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AQUEOUS EQUILIBRIA IIIUnit 5

• SOLUBILITY CONSTANTS• COMPLEX IONS• NET IONIC EQUATIONS• Chapter 4.3 & 19.3-19.4 (Silberberg)

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Chapter 19

Ionic Equilibria in Aqueous Systems

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Ionic Equilibria in Aqueous Systems

19.3 Equilibria of Slightly Soluble Ionic Compounds

19.4 Equilibria Involving Complex Ions

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Goals & Objectives

• See the following Learning Objectives on page 870.

• Understand These Concepts:• 19.9-17

• Master these Skills:• 19.7-11, 13-16.

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Equilibria of Slightly Soluble Ionic Compounds

Any “insoluble” ionic compound is actually slightly soluble in aqueous solution.We assume that the very small amount of such a compound that dissolves will dissociate completely.

For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions.

PbF2(s) Pb2+(aq) + 2F-(aq)

Qc =[Pb2+][F-]2

[PbF2]Qsp = Qc[PbF2] = [Pb2+][F-]2

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Solubility Product Principle

• Although all compounds dissolve in water to some extent, some compounds are classified as “insoluble”. The solubility of these insoluble compounds can be described in terms of their solubility product constant, Ksp.

• Memorize the Solubility generalization handout.

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Solubility Product Principle

• Consider AgCl dissolved in water• AgCl(s) = Ag+ + Cl-

• Ksp = [Ag+][Cl-]

• Consider Ag2S dissolved in water

• Ag2S(s) = 2Ag+ + S2-

• Ksp = [Ag+]2[S2-]

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Solubility Product Principle

• Consider Ca3(PO4)2 dissolved in water

• Ca3(PO4)2 = 3Ca2+ + 2PO43-

• Ksp = [Ca2+]3[PO43-]2

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Qsp and Ksp

Qsp is called the ion-product expression for a slightly soluble ionic compound.

For any slightly soluble compound MpXq, which consists of ions Mn+ and Xz-,

Qsp = [Mn+]p[Xz-]q

When the solution is saturated, the system is at equilibrium, and Qsp = Ksp, the solubility product constant.

The Ksp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation).

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Metal Sulfides

Metal sulfides behave differently from most other slightly soluble ionic compounds, since the S2- ion is strongly basic.

We can think of the dissolution of a metal sulfide as a two-step process:

MnS(s) Mn2+(aq) + S2-(aq)

S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)

MnS(s) + H2O(l) Mn2+(aq) + HS-(aq) + OH-(aq)

Ksp = [Mn2+][HS-][OH-]

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Sample Problem 19.5 Writing Ion-Product Expressions

SOLUTION:

PROBLEM: Write the ion-product expression at equilibrium for each compound:(a) magnesium carbonate (b) iron(II) hydroxide(c) calcium phosphate (d) silver sulfide

PLAN: We write an equation for a saturated solution of each compound, and then write the ion-product expression at equilibrium, Ksp. Note the sulfide in part (d).

(a) MgCO3(s) Mg2+(aq) + CO32-(aq) Ksp = [Mg2+][CO3

2-]

(b) Fe(OH)2(s) Fe2+(aq) + 2OH-(aq) Ksp = [Fe2+][OH-]2

(c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO4

3-]2

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Sample Problem 19.5

(d) Ag2S(s) 2Ag+(aq) + S2-(aq) S2-(aq) + H2O(l) → HS-(aq) + OH-(aq)

Ksp = [Ag+]2[HS-][OH-]

Ag2S(s) + H2O(l) 2Ag+(aq) + HS-(aq) + OH-(aq)

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Determination of Ksp

• One liter of saturated silver chloride solution contains 0.00192g of dissolved AgCl at 25oC. Calculate the molar solubility of , and Ksp for, AgCl.

• The molar solubility of CaF2 is 2.14x10-4 at 25oC. Determine Ksp for CaF2.

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Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic

Compounds at 25°C

Name, Formula Ksp

Aluminum hydroxide, Al(OH)3

Cobalt(II) carbonate, CoCO3

Iron(II) hydroxide, Fe(OH)2

Lead(II) fluoride, PbF2

Lead(II) sulfate, PbSO4

Silver sulfide, Ag2S

Zinc iodate, Zn(IO3)2

3x10-34

1.0x10-10

4.1x10-15

3.6x10-8

1.6x10-8

4.7x10-29

8x10-48

Mercury(I) iodide, Hg2I2

3.9x10-6

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Sample Problem 19.6 Determining Ksp from Solubility

PROBLEM: (a) Lead(II) sulfate (PbSO4) is a key component in lead-acid car batteries. Its solubility in water at 25°C is 4.25x10-3 g/100 mL solution. What is the Ksp of PbSO4?

(b) When lead(II) fluoride (PbF2) is shaken with pure water at 25°C, the solubility is found to be 0.64 g/L. Calculate the Ksp of PbF2.

PLAN: We write the dissolution equation and the ion-product expression for each compound. This tells us the number of moles of each ion formed. We use the molar mass to convert the solubility of the compound to molar solubility (molarity), then use it to find the molarity of each ion, which we can substitute into the Ksp expression.

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Sample Problem 19.6

Ksp = [Pb2+][SO42-]

Ksp = [Pb2+][SO42-] = (1.40x10-4)2

SOLUTION:

= 1.96x10-8

(a) PbSO4(s) Pb2+(aq) + SO42-(aq)

Converting from g/mL to mol/L:

4.25x10-3g PbSO4

100 mL solnx 1000 mL

1 Lx 1 mol PbSO4

303.3 g PbSO4 = 1.40x10-4 M PbSO4

Each mol of PbSO4 produces 1 mol of Pb2+ and 1 mol of SO42-, so

[Pb2+] = [SO42-] = 1.40x10-4 M

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Sample Problem 19.6

Ksp = [Pb2+][F-]2

Ksp = [Pb2+][F-]2 = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8

(b) PbF2(s) Pb2+(aq) + F-(aq)

Converting from g/L to mol/L:

0.64 g PbF2

1 L solnx 1 mol PbF2

245.2 g PbF2 = 2.6x10-3 M PbF2

Each mol of PbF2 produces 1 mol of Pb2+ and 2 mol of F-, so [Pb2+] = 2.6x10-3 M and [F-] = 2(2.6x10-3) = 5.2x10-3 M

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Sample Problem 19.7 Determining Solubility from Ksp

PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a strong, inexpensive base. Calculate the molar solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.

PLAN: We write the dissolution equation and the expression for Ksp. We know the value of Ksp, so we set up a reaction table that expresses [Ca2+] and [OH-] in terms of S, the molar solubility. We then substitute these expressions into the Ksp expression and solve for S.

SOLUTION:

Ksp = [Ca2+][OH-]2 = 6.5x10-6 Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)

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Sample Problem 19.7

-Initial 0 0

Change - +S + 2S

Ksp = [Ca2+][OH-]2 = (S)(2S)2 = 4S3 = 6.5x10-6

= 1.2x10-2 M

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (M)

Equilibrium - S 2S

S = =3 Ksp4 3 6.5x10-6

4

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Uses of Ksp

• Determine the molar solubility of BaSO4 in pure water at 25oC. Ksp=1.1x10-10

• Calculate the molar solubility and pH of a saturated solution of Mg(OH)2 at 25oC. Ksp= 1.5x10-11

• Determine the molar solubility of BaSO4 in 0.010M sodium sulfate solution at 25oC. Ksp= 1.1x10-10

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Table 19.3 Relationship Between Ksp and Solubility at 25°C

No. of Ions Formula Cation/Anion Ksp Solubility (M)

2 MgCO3 1/1 3.5x10-8 1.9x10-4

2 PbSO4 1/1 1.6x10-8 1.3x10-4

2 BaCrO4 1/1 2.1x10-10 1.4x10-5

3 Ca(OH)2 1/2 6.5x10-6 1.2x10-2

3 BaF2 1/2 1.5x10-6 7.2x10-3

3 CaF2 1/2 3.2x10-11 2.0x10-4

3 Ag2CrO4 2/1 2.6x10-12 8.7x10-5

The higher the Ksp value, the greater the solubility, as long as we compare compounds that have the same total number of ions in their formulas.

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Figure 19.12 The effect of a common ion on solubility.

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

If Na2CrO4 solution is added to a saturated solution of PbCrO4, it provides the common ion CrO4

2-, causing the equilibrium to shift to the left. Solubility decreases and solid PbCrO4 precipitates.

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Sample Problem 19.8 Calculating the Effect of a Common Ion on Solubility

PROBLEM: In Sample Problem 19.7, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10 M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.

PLAN: The addition of Ca2+, an ion common to both solutions, should lower the solubility of Ca(OH)2. We write the equation and Ksp expression for the dissolution and set up a reaction table in terms of S, the molar solubility of Ca(OH)2. We make the assumption that S is small relative to [Ca2+]init because Ksp is low. We can then solve for S and check the assumption.

SOLUTION:

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2

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Sample Problem 19.8

Change - +S + 2S

Ksp = [Ca2+][OH-]2 = 6.5x10-6 ≈ (0.10)(2S)2 = (0.10)(4S2)

= 4.0x10-3 M

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (M)

Equilibrium - 0.10 + S 2S

[Ca2+]init = 0.10 M because Ca(NO3)2 is a soluble salt, and dissociates completely in solution.

4S2 ≈ 6.5x10-6

0.10so S ≈6.5x10-5

4

Checking the assumption: 4.0x10-3 M

0.10 Mx 100 = 4.0% < 5%

-Initial 0.10 0

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Effect of pH on Solubility

Changes in pH affects the solubility of many slightly soluble ionic compounds.

The addition of H3O+ will increase the solubility of a salt that contains the anion of a weak acid.

CaCO3(s) Ca2+(aq) + CO32-(aq)

CO32-(aq) + H3O+(aq) → HCO3

-(aq) + H2O(l)

HCO3-(aq) + H3O+(aq) → [H2CO3(aq)] + H2O(l) → CO2(g) + 2H2O(l)

The net effect of adding H3O+ to CaCO3 is the removal of CO3

2- ions, which causes an equilibrium shift to the right. More CaCO3 will dissolve.

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Sample Problem 19.9 Predicting the Effect on Solubility of Adding Strong Acid

PROBLEM: Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of each ionic compound:

PLAN: We write the balanced dissolution equation for each compound and note the anion. The anion of a weak acid reacts with H3O+, causing an increase in solubility.

SOLUTION:

(a) lead(II) bromide (b) copper(II) hydroxide (c) iron(II) sulfide

(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)

Br- is the anion of HBr, a strong acid, so it does not react with H3O+. The addition of strong acid has no effect on its solubility.

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Sample Problem 19.9

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

OH- is the anion of H2O, a very weak acid, and is in fact a strong base. It will react with H3O+:

The addition of strong acid will cause an increase in solubility.

OH-(aq) + H3O+(aq) → 2H2O(l)

(c) FeS(s) Fe2+(aq) + S2-(aq)

S2- is the anion of HS-, a weak acid, and is a strong base. It will react completely with water to form HS- and OH-. Both these ions will react with added H3O+:

The addition of strong acid will cause an increase in solubility.

HS-(aq) + H3O+(aq) → H2S(aq) + H2O(l)OH-(aq) + H3O+(aq) → 2H2O(l)

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Figure 19.14 Limestone cave in Nerja, Málaga, Spain.

Limestone is mostly CaCO3 (Ksp = 3.3x10-9).

CO2(g) CO2(aq) H2O

CO2(aq) + 2H2O(l) H3O+(aq) + HCO3-(aq)

Ground water rich in CO2 trickles over CaCO3, causing it to dissolve. This gradually carves out a cave.

CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3-(aq)

Water containing HCO3- and Ca2+ ions

drips from the cave ceiling. The air has a lower P than the soil, causing CO2 to come out of solution. A shift in equilibrium results in the precipitation of CaCO3 to form stalagmites and stalactites.

CO2

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Predicting the Formation of a Precipitate

For a saturated solution of a slightly soluble ionic salt, Qsp = Ksp.

When two solutions containing the ions of slightly soluble salts are mixed,

If Qsp = Ksp, the solution is saturated and no change will occur.If Qsp > Ksp, a precipitate will form until the remaining solution is saturated.

If Qsp =< Ksp, no precipitate will form because the solution is unsaturated.

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Sample Problem 19.10 Predicting Whether a Precipitate Will Form

PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions containing the component ions. Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

PLAN: First we need to decide which slightly soluble salt could form, look up its Ksp value in Appendix C, and write the dissolution equation and Ksp expression. We find the initial ion concentrations from the given volumes and molarities of the two solutions, calculate the value for Qsp and compare it to Ksp.

SOLUTION:

The ions present are Ca2+, NO3-, Na+, and F-. All Na+ and NO3

- salts are soluble, so the only possible precipitate is CaF2 (Ksp = 3.2x10-11).

CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = [Ca2+][F-]2

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Sample Problem 19.10

Ca(NO3)2 and NaF are soluble, and dissociate completely in solution.

We need to calculate [Ca2+] and [F-] in the final solution.

Amount (mol) of Ca2+ = 0.030 M Ca2+ x 0.100 L = 0.030 mol Ca2+.

[Ca2+]init =0.030 mol Ca2+

0.100 L + 0.200 L= 0.10 M Ca2+

Amount (mol) of F- = 0.060 M F- x 0.200 L = 0.012 mol F-.

[F-]init =0.012 mol F-

0.100 L + 0.200 L= 0.040 M F-

Qsp = [Ca2+]init[F-]2init = (0.10)(0.040)2 = 1.6x10-4

Since Qsp > Ksp, CaF2 will precipitate until Qsp = 3.2x10-11.

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Uses of Ksp

• Predict whether a precipitate of PbSO4 will form in a solution having [Pb2+] = 0.050M and [SO4

2-] = 0.0050M. Ksp = 1.8x10-8

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Selective Precipitation

Selective precipitation is used to separate a solution containing a mixture of ions.

A precipitating ion is added to the solution until the Qsp of the more soluble compound is almost equal to its Ksp.

The less soluble compound will precipitate in as large a quantity as possible, leaving behind the ion of the more soluble compound.

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Sample Problem 19.12 Separating Ions by Selective Precipitation

PROBLEM: A solution consists of 0.20 M MgCl2 and 0.10 M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20.

PLAN: Both compounds have 1/2 ratios of cation/anion, so we can compare their solubilities by comparing their Ksp values. Mg(OH)2 is 1010 times more soluble than Cu(OH)2, so Cu(OH)2 will precipitate first. We write the dissolution equations and Ksp expressions. Using the given cation concentrations, we solve for the [OH-] that gives a saturated solution of Mg(OH)2. Then we calculate the [Cu2+] remaining to see if the separation was successful.

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Sample Problem 19.12

SOLUTION:

Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+][OH-]2 = 6.3x10-10

Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = [Cu2+][OH-]2 = 2.2x10-20

Ksp

[Mg2+][OH-] = = 6.3x10-10

0.20= 5.6x10-5 M

This is the maximum [OH-] that will not precipitate Mg2+ ion.

Calculating the [Cu2+] remaining in solution with this [OH-]

Ksp

[OH-]2[Cu2+] = =

2.2x10-20

(5.6x10-5)2= 7.0x10-12 M

Since the initial [Cu2+] is 0.10 M, virtually all the Cu2+ ion is precipitated.

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Figure B19.1 Formation of acidic precipitation.

Chemical Connections

Since pH affects the solubility of many slightly soluble ionic compounds, acid rain has far-reaching effects on many aspects of our environment.

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Complex Ion Equilibria

• Many complex ions are known to exist. The majority consist of a metal ion with several anions or molecules coordinated to it. The class of compounds is called coordination compounds.

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Figure 19.15 Cr(NH3)63+, a typical complex ion.

A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands.

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Figure 19.16 The stepwise exchange of NH3 for H2O in M(H2O)42+.

The overall formation constant is given by

Kf = [M(NH3)4

2+]

[M(H2O)42+][NH3]4

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Complex Ion Equilibria• Some examples of complex ions:• NH3 complexes

– Ag(NH3)2+

– Cu(NH3)42+

– Zn(NH3)42+

• OH- complexes– Al(OH)4¯

– Cr(OH)4¯

– Zn(OH)42-

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Table 19.4 Formation Constants (Kf) of Some Complex Ions at 25°C

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Sample Problem 19.13 Calculating the Concentration of a Complex Ion

PROBLEM: An industrial chemist converts Zn(H2O)42+ to the more

stable Zn(NH3)42+ by mixing 50.0 L of 0.0020 M

Zn(H2O)42+ and 25.0 L of 0.15 M NH3. What is the final

[Zn(H2O)42+] at equilibrium? Kf of Zn(NH3)4

2+ is 7.8x108.

