9-1

Chapter 2 Precipitation and Solubility

Equilibrium(沉淀-溶解平衡及其应用)

2.1 Precipitation-Solubility Equilibrium

(沉淀-溶解平衡)

2.2 Factors of affecting the solubility

(影响溶解度的因素)

2.3 Shift of Precipitation and solubility

Equilibria (沉淀-溶解平衡移动)

2.4 Application of precipitation reaction

(沉淀反应在分析化学中的应用)

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2.1 Precipitation and Solubility

Equilibrium(沉淀-溶解平衡)

2.1.1 Precipitation Reactions(沉淀反应)

Many chemical reactions occur in aqueous solution. A

Precipitation Reactions (沉淀反应) is one that occurs in

solution and results in the formation of an insoluble product. For

example, an aqueous solution of the soluble ionic compound,

sodium sulfate(Na2SO4), can be mixed with an aqueous solution

of the soluble ionic compound, barium hydroxide (Ba(OH)2).

The result is the formation of a Precipitation (沉淀), the

insoluble ionic compound barium sulfate (BaSO4).

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2.2.2 The Solubility-Product Constant(溶度积常数), Ksp

By definition, a saturated solution of a salt is in equilibrium with

the undissolved solid. The equilibrium for the dissolution of

calcium carbonate in water is represented by the equation

The Kc for this equilibrium is given a special subscript(下标)

to denote that the process in question is the dissolving of a salt.

The subscript (下标) sp stands for solubility product (溶度积).

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The procedure for writing the equilibrium expression for such

processes is the same for writing equilibrium expressions for any

other equilibria. The Ksp for the dissolution of calcium carbonate

is 2.8×10–9. The very small magnitude(数值) of the equilibrium

constant (also called the Solubility-Product Constant (溶度积常数)

tells us the same thing as before: The equilibrium lies far to the

left. Undissolved solid is favored over aqueous ions. This

corresponds to calcium carbonate's having a very low solubility.

AmBn (s) mAn+(aq) + nBm-(aq)

Ksp= [An+]m[Bm-]n

You can find the Ksp values in the table

9-5

Example

BaSO4(s) Ba2+(aq) + SO42-(aq)

Ksp = [Ba2+ ][SO42- ]

Ag2SO4 (s) 2Ag+(aq) + SO42-(aq)

PbCrO4(s) Pb2+(aq) + CrO42-(aq)

Ba(OH)2(s) Ba2+(aq) + 2OH-(aq)

Ksp = [Ag+ ]2[SO42- ]

Ksp = [Pb2+ ][CrO42- ]

Ksp = [Ba2+ ][OH -] 2

9-6

The solubility of a salt can be determined from its solubility

product constant. When solid barium sulfate dissolves (to the

extent that it will dissolve) in water, the undissolved solid is in

equilibrium with aqueous barium ions and aqueous sulfate ions.

For barium sulfate, the Ksp expression is

Ksp = [Ba2+ ][SO42- ]

The concentrations of barium ions and sulfate ions are equal if

the dissolution of barium sulfate is the only source of the ions,

and therefore,

The value of Ksp for BaSO4 at 25℃ is 1.1×10–10.

2.2.3 The Solubility and Ksp(溶解度与Ksp的关系)

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The concentration of the salt at equilibrium is its molar

solubility. Thus, the aqueous solubility of barium sulfate at

25℃ is 1.05×10–5 M

Questions

1. What is the solubility of zinc carbonate? (Ksp = 1.4×10–11)

(1) 3.7×10–6 M (2) 7.0×10–12 M

(3) 1.5×10–4 M (4) 1.4×10–11 M

2. The molar solubility of gold(III) chloride is 3.3×10–7 M.

What is its Ksp?

(1) 1.1×10–13 (2) 1.4×10–19

(3) 3.2×10–25 (4) 3.3×10–7

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Example

Determine the molar solubility of BaF2. (Ksp = 1.0×10–6)

When barium fluoride dissolves, it does so via the equation

BaF2(s) Ba2+(aq) + 2F-(aq)

For every mole of barium fluoride that dissolves, one mole of

barium ions and two moles of fluoride ions are produced. We

can call the unknown concentration of barium ions x and the

unknown concentration of fluoride ions 2x.

