Ionic Equilibria III: The Solubility Product Principle

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2020Ionic Equilibria III:Ionic Equilibria III:

The Solubility The Solubility Product PrincipleProduct Principle

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Chapter GoalsChapter Goals• Solubility Product Constants溶解度積常數• Determination of Solubility Product

Constants 決定溶解度積常數• Uses of Solubility Product Constants• Fractional Precipitation分級沉澱• Simultaneous Equilibria Involving Slightly

Soluble Compounds• Dissolving Precipitates

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Insoluble

soluble

Aqueous Solutions: An Aqueous Solutions: An IntroductionIntroduction

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Solubility Product Solubility Product ConstantsConstants

• Silver chloride, AgCl, is rather insoluble in water.

• Careful experiments show that if solid AgCl is placed in pure water and vigorously stirred, a small amount of the AgCl dissolves in the water.

Ag+Cl- Ag+(aq) + Cl-

(aq)

• The equilibrium constant expression for this dissolution is called a solubility product constant 溶解度積常數 .– Ksp = solubility product constantKsp = [Ag+][Cl-] = 1.8x10-10

當溶液解離達平衡後，反應物以及生成物離子的濃度皆不改變，此離子濃度會達成一個固定的比例即稱溶解度積常數

Appendix H

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• Consider the dissolution of silver sulfide in water.

Ag2 S(aq)

2Ag+ + S2-

H2O

100%

• The solubility product expression for Ag2S is:Ksp = [Ag+]2[S2-] = 1.0x10-49

Ksp的數值愈大，表示溶液中離子的濃度也愈大；也就是水溶解化合物時，化合物在水中分解成離子的數量也就愈多。

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• The dissolution of solid calcium phosphate in water is represented as:

The solubility product constant expression is:

Ca22+(PO4

3-)2(s) 2Ca2+

+ 2PO43-

H2O

100%

Ksp = [Ca2+]3[PO43-]2 = 1.0x10-25

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• In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:

MrYs(s) rMs+

+ sYr-

H2O

100%

Ksp = [Ms+]r[Yr-]s

溶解度的另一表示方法 :•The solubility of a compound molar solubility莫耳溶解度The number of moles that dissolve to one liter of saturated solution (M) 每公升飽和水溶液中所含溶質的莫耳數

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• The same rules apply for compounds that have more than two kinds of ions.

• One example of a compound that has more than two kinds of ions is calcium ammonium phosphate.

CaNH4PO4(s) Ca2+

(aq) + NH4+

(aq)+PO43-

(aq)

H2O

100%

Ksp = [Ca2+][NH4+][PO4

3-]

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Determination of Determination of Solubility Product Solubility Product

ConstantsConstantsExample 20-1: One liter of saturated silver chloride

solution contains 0.00192 g of dissolved AgCl at 25oC. Calculate the molar solubility of, and Ksp for, AgCl.

The molar solubility can be easily calculated from the data:

? M AgCl= 1.0L

0.00192g AgCl143g AgCl

= 1.34x10-5 M

AgCl(s) Ag+ + Cl-

1.34x10-5 M 1.34x10-5 M

Ksp = [Ag+][Cl-]= (1.34x10-5) x (1.34x10-5)= 1.8x10-10 M

1.34x10-5 M

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Determination of Determination of Solubility Product Solubility Product

ConstantsConstantsExample 20-2: One liter of saturated calcium fluoride

solution contains 0.0167 gram of CaF2 at 25oC. Calculate the molar solubility of, and Ksp for, CaF2.

? M CaF2= 1.0L

0.0167g CaF278.1g = 2.14x10-4 M

CaF2 Ca2+ + 2F-

2.14x10-4 2x(2.14x10-4)

Ksp = [Ca2+][F-]2= (2.14x10-4) x (4.28x10-4)2

= 3.92x10-11 mol/L

2.14x10-4 M

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Uses of Solubility Uses of Solubility Product ConstantsProduct Constants

Example 20-3: Calculate the molar solubility of barium sulfate, BaSO4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25oC. For barium sulfate, Ksp= 1.1 x 10-10.

