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Chapter Fifteen:

APPLICATIONS OF AQUEOUSEQUILIBRIA

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Contentsp680

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15-1 Solution of Acids or BasesContaining a Common Ion

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Le Chatelier’s Principle

Common ion effect

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The equilibrium concentration of OH- ionsis reduced by the common ion effect.

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The position of ammonia-waterequilibrium is shifted to the left.

Ex 15.1 Acidic Solutions ContainingCommon Ions

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Calculate [H+] and the percent of HF in a solution

containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF.

Solution:

1.0 M HF [H+] = 2.7 x 10 –2 M

The percent dissociation of 1.0 M HF is

2.7% ( 2.7 x 10-2/1.0)

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15-2 Buffered Solutionsp684

The most important application of acid-base solutions

containing a common ion is for buffering. A buffered

solution is one that resists a change in its pH when either

hydroxide ions pr protons are added. The most important

practical example of a buffered solution is our blood, which

can absorb the acids and bases produced in biologic reactions

without changing its pH. A constant pH for blood is vital

because cells can survive only in a very narrow pH range.

Ex 15.2 The pH of a Buffered Solution I

Solution:

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The major species in the the solution are

HC2H3O2, Na+, C2H3O2-, and H2O

A buffered solution contains 0.50 M Acetic acid (HC2H3O2,

Ka = 1.8 x 10-5) and 0.050 M sodium acetate (NaC2H3O2).

Calculate the pH of this solution.

weak neither base very weakacid acid nor (conjugate baseor

base of HC2H3O2)base

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Solving Problems with Buffered Solutions

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Buffering: How Does It Work?p687

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Buffering: How Does It Work? p688

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Ex 15.4 The pH of a Buffered Solution II p689

Calculate the pH of a solution containing 0.75 M lacticacid (ka = 1.4 x 10-4) and 0.25 M sodium lactate. Lacticacid ) HC3H5O3) is a common constituent of biologicsystem. For example, it is found in milk and is presentin human muscle tissue during exertion.Solution:

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Buffers contain relatively large amounts of weakacid and corresponding conjugate base.

Added H+ reacts to completion with theconjugate base.

Added OH- reacts to completion with the weakacid.

The pH is determined by the ratio of theconcentrations of the weak acid and conjugatebase.

p692Summary of the Most ImportantCharacteristic of Buffered Solutions

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15-3 Buffering Capacityp693

The capacity of a buffered solution isdetermined by the magnitudes of [HA] and [A-]

Ex15.7 Adding strong Acid to a BufferedSolution II

Solution:

For A:

Initial:

For B:

The optimal buffering system has a

pKa value close to the desired pH.

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15-4 Titrations and pH Curves p696

Strong acid-Strong Base Titrations

A. No NaOH has been added.

[H+] = 0.200 M pH = 0.699

B. 10.0 mL of 0.100 M has been added.

H+(aq) + OH-

(aq) → H2O(l)

pH = - log(0.15) = 0.82

17Na+, NO3-, H2O

C. 20.0 mL (total) of 0.100 M NaOH has been added.

pH = 0.942

D. 50.0 mL (total ) of 0.100 mL M NaOH has been added.

Preceding exactly as for points B and C, the pH is foundto be 1.301.

Enough OH- has been added to react exactly with H+

from the nitric acid. This is the stoichiometric point, orequivalence point, of the titration. At this point the majorspecies in solution are

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Titration p696

F. 150.0 mL (total) of 0.100 M NaOH has been added.

pH = 12.40

Proceeding as for point F,the pH is found to be 12.60

G. 200.0 mL (total) of0.100 M NaOH hasbeen added.

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Titration curve p699

Figure 15.1The pH curve for the titration of 50.0 mL of 0.200

M HNO3 with 0.100 M NaOH. Note that the equivalence

point occurs at 100.0 mL of NaOH added, the point where

exactly enough OH- has been added to react with all the H+

originally present. The pH of 7 at the equivalence point is

characteristic of a strong acid-base titration.

