c (WO 2-

4)

c (OH -)

=10-

50.2=5×10-5

8.常温下,K sp [Fe(OH)2]=1.64×10

-14

,则求得反应:Fe 2+

+2H 2O Fe(OH)2+2H +

的平衡

常数为________________(保留1位小数)。 答案 6.1×10

-15

解析 Fe 2+

+2H 2O Fe(OH)2+2H +

的平衡常数K =c 2(H +

)c (Fe 2+)=c 2(H +

)c 2(OH -

)c (Fe 2+)c 2(OH -)=(1×10-

14)2

1.64×10-

14≈6.1×10

-15

9.甲烷重整可选氧化物NiO -Al 2O 3作为催化剂。工业上常用Ni(NO 3)2、Al(NO 3)3混合液加入氨水调节pH =12(常温),然后将浊液高压恒温放置及煅烧等操作制备。加入氨水调节pH =12时,c (Ni 2+

)为________。已知:K sp [Ni(OH)2]=5×10

-16

答案 5×10

-12

mol·L -1

解析 pH =12,c (OH -

)=0.01mol·L -

1, K sp [Ni(OH)2]=5×10

-16

,c (Ni 2+

)=5×10-

16(10-2)

2mol·

L -1=5×10-12mol·L -

1。

10.一定温度下,K sp[Mg3(PO4)2]=6.0×10-29,K sp[Ca3(PO4)2]=6.0×10-26。向浓度均为0.20mol·L-1的MgCl2和CaCl2混合溶液中逐滴加入Na3PO3,先生成________________沉淀(填化学式);当测得溶液其中一种金属阳离子沉淀完全(浓度小于10-5mol·L-1)时,溶液中的另一种金属阳离子的物质的量浓度c=________mol·L-1。

答案Mg3(PO4)210-4

解析难溶物溶度积常数小的先沉淀,根据溶度积常数相对大小知,Mg3(PO4)2先生成沉淀,当溶液中c(Mg2+)=10-5mol·L-1时,认为镁离子完全沉淀,则溶液中的

c2(PO3-4)=K sp[Mg3(PO4)2]

c3(Mg2+)

6.0×10-29

(10-5)3

=6.0×10-14,所以c(Ca2+)=

36×10-26

6×10-14

mol·L-1=

10-4mol·L-1。

(完整版)高中英语语法填空专题训练和答案

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