Lecture 9 Solubility and Complex Ion Equilibria - Chap 16

Solubility and Complex Ion Equilibria

Chapter 16

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SOLUBILITY PRODUCT

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AgCl(s) Ag+(aq) + Cl¯(aq)⇄

Ksp = 1.6x10-10 = [Ag+(aq)][Cl¯(aq)]

This is the reaction for dissolving solid silver chloride in water. The equilibrium constant for reactions of this

type are called solubility products – Ksp.

Can use the Ksp to calculate solubility.

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AgCl(s) Ag⇄ +(aq) + Cl¯(aq)at equilibrium x x

Ksp = 1.6x10-10 = [Ag+(aq)][Cl¯(aq)] = x2

So a saturated solution of AgCl is 1.3x10-5M

Ksp of CaF2 is 4.0x10-11

CaF2(s) Ca⇄ +2(aq) + 2F¯(aq) at equilibrium x 2x

4x3= 4x10-11, x = 2.2x10-4M

Mx 510 103.1106.1

11222 100.4)2)((]][[ xxFCaK sp

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Your turn to try one: what is the molar solubility of iron(II) sulfide?

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What is the molar solubility of lead(II) bromide? Ksp = 4.6x10-6

1.0.0021M2.0.013M3.0.017M4.0.010M5.I’m lost

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Predicting if a precipitate will form:The Ksp for PbF2 is 4x10-8. If you add a 50mL of a 1.0mM solution of Ba(NO3)2 to

50mL of a 0.10M solution of KF, will a precipitate form?

1.Write the reaction.PbF2(s) Pb⇄ +2(aq) + 2F¯(aq)2.Write Q and calculate its value:Q = [Pb+2(aq)][F¯(aq)]2=(.001/2)(0.10/2)2) = 1.25x10-6

3.Compare Q to K, and determine which direction the reaction will proceed.

Q>K, so the reaction proceeds left. 7

Will a precipitate form if 0.10L of 8.0x10-3M lead(II) nitrate is added to 0.40L of 5.0x10-3M sodium

sulfate?

1. Determine what will precipitate

2. Determine the concentration of relevant ions

3. Calculate Q, and compare it to Ksp

From table of Ksp values, find Ksp of lead(II) sulfate = 1.3x10-8

MiVi=MfVf, final V = 0.50L, so Mf=Mi(Vi/Vf)[Pb2+]= 0.008M(0.1/0.5)=1.6x10-3M[SO4

-2] = 0.005M(0.4/0.5) = 0.004M

Q = [Pb2+][SO4-2] = [1.6x10-3][0.004] = 6.4x10-6

So Q > Ksp=1.3x10-8 Since Q > K, precipitate will form.

Will a precipitate form when 150mL of 0.010M Sr(NO3)2 is added to 450mL of 0.016M K2CrO4? 8

Which is less soluble:

1.Sr3(PO4)2 Ksp = 1x10-31

2.AgI Ksp = 1.5x10-16

Sr3(PO4)2(s) 3Sr⇄ +2(aq) + 2PO4-3(aq)

3x 2xKsp(Sr3(PO4)2) = (3x)3(2x)2 = 1x10-31, x = 3x10-7M

AgI(s) Ag⇄ +(aq) + I¯(aq) x xKsp(AgI) = x2 = 1.5x10-6, x = 1.2x10-8M

Sr3(PO4)2 is more soluble than AgI! 9

It is true that solubility decreases in the series• PbSO4 Ksp = 1.3x10-8

• AgCl Ksp = 1.6x10-10

• BaCrO4 Ksp = 8.5x10-11

But is the same thing true for this series?• AgI Ksp = 1.5x10-16

• Zn(OH)2 Ksp = 4.5x10-17

• Sr3(AsO4)2 Ksp = 4.3x10-19

1. Yes2. No3. lost

• 1.2x10-8M• 2.2x10-6M• 1.0x10-4M

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Things that can alter solubility

1. Common ion2. pH3. Complex ions

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Consider the solubility of lead(II) carbonate, Ksp = 4.6x10-6

PbBr2(s) Pb⇄ +2(aq) + 2Br¯(aq)

Ksp = [Pb+2(aq)][Br¯(aq)]2 = (x)(2x)2 = 4.6x10-6 x = 0.01M

What is the solubility of PbBr2 in the presence of 0.10M KBr(aq)?

Ksp = [Pb+2(aq)][Br¯(aq)]2 = (x)(2x+0.10)2 = (x)(0.10)2 = 4.6x10-6 Approximation that 2X << 0.10, Solving, x = 4.6x10-4M

In the presence of a common ion, the solubility is greatly reduced!

Le Chatelier’s principle!

COMMON ION EFFECT

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Consider the solubility of magnesium fluoride, Ksp = 6.8x10-6

MgCO3(s) Mg⇄ +2(aq) + CO3-2(aq)

Ksp = [Mg+2(aq)][ CO3-2(aq)] = (x)(x) = 6.4x10-9 x = 1.2x10-3M

However, it is freely soluble in 1.0M HCl(aq). WHY?

Acid alters the concentration of CO3-2(aq) by driving the

equilibrium towards protonated species.

The solubility of any insoluble salt containing a basic anion will be affected by pH: F¯, CO3

-2, S-2, OH¯, PO4-3, SO4

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pH EFFECTS

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Metal ions can act as Lewis acids, reacting with Lewis Bases:

Formation Constant

Complex Ions

Be2+(aq) + F–(aq) ⇄ BeF+(aq) K1 = 7.9 x 104

BeF+(aq) + F–(aq) ⇄ BeF2(aq) K2 = 5.8 x 103

BeF2(aq) + F–(aq) ⇄ BeF3–

(aq) K3 = 6.1 x 102

BeF3–

(aq) + F–(aq) ⇄ BeF42–

(aq) K4 = 2.7 x 101

The BeF42–

(aq) is called a complex ion. The Lewis base is called the ligand. 14

Metal ions can act as Lewis acids, reacting with Lewis Bases:

Formation Constant

Complex Ions

Ag+ + NH3 ⇄ AgNH3+ Kf1 = 2.1 x 103

AgNH3+ + NH3 ⇄ Ag(NH3)2

+ Kf2 = 8.2 x 103

Because these equilibria lie to the right, they reduce the concentration of free metal ion, and can increase the solubility of insoluble salts.

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Metal ions can act as Lewis acids, reacting with Lewis Bases:

Formation Constant

Complex Ions

Ag+ + NH3 ⇄ AgNH3+ Kf1 = 2.1 x 103

AgNH3+ + NH3 ⇄ Ag(NH3)2

+ Kf2 = 8.2 x 103

AgI(s) Ag⇄ +(aq) + I¯(aq) Ksp = 1.5x10-16, solubility 1.2x10-8M

In presence of 1M NH3(aq):AgI(s) + 2NH3(aq) Ag(NH⇄ 3)2

+(aq) + I¯(aq) K = Kf1xKf2xKsp=2.6x10-9

223

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1

))((

][

]][)([ xx

NH

INHAgK

59 101.5106.2 x

Note approximation: [NH3] = 1-2x, but x is very small. 16

Bottom lines:• Nomenclature: complex ion, ligand• Lewis acid/base reaction• Idea of formation constant• General effect of increasing solubility of ‘insoluble’ ions

Complex Ions

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