PLAN: We write the reaction equation and the Kf expression, and use a reaction table to calculate equilibrium concentrations. To set up the table, we must first find [Zn(H2O)4

2+]init and [NH3]init using the given volumes and molarities. With a large excess of NH3 and a high Kf, we assume that almost all the Zn(H2O)4

2+ is converted to Zn(NH3)42+.

SOLUTION:

Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)4

2+(aq) + 4H2O(l)

Kf =[Zn(NH3)4

2+]

[Zn(H2O)42+][NH3]4

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Sample Problem 19.13

[Zn(H2O)42+]initial = = 1.3x10-3 M50.0 L x 0.0020 M

50.0 L + 25.0 L

[NH3]initial = = 5.0x10-2 M25.0 L x 0.15 M

50.0 L + 25.0 L

Initial 1.3x10-3 5.0x10-2 0 -

Change ~(-1.3x10-3) ~(-5.2x10-3) ~(+1.3x10-3) -

Zn(H2O)42+(aq) + 4NH3(aq) Zn(NH3)4

2+(aq) + 4H2O(l)Concentration (M)

Equilibrium x 4.5x10-2 1.3x10-3 -

4 mol of NH3 is needed per mol of Zn(H2O4)2+, so[NH3]reacted = 4(1.3x10-3 M) = 5.2x10-3 M and [Zn(NH3)4

2+] ≈ 1.3x10-3 M

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Kf =[Zn(NH3)4

2+]

[Zn(H2O)42+][NH3]4

= 7.8x108 =(1.3x10-3)

x(4.5x10-2)4

x = [Zn(H2O)42+ = 4.1x10-7 M

Sample Problem 19.13

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Complex Ion Equilibria

• Dissociation of complex ions

• Ag(NH3)2+ = Ag+ + 2NH3

• Kd = [Ag+][NH3]2

• ___________• [Ag(NH3)2

+]

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Complex Ion Equilibria

• Cr(OH)4¯ = Cr3+ + 4OH¯

• Kd = [Cr3+][OH¯]4

• _________________

• [Cr(OH)4-1]

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Complex Ion Equilibria

Determine the concentration of silver ions in a solution that is 0.10M in [Ag(NH3)2]+. Kd = 6.3x10-8

Determine the [Ag+] in a solution that is 0.10M in Ag(NH3)2NO3 and 0.10M in NH3.

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Complex Ion Equilibria

Ag+

Cu2+

Zn2+

Al3+

Cr3+

Zn2+

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Net Ionic Equations

• Rules for writing:– A. List predominant species

• All soluble salts, strong acids and strong bases are written as their component ions. All others are written as the molecule.

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Net Ionic Equations

• Rules (continued):– B. Combine ions of opposite charge and

look for one or more of the following:• 1. formation of a weak acid• 2. formation of a weak base• 3. formation of water• 4. formation of an insoluble substance• 5. formation of a complex ion

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Net Ionic Equations

• Rules (continued):– C. If no reaction in B. above, then look at

each of the following in the order given as a source of secondary species:• 1. solubility equilibria• 2. complex equilibria• 3. weak acid equilibria• 4. weak base equilibria• 5. hydrolysis equilibria (only if necessary)

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Net Ionic Equations

• Rules (continued):– D. If no reaction in C, then try secondary

species reacting with secondary species again leaving hydrolysis to last.

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Writing Net Ionic Equations

• Give net ionic equations for each of the following reactions occurring in aqueous solution. Also indicate the form of K in terms of other constants such as Ka, Kb, Kw, Ksp or Kd.– 1. AgNO3 + NaCl

– 2. BaCl2 + Na2SO4

– 3. FeCl3 + NaOH

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Writing Net Ionic Equations

– 4. CuS + HNO3

– 5. AlCl3 + NH3

– 6. Cr(NO3)3 + NaOH

– 7. Cr(NO3)3 + NaOH(xs)

– 8. Cr(NO3)3 + NH3(xs)

– 9. ZnCl2 + NH3(xs)

– 10. Cu(OH)2 + NH3(xs)

– 11. K2Zn(OH)4 + HNO3(xs)

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Writing Net Ionic Equations

– 12. AgCl + NH3(xs)

– 13. Ni(CH3COO)2 + H2O

– 14. Cr(OH)3 + NaOH(xs)

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