BaSO4 Ba2+ + SO4

2-

x M x M Ksp = [Ba2+][SO4

2-]= 1.1x10-10

x= 1.0x10-5 M

x M

(x)(x) = 1.1x10-10

[Ba2+]=[SO42-]= 1.0x10-5 M

1.0x10-5M x 1L = ?g BaSO4= 1.0x10-5 mol = 1.0x10-5 x 234

= 2.34x10-3 g in 1L

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Uses of Solubility Uses of Solubility Product ConstantsProduct Constants

Example 20-4: The solubility product constant for magnesium hydroxide, Mg(OH)2, is 1.5 x 10-11. Calculate the molar solubility of magnesium hydroxide and the pH of a saturated magnesium hydroxide solution at 25oC.Mg(OH)2 Mg2+

+ 2 OH-

x M 2x M Ksp = [Mg2+][OH-]2= 1.5x10-11

x= 1.6x10-4 M

x M

(x)(2x)2 = 1.5x10-11

[OH-]= 2x (1.6x10-4) =3.2x10-4 M

pOH=3.49, pH =10.51

4x3= 1.5x10-11 M x3= 3.75x10-12 M

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The Common Ion Effect The Common Ion Effect in Solubility Calculationsin Solubility Calculations

Example 20-5: Calculate the molar solubility of barium sulfate, BaSO4, in 0.010 M sodium sulfate, Na2SO4, solution at 25oC. Compare this to the solubility of BaSO4 in pure water. (Example 20-3). (What is the common ion? How was a common ion problem solved in Chapter 19?)Na2SO4 2 Na+

+ SO42-

2x(0.010)M 0.010 M 0.010 M

BaSO4 Ba2+ + SO4

2-

x M x M x M

Ksp = [Ba2+][SO42-]= 1.1x10-10= (x)(0.01+x)

0.01+x 0.01

(0.01)x = 1.1x10-10 x = 1.1x10-8Molar solubility of BaSO4

Soluble ionic salt完全解離 Slightly Soluble salt溶解度很低

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The Common Ion Effect The Common Ion Effect in Solubility Calculationsin Solubility Calculations• The molar solubility of BaSO4 in 0.010 M Na2SO4 solution is 1.1 x 10-8 M.• The molar solubility of BaSO4 in pure water is 1.0 x 10-5 M.

– BaSO4 is 900 times more soluble in pure water than in 0.010 M sodium sulfate!

– Adding sodium sulfate to a solution is a fantastic method to remove Ba2+ ions from solution!

• If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.

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The Reaction Quotient in The Reaction Quotient in Precipitation ReactionsPrecipitation Reactions

The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form.

If Qsp< Ksp Forward process is favored ( 反應向右 ) No precipitation occurs; if solid is

present, more solid can dissolve ( 不會產生沉澱 )

If Qsp= Ksp Solution is just saturated ( 正好飽和 ) Solid and solution are in

equilibrium; neither forward nor reverse process is favored

If Qsp> Ksp Reverse process is favored ( 反應向左 ) precipitation occurs to form more

solid ( 產生沉澱 )

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Example 20-6: We mix 100 mL of 0.010 M potassium sulfate, K2SO4, and 100 mL of 0.10 M lead (II) nitrate, Pb(NO3)2 solutions. Will a precipitate form?

K2SO4 2 K+ + SO4

2-

Pb(NO3)2 Pb2+ + 2 NO3

-Will PbSO4 precipitate??

MPb(NO3)2= 100mL x 0.10M100mL+100mL

=0.050 M Pb(NO3)2

MK2SO4= 100mL x 0.10M100mL+100mL=0.050 M K2SO4

Qsp = [Pb2+][SO42-]= (0.050)(0.050)= 2.5x10-4

Ksp= 1.8x10-8 for PbSO4

Qsp> Ksp therefore solid forms

PbSO4 Pb2++ SO42-

0.05M 0.05x2M 0.05M

0.05M 0.05M 0.05x2M

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Example 20-7: Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO3)2. We can do this by precipitating mercury (II) ions as the insoluble compound HgS. What concentration of sulfide ions, from a soluble compound such as Na2S, is required to reduce the Hg2+ concentration to 1.0 x 10-8 M? For HgS, Ksp=3.0 x 10-53.

Ksp = [Hg2+][S2-]= 3.0x10-53

HgS Hg2++ S2-

= [S2-]x(1.0x10-8) [S2-] = 3.0x10-45

If enough S2-, in th form of Na2S, is added to just slightly exceed 3.0x10-45 M the mercury will precipitate


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Example 20-8: Refer to example 20-7. What volume of the solution (1.0 x 10-8 M Hg2+ ) contains 1.0 g of mercury?

?L =1.0g Hg2+ x 1 mol Hg2+

201g Hg2+ x 1.0 L1.0x10-8 mol Hg2+

?L =5.0x105 L

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Fractional PrecipitationFractional Precipitation• The method of precipitating some ions from

solution while leaving others in solution is called fractional precipitation 分級沉澱 .– If a solution contains Cu+, Ag+, and Au+, each ion

can be precipitated as chlorides.

Ksp = [Cu+][Cl-]=1.9x10-7CuCl Cu++ Cl-

Ksp = [Ag+][Cl-]=1.8x10-10AgCl Ag++ Cl-

Ksp = [Au+][Cl-]=2.0x10-13AuCl Au++ Cl-

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Example 20-9: If solid sodium chloride is slowly added to a solution that is 0.010 M each in Cu+, Ag+, and Au+ ions, which compound precipitates first? Calculate the concentration of Cl- required to initiate precipitation of each of these metal chlorides.

AuCl has the small Ksp, so it precipitates firstBy the same reasoning, CuCl will precipitate last

•Calculate the concentration of Cl- required to precipate AuCl

Ksp = [Au+][Cl-]=2.0x10-13= (0.010)x[Cl-][Cl-]=2.0x10-11

•Repeat the calculation for silver chloride.Ksp = [Ag+][Cl-]=1.8x10-10= (0.010)x[Cl-]

[Cl-]=1.8x10-8

•Finally, for copper (I) chloride to precipitate.Ksp = [Cu+][Cl-]=1.9x10-7= (0.010)x[Cl-]

[Cl-]=1.9x10-5

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Fractional PrecipitationFractional Precipitation• These three calculations give the [Cl-]

required to – precipitate AuCl ([Cl-] >2.0 x 10-11 M), – precipitate AgCl ([Cl-] >1.8 x 10-8 M),– precipitate CuCl ([Cl-] >1.9 x 10-5 M).

• It is also possible to calculate the amount of Au+ precipitated before the Ag+ begins to precipitate, as well as the amounts of Au+ and Ag+ precipitated before the Cu+ begins to precipitate.

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Example 20-10: Calculate the percentage of Au+ ions that precipitate before AgCl begins to precipitate.– Use the [Cl-] from Example 20-9 to determine the

[Au+] remaining in solution just before AgCl begins to precipitate.

Ksp = [Au+][Cl-]=2.0x10-13

[Au+]x(1.8x10-8)=2.0x10-13

[Au+]=1.1x10-5 unprecipitated

• The percent of Au+ ions unprecipitated just before AuCl precipitates is:

• Therefore, 99.9% of the Au+ ions precipitates before AgCl begins to precipitate.

% Au+unprecipitated=

[Au+]unprecipitated

[Au+]originalx100%

= 1.1x10-5

0.01 x100%

=0.1% unprecipitated

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Fractional PrecipitationFractional Precipitation• A similar calculation for the concentration of Ag+ ions

unprecipitated before CuCl begins to precipitate is:

Ksp = [Ag+][Cl-]=1.8x10-10

[Ag+]x(1.8x10-8)=1.9x10-5

[Ag+]=9.5x10-6 unprecipitated

• The percent of Ag+ ions unprecipitated just before AgCl precipitates is:

• Therefore, 99.905% of the Ag+ ions precipitates before CuCl begins to precipitate.

% Ag+unprecipitated=

[Ag+]unprecipitated

[Ag+]originalx100%

= 9.5x10-6

0.01 x100%

=0.095% unprecipitated

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Simultaneous Equilibria Simultaneous Equilibria Involving Slightly Soluble Involving Slightly Soluble

CompoundsCompounds

• Many weak acids and bases react with many metal ions to form insoluble compounds.許多弱酸和弱鹼可以和金屬反應形成不可溶的化合物– The weak acid or weak base

equilibrium as well as the solubility equilibrium

• Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

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CompoundsCompoundsExample 20-12: If 0.10 mole of ammonia and 0.010 mole of

magnesium nitrate, Mg(NO3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? – For Mg(OH)2, Ksp = 1.5 x 10-11. Kb for NH3 = 1.8 x 10-5.