Ksp = 1.0×10–6 = [Ba2+ ][F- ]2 = (x)(2x)2 = 4x3

[Ba2+] = 0.0063 M

The solubility of barium fluoride BaF2(s) is 0.0063 M.

x = 0.0063 [F-] = 0.013 M

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2.2 Factors of affecting the solubility (影响溶解度的因素)

2.2.1 Common-Ion Effect(同离子效应)

The solubility of a substance can be affected by the presence of

other solutes. For instance, based upon the Common-Ion Effect,

we would expect gold(I) chloride to be less soluble in a 0.10 M

solution of sodium chloride than in pure water. The presence of

more chloride (the common ion) would drive the equilibrium to

the left, corresponding to the dissolution of less AuCl solid.

the product of the gold(I) ion concentration and the chloride ion

concentration must be equal to the Ksp.

Ksp = 2.0×10–13 = [Au+ ][Cl- ]

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determine the solubility of gold(I) chloride in 0.10 M sodium

chloride. In this case, the concentrations of gold(I) ion and

chloride ion are not equal to each other because there is an

independent source of chloride ion (sodium chloride).

2.0×10–13 = [Au+ ][Cl- ] = (x)(0.1+x)

Because the Kspis so small, we can assume that x will be

insignificant compared to 0.10 M. Solving for x then, we get x

= 2×10–12. The solubility of gold(I) chloride in 0.10 M sodium

chloride is

Gold(I) chloride is significantly less soluble in 0.10 M sodium

chloride than it is in pure water. This is what we expected.

= [Au+ ] = [Cl- ] = 4.5 ×10–7 M 13100.2

[Au+ ] = 2×10–12 M

9-11

2.2.2 Acidity or pH(酸度或pH)

Because some ionic solids produce hydroxide ion(OH-) on

dissociation, pH can also affect solubility. As seen in the

following example, tin(II) hydroxide, is less soluble in a basic

solution than in an acidic one.

The solubility of tin(II) hydroxide in water at 25 ℃ is 3.3×10–

10 M (Ksp = 1.4×10–28). What is the solubility of tin(II)

hydroxide (Sn(OH)2 ) in a solution with a pH of 10?

Determine [OH–] from the pH.

[OH- ] = 10- pOH = 1.0×10–4 M

pOH = 14 – pH = 4

Sn(OH)2(s) Sn2+(aq) + 2OH-(aq)

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1.4×10–28 = [Sn2+][OH-]2 = (x)(1.0×10–4 +2x)2

Because Ksp is so small, we can assume that x (and 2x) would

be very small compared to 1.0×10–4. The above expression

then simplifies to

1.4×10–28 = (x)(1.0×10–4)2

The solubility of tin(II) hydroxide in a solution of pH 10 is

1.4×10–20 M.

x = 1.4×10–20

What is the molar solubility of silver chloride (Ksp =

1.8×10–10) in a solution of 0.10 M calcium chloride?

(1) 9.0×10–10 M (2) 1.8×10–9 M

(3) 1.3×10–5 M (4) 4.5×10–9 M

Questions

9-13

2.2.3 Formation of Complex Ions(配离子的生成)

Salts that contain basic anions(指可生成碱性离子的盐) are

also made less soluble by an increased concentration of

hydroxide ion. Fluoride salts, for example, produce the

fluoride ion in solution.

Any additional hydroxide will drive the second equilibrium to

the left, increasing the fluoride ion concentration and driving

the original equilibrium (the dissolution of CaF2) to the left.

Fluoride is a strong conjugate base and will itself contribute to

the hydroxide ion concentration.

9-14

Many metal ions can form complex ions when combined with

the appropriate substance. Copper(II) (Cu2+) ion combines with

cyanide(氰) ions(CN-) to produce Cu(CN)42–.

The equilibrium constant for the formation of this complex ion

(配合离子生成常数), Kf, (the f stands for "formation") is

1×1025. The equilibrium lies far to the right.

Copper(II) carbonate(CuCO3) is not very soluble in water; its

Ksp= 1.4×10–10

If a source of cyanide ion is added to a saturated solution of

copper(II) carbonate, the formation of the Cu(CN)42– ion removes

a product of the dissolution of CuCO3. This draws the

equilibrium to the right, increasing the solubility of CuCO3.

9-15

The sum of the two equilibria is

The equilibrium constant for the net reaction is the product of the

equilibrium constants for the individual steps. Therefore, the Kc

for the dissolution of copper(II) carbonate in solution of aqueous

cyanide ions is

The overall equilibrium lies far to the right.

What is the equilibrium constant for the dissolution of copper(II)

sulfide(CuS) solid (Ksp= 8.5×10–45, Kf = 1.38×1012) in aqueous

ammonia?

(1) 5×1012 (2) 1.2×10–32 (3) 8×1048 (4) 8×10–19

Questions

9-16

2.3 Shift of Precipitation and solubility

Equilibria (沉淀-溶解平衡移动)

The salt precipitates when the solubility product exceeds the

solubility product constant. Because salts have different Ksp values,

they can be selectively(选择) precipitated as a means of separation.

Q＜Ksp，Unsaturated solutions(不饱和溶液)，no

precipitate or precipitation solve

Q has the same form as K sp, . . . but uses existing concentrations(现有的浓度)

Q>K sp，supersaturated solution(过饱和溶液),

precipitation produce in solution

Q=Ksp at EQUILIBRIUM

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2.3.1 Formation of Precipitation (沉淀的生成)

Q>K sp precipitation produce in solution

Example

A solution is 20mL,0.002 M in Na2SO4. Could precipitation

form as 10mL,0.020 M BaCl2 is added to the solution?

(BaSO4 Ksp=1.08×10–10)

Q = [Ba2+] [SO42-] = 0.0067 × 0.0013 = 8.7×10–6

[Ba2+] = 0.020 M ×10mL/(10mL + 20mL) = 0.0067 M

[SO42-] = 0.0020 M ×20mL/(10mL + 20mL) = 0.0013 M

Q = 8.7×10–6 >K sp =1.08×10–10

So, precipitation form in solution

9-18

2.3.2 Fractional Precipitation of Ions(分步沉淀)

If a ion can precipitate a number of other ions, the order of

precipitation will be different because of the difference of Ksp.

A solution contains 0.15 M Hg22+ and 0.35 M Ag+ ions. Which

ion will be precipitated first as sodium chloride is added to the

solution? When Hg22+ begins to precipitate, how much is the

Ag+ ion concentration?

Consider the solubility product expressions of the two salts that

could get the result of chloride ion needed to precipitate two

cation.

Example

Step 1 Ksp(Hg2Cl2) = 1.3×10–18 and Ksp(AgCl) = 1.8×10–10

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Taking into account the concentrations of metal ions in solution,

determine the maximum chloride ion concentration that could

exist in solution without exceeding the Ksp

1.8×10–10 = (0.35)[Cl- ] 1.3×10–18 = (0.15)[Cl- ]2

For the precipitation of Hg2Cl2 to begin, the chloride ion

concentration must exceed 2.9×10–9 M. For the precipitation

of AgCl to begin, the chloride ion concentration must exceed

5.1×10–10 M.

Conclusion: AgCl will be precipitated first.

[Cl- ] = 2.9×10–9 M [Cl- ] = 5.1×10–10 M

Ksp(Hg2Cl2) = 1.3×10–18 Ksp(AgCl) = 1.8×10–10

9-20

A solution is 0.10 M in Ag+ and 0.10 M in Au3+. Which one will

precipitate first as sodium chloride is added?

(AgCl Ksp=1.8×10–10; AuCl3 Ksp= 3.2×10–25)

(1) Ag+ will precipitate first. (2) Au3+ will precipitate first

Questions

Step 2 When Hg22+ begins to precipitate, [Cl- ] = 2.9×10–9 M

1.8×10–10 = [Ag+ ][Cl- ] [Ag+ ] = 6.2×10–2 M

When Hg22+ begins to precipitate, Ag+ is 6.2×10–2. Then

Hg22+ and Ag+ will precipitate together to form co-precipitation

(共沉淀) because Ag+ is not precipitated completely(沉淀完

全) . So we cannot separate the two ions in this consequence.

In general, the ion can be consider precipitating completely

when this ion concentration is lower than 10–5 M or 10–6 M

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2.3.3 Precipitation separation of Ions (沉淀与离子分离)

What is the suitable pH arrange for separation of Mn2+ and Fe3+

at 0.10M of ion’s concentration ?

Mn(OH)2 = Mn2+ + 2OH-

Ksp[Mn(OH)2] = 4.0×10–14

[Mn2+]·[OH-]2 =4.0×10–14

Step 1—Which ion will precipitate first?

Fe(OH)3 = Fe3+ + 3OH-

Ksp[Fe(OH)3] = 1.1×10–36

[Fe3+]·[OH-]3 =1.1×10–36

[OH-] =6.32×10–7 M [OH-] =2.22×10–12 M

Conclusion: Fe3+ will be precipitated first.

pH = 7.80 pH = 2.35

Solution

9-22

Fe(OH)3 = Fe3+ + 3OH-

Ksp[Fe(OH)3] = 1.1×10–36

[Fe3+]·[OH-]3 = 1.0×10-5×[OH-]3 = 1.1×10–36

[OH-] =4.80×10–11 M

Answer: the suitable pH arrange for

separation of Mn2+ and Fe3+ at 0.10M of

ion’s concentration is 3.68 ～7.80 .

pH = 3.68

Step 2— How much of [OH]- is when Fe3+ precipitated

completely (the Fe3+ is 10-5 M at this time) ?

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2.3.4 Multiple equilibrium (多重平衡)

A solution of 0.10 M ZnCl2 saturate with H2S. How much the

H+ ions would be controlled to permit the ZnS to precipitate in

the solution? ( ZnS Ksp= 1.2×10–28 and H2S saturated aqueous

[H2S] = 0.10M)

Example

Solution

(1) Zn2+ + S2- ZnS K1=1/Ksp(ZnS) =1/1.2×10–28

K2=Ka1(H2S) =1.1×10–7 (2) H2S H+ + HS-

(3) HS- H+ + S2- K3=Ka2(H2S) =1.0×10–14

Total reaction Zn2+ + H2S ZnS(s) + 2H+

(1)+(2)+(3) Zn2+ + H2S ZnS(s) + 2H+

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Ktotal = 92 = [H+]2 /([Zn2+ ][H2S]) [H+] = 0.96M

Ktotal = K1· K2 · K3 = Ka1 · Ka2 /Ksp(ZnS) = 92

So, when [H+] < 0.96 M , ZnS would be precipitated in the

solution.

2.3.5 Transformation of precipitation(沉淀的转化)

One precipitation could be transform to another

precipitation when adding a suitable precipitant(沉淀剂).

CaSO4 (s) + CO32- (aq) CaCO3 (s) + SO4

2- (aq)

Ktotal = [SO42- ]/[CO3

2- ] = [SO42- ]·[Ca2+]/[CO3

2- ]·[Ca2+]

= Ksp(CaSO4) /Ksp(CaCO3)

= (2.45×10–5)/(8.7×10–9) = 2.8×103

The greater of Ktotal is, the easier of transformation achieves.

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2.3.6 Solution of precipitation(沉淀的溶解)

Q＜K sp ， precipitation will solve

CaC2O4 Ca2+ + C2O42-

HCl Cl- + H+

HC2O4- + H+ H2C2O4

+

Ka2

Ka1

Ksp Sample

The way generally is to add suitable ions which can react

with ions of precipitation and form weak electrolyte, e.x.

weak acid, weak base and complex compound etc.