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Neutralization of a Strong Acid with a Strong Base

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The pH Curve for the Titration of 50.0 mL of0.200 M HNO3 with 0.100 M NaOH

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Figure 15.2 The pH Curve for the Titrationof 100.0 mL of 0.50 M NaOH with 1.0 M HCI.

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Figure 15.3 The pH Curve for the Titrationof 50.0 mL of 0.100 M HC2H3O2 with 0.100M NaOH.

Titrations of Weak Acids with Strong Basesp704

The shapes of the strong and weak acid

curves are the same after the equivalence

points because excess OH- controls the pH in

this region in both cases.

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Figure 15.4 The pH Curves for the Titrationsof 50.0-mL Samples of 0.10 M Acids withVarious Ka Values with 0.10 M NaOH.

The pH curves for the titrations ofsamples with various Ka

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Weak Acid - Strong Base Titration

Step 1: A stoichiometry problem (reaction is

assumed to run to completion) then

determine remaining species.

Step 2: An equilibrium problem (determine

position of weak acid equilibrium and

calculate pH).

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Figure 15.5 The pH Curve for the Titrations of100.0mL of 0.050 M NH3 with 0.10 M HCl.

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15-5 Acid-Base Indicators

Marks the end point of a titration by changing color.

The equivalence point is not necessarily the same

as the end point.

indicator

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The acid and base forms of the indicatorphenolphthalein. The indicatorphenolphthalein is colorless in acidicsolution and pink in basic solution.

Figure 15.6

Relative Solubilities

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CuI(s)

CaSO4(s)

16105.1 spK

12100.5 spK

5101.6 spK

AgI(s)

15.6 Solubility Equilibria andthe Solubility Product

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Common Ion Effect p722

Ex 15.15 Solubility and Common IonsCalculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a

0.025 M NaF solution.Solution:

Thus mol solid CaF2 dissolves per liter of the0.025 M NaF solution.

We will now see what happen the water containsan ion in common with the dissolving salt.

8104.6

15-7 Precipitation and QualitativeAnalysis

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Ex 15.16 Determining Precipitation ConditionsA solution is prepares by adding 750.0 mL of 4.00 x 10-3

M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M KIO3. Will

Ce(IO3)3 (Ksp = 1.9 x 10-10) precipitate from this solution?

Solution:

Ex 15.18 Selective Precipitation p728

A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 MPb2+. If a source of I- is added gradually to thissolution, will PbI2 (Ksp = 1.4 x 10-8) or CuI ( Ksp = 5.3 x10-12) precipitate first? Specify the concentration of I-necessary to begin precipitation of each salt.Solution:

As I- is added to the mixed solution, CuI willprecipitate first, since the [I-] required is less.

15-8 Equilibria Involving Complex Ions

]][NH)[Ag(NH])[Ag(NH

102.833

2332

K

mL)(200.0)mL)(2.0(100.0

][NH 03M

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Ag+ + 2NH3 → Ag(NH3)2+

Before reaction: 5.0 x 10-4 M 1.0 M 0 M

After reaction: 0 1.0-2(5.0 x 10-4) = 1.0 M 5.0 x 10-4 M

mL)(200.0)10mL)(1.0(100.0

][Ag3

0M

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Complex Ion Equilibria p731

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Nickel(II) Complexes

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Ex 15.19 Complex Ions

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232 104.7)OAg(SOSAg K

Calculate the concentrations of Ag+, Ag(S2O3)-, and

Ag(S2O3)3- in a solution prepared by mixing 150.0 mL of

1.00 x 10-3 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3.

The stepwise formation equilibria are:

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3232

23232 109.3)OAg(SOS)OAg(S K

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Solution:

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Complex Ions and Solubility p734

The solubility in pure water is

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Separation Scheme