Calculate Qsp for Mg(OH)2 and compare it to Ksp.–Mg(NO3)2 is a soluble ionic compound so [Mg2+] = 0.010 M.–Aqueous ammonia is a weak base that we can calculate

[OH-]. NH3 + H2O NH4++ OH-

0.1M-x +x +x

Kb=[NH4

+] [OH-][NH3]

= 1.8x10-5=(x)(x)

(0.10-x)

x2=1.8x10-6

x=1.3x10-3

=[OH-]

Qsp = [Mg2+][OH-]2= (0.010)(1.3x10-3)2= 1.7x10-8

Qsp> Ksp thus Mg(OH)2 will precipitate

Mg(NO3) 2 Mg2++ 2NO3-可溶性鹽類

0.01M +0.01M+0.01x2M

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Example 20-13: How many moles of solid ammonium chloride, NH4Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0.10 M in aqueous ammonia and 0.010 M in magnesium nitrate, Mg(NO3)2 ? (Note the similarity between this problem and Example 20-12.)

Calculate the maximum [OH-] that can exist in a solution that is 0.010 M in Mg2+.

Ksp = [Mg2+][OH-]2=1.5x10-11

[OH-]2x(0.010)=1.5x10-11

[OH-]2=1.5x10-9 [OH-] 3.9x10-5 M

Using the maximum [OH-] that can exist in solution, determine the number of moles of NH4Cl required to buffer 0.10 M aqueous ammonia so that the [OH-] does not exceed 3.9 x 10-5 M.NH4Cl → NH4

++ Cl-

NH3 + H2O NH4++ OH-

xM xM

(0.1M) (3.9x10-5)

[OH-]= [base][salt] =1.8x10-5 (x)

(0.10)x

x= 0.046 =[NH4Cl]

xM

Because there is 1.0L of solutionThere are 0.046 mol NH4Cl

Kb x

Buffer solution

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CompoundsCompounds• Check these values by calculating Qsp for

Mg(OH)2.Qsp = [Mg2+][OH-]2= (0.010)(3.9x10-5)2= 1.5x10-11

Qsp = KspThus this system is at equilibrium

• Use the ion product for water to calculate the [H+] and the pH of the solution.

[H+][OH-] =1.0x10-14

[H+]x(3.9x10-5)=1.0x10-14

[H+] =2.6x10-10

pH=9.59

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Dissolving PrecipitatesDissolving Precipitates• For an insoluble solid, if the ion

concentrations (of either the cation or anion) are decreased, a solid precipitate can be dissolved.(在不可溶固體中 ,若任一離子濃度降低 ,則此沉澱的固體會被再溶解 )– The trick is to make Qsp < Ksp.

• One method is to convert the ions into weak electrolytes.採用的方法乃將離子轉成弱電解質– Make these ions more water soluble. 讓此離子更水溶

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Dissolving PrecipitatesDissolving Precipitates• If insoluble metal hydroxides are

dissolved in strong acids, they form soluble salts and water. ( 不可溶金屬氫氧化合物可溶於強酸而形成可溶性鹽類及水 )

• For example, look at the dissolution of Mg(OH)2 in HCl.Mg(OH)2(s) + 2 HCl(aq) MgCl2(aq) + 2H2O(l)

orMg(OH)2(s) + 2 H+

(aq) Mg2+(aq) + 2H2O(l)

Notice that the insoluble Mg(OH)2 is converted to the more soluble MgCl2

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Dissolving PrecipitatesDissolving Precipitates• A second method is to dissolve

insoluble metal carbonates in strong acids.(另一方法是將不可溶金屬碳酸化合物溶於強酸中 )– The carbonates will form soluble salts,

carbon dioxide, and water.CaCO3(s) + 2 HCl(aq) CaCl2(aq) + CO2(g) + 2H2O(l)

CaCO3(s) + 2 H+(aq) Ca2+

(aq) + CO2(g) + 2H2O(l)

or

Notice that the formation of carbon dioxide removes ions and makes the carbonate dissolve

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Dissolving PrecipitatesDissolving Precipitates• A third method is to convert an ion to another

species by an oxidation-reduction reaction.(第三種方法乃利用氧化還原反應改變離子態 )

• For example, the dissolution of insoluble metal sulfides in hot nitric acid causes the sulfide ions to be oxidized to elemental sulfur.

3PbS(s) + 8H+ + 2NO3

- 3Pb2+ + 3S0(s) +2NO(g) + 4H2O(l)

[Pb2+][S2-]=8.4x10-28=Ksp

The NO3- oxidizes S2- to S0 making the Qsp<Ksp

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Dissolving PrecipitatesDissolving Precipitates• A fourth method is complex ion formation.(第四種方法是形成錯離子 )

• The cations in many slightly soluble compounds will form complex ions.– This is the method used to dissolve unreacted

AgBr and AgCl on photographic film.• Photographic “hypo” is Na2S2O3.

2AgBr(s) 2Ag(s) + Br2(l) h

In the unexposed portion of the film AgBr(s) is leftAgBr(s) + 2 Na2S3O2(aq) Na3Ag(S2O3)2 (aq) + NaBr(aq)

AgBr(s) + 2 Na2S32-

[Ag(S2O3)2]3-(aq) + Br-

(aq)

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Dissolving PrecipitatesDissolving Precipitates• Copper(II) hydroxide, which is light blue

colored, dissolves in aqueous ammonia to form dark blue, [Cu(NH3)4]2+.

Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH-

Cu(OH)2(s) Cu2+(aq) + 2OH-

(aq)

Cu2+(aq) + 4NH3 [Cu(NH3)4]2+

(aq)

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Complex Ion EquilibriaComplex Ion Equilibria• A metal ion coordinated to several

neutral molecules or anions forms compounds called complex ions.

• Familiar examples of complex ions include:

[Cu(NH3)4]2+[Ag(NH3)2]+

[Pt(NH3)2]+Co(NH3)2Cl2

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Complex Ion EquilibriaComplex Ion Equilibria• The dissociation of complex ions can

be represented similarly to equilibria.• For example:

[Cu(NH3)4]2+ Cu2+ + 4NH3

Kd=[Cu2+] [NH3]4

[[Cu(NH3)4]2+]

[Ag(NH3)2]+ Ag+ + 2NH3

Kd=[Ag+] [NH3]2

[[Ag(NH3)2]+]

Complex ion equilibrium constants are called dissociation constants Kd.

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Complex Ion EquilibriaComplex Ion EquilibriaExample 20-14: Calculate the concentration of silver

ions in a solution that is 0.010 M in [Ag(NH3)2]+. Kd = 6.3 x 10-8

Write the dissociation reaction and equilibrium concentrations.

Substitute the algebraic quantities into the dissociation expression.

[Ag(NH3)2]+ Ag+ + 2NH3

Kd=[Ag+] [NH3]2

[[Ag(NH3)2]+]

(0.01-x)M (x)M (2x)M

= 6.3x10-8 =(x)(2x)2

(0.010-x)

4x3= 6.3x10-10 x3= 1.6x10-10 x= 5.4x10-4 M

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Example 20-15: How many moles of ammonia must be added to 2.00 L of water so that it will just dissolve 0.010 mole of silver chloride, AgCl?

AgCl(s) + 2NH3 [Ag(NH3)2]++ Cl-? M AgCl = 0.01mol/2.0L = 5x10-3 M

2(5x10-3) M5x10-3 M 5x10-3 M 5x10-3 M

[Ag(NH3)2]+ Ag+ + 2NH3 Kd=[Ag+] [NH3]2

[[Ag(NH3)2]+]

AgCl Ag+ + Cl- Ksp=[Ag+][ Cl-] =1.8x10-10

=6.3x10-10

The [Ag+] in the solution must satisfy both equilibrium constant expressions. Because the [Cl-] is known, the equilibrium concentration of Ag+ can be calculated from Ksp for AgCl.

Ksp=[Ag+][ Cl-] =1.8x10-10=[Ag+](5x10-3)

[Ag+] =3.6x10-8Which is the maximum [Ag+] possible

Kd=[Ag+] [NH3]2

[[Ag(NH3)2]+]=6.3x10-10 =

(3.6x10-8)[NH3]2

(5x10-3)[NH3]2= 8.75x10-3 [NH3]= 0.094 M

[NH3]total= 0.094 M + 2x(5.0x10-3) =0.104M? mol NH3= 2.0L x 0.104M = 0.21

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Ionic Equilibria III: The Solubility Product Principle