9-26

Example

What the lowest concentration of 1.0 L HCl is need if that can

dissolve 0.1 mol MnS or 0.1 mol CuS? (MnS Ksp= 1.4×10–15,

CuS Ksp= 8.5×10–45 )

(1) MnS Mn2+ + S2- K1 = Ksp(MnS) = 1.4×10–15

K3 = 1/Ka1(H2S) = 1/(1.1×10–7) (3) H+ + HS- H2S

(2) H+ + S2- HS- K2 = 1/Ka2(H2S) = 1/(1.0×10–14)

Ktotal = [Mn2+ ][H2S] / [H+]2 [H+] = 8.8×10–5 M

Ktotal = K1· K2 · K3 = Ksp(MnS) / (Ka1 · Ka2) = 1.3×106

Total reaction MnS(S) + 2H+ Mn2+ + H2S

(1)+(2)+(3) MnS(s) + 2H+ Mn2+ + H2S

The total [H+] = 0.20 + 8.8×10–5 ≈ 0.20 M

9-27

Calculate again for the 0.1mol CuS, you will get,

[H+] = 3.6×1010 M ------(DIY)

Ksp(CuS) is so small that it cannot be dissolved in

HCl. (The highest concentration of HCl is 12 M .)

We can not dissolve the CuS by using HCl and

can separate MnS and CuS in this way.

9-28

Precipitation titration: A titration in which the analyte and

titrant form an insoluble precipitate.

To be applied in the precipitation titration, the precipitation

reactions must have the following characteristics:

(1) The precipitation form must have invariable composition and

little solubility;

(2) The precipitation reaction must be fast and quantitatively

completed;

(3) There are suitable indicators or other methods to indicate

the end point of titration.

2.4 Application of precipitation reaction

沉淀反应在分析化学中的应用

9-29

沉淀容量分析（银量法）

1. 银量法的滴定曲线

银量法中, 随着滴定的进行, 溶液中Ag+或X-离子的浓度不

断发生变化, 即pAg或pX不断发生变化, 在化学计量点前后,

pAg或pX会出现突跃。如果以滴定百分数或滴定剂体积为横

坐标, 以pAg或pX为纵坐标作图, 所得曲线即为沉淀滴定的

滴定曲线。

In a argentometric titration, as the titration carrying on,

the concentration of Ag+ and X

- is changing. Near the

equivalence point, the changing is rapidly and there is a

jump.

9-30

Calculating the Titration Curve

Now let’s calculate the titration curve for the titration of

20.00 mL of 0.1000 M Cl– with 0.1000 M Ag+. The

equilibrium constant for the reaction is:

θ θ -1 -10 -1 9

sp( ) =(1.77 10 ) =5.6 10K K

c(Cl-)= 0.1000 (20.00-19.98)/(20.00+19.98)

= 5.010-5 mol·L-1

1. Before the equivalence point, titration percent is 99.9%,

RE=-0.10%:

6

5

10

θ

θ

spθ 105.3100.5

1077.1

/)(Cl

(AgCl))/(Ag

cc

Kcc

pAg=5.45 pCl=4.30

9-31

2. At the equivalence point, RE= 0% (sp) :

5θ

sp

θθ 103.1(AgCl))/(Cl/)(Ag Kcccc

pAg=4.30 pCl=5.45

pAg=4.88=pCl

3. After the equivalence point, titration percent is 100.1%,

RE=+0.10%:

c(Ag+)= 0.1000 (20.02-20.00)/(20.00+20.02)

= 5.010-5 mol·L-1

9-32

Data for Titration of 20.00 mL of 0.1000 M Cl- with

0.1000 M Ag+

Volume (mL) of AgNO3 pAg pCl

0.00 8.75 1.00

18.00 7.47 2.28

19.98 5.45 4.30

20.00 4.88 4.88

20.02 4.30 5.45

22.00 2.28 7.47

40.00 1.47 8.28

jump range

9-33

突跃范围 pAg(sp)

Cl- 5.45-4.30 4.88

Br- 7.97-4.30 6.14

I- 11.77-4.30 8.04

0.1mol·L-1的硝酸银溶液滴定0.1mol·L-1的Cl-, Br-, I-

影响沉淀滴定突跃的因素:

滴定突跃，

有关：和与

SP

SP

Kc

Kc

X

X

由右图可见,沉淀滴定的滴定

曲线与强酸强碱互相滴定的滴定曲线非常相似。

9-34

2. Application 应用

直接银量法测定Cl-和Br-

间接银量法测定Ag+