ce 541 - equilibrium chemistry

7/29/2019 CE 541 - Equilibrium Chemistry

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Equilibrium Chemistry

CE 541


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Important to:

Determine the relationship between constituents in water

Understand the effect of alterations of water on the differentchemical species present

Limitations of Equilibrium Calculations

Dynamic changes (in wastewater and surface water)

Due to exposure to sun Due to exposure to pollution (organic and inorganic)

Rapid reactions (reaction between acids and bases)

Very slow reactions (oxidation-reduction in natural waters)

Precipitation reactions

Lack of information on accurate equilibrium constants for

many of the reactions taking place in natural waters


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Ion Activity Coefficients

The activity of ion or molecule can be found bymultiplying its molar concentration by an activitycoefficient,

{A} = [A]{A} = activity

[A] = concentration

= activity coefficientFor practical reasons and rough calculations [A] isused in place of {A}


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Some investigators found that the activitycoefficients for ions in an electrolyte were relatedto the concentration of charged particles in the

solution. They introduced the ionic strength asan empirical measure of the interactions among allthe ions in a solution.

= ionic strength Ci = molar concentration of the ith ion

Zi = charge of the ith ion

Langelier estimated as (TDS 2.5 10-5)

i

iiZC2

2

1


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Other investigators found that, for dilute solution,there is a relationship between and as follows:

This relationship is used for dilute solutions with ionicstrength < 0.1 (in Environmental Engineering, most waters ofinterest are more dilute than this, except seawater)

This relationship is used for solutions with up to0.5 M.

)(1

5.0log 2 GuntelbergZ

)(2.01

5.0log 2 DavisZ


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Conclusion

There is no good relationship that provides asatisfactory estimate of for > 0.5 M. In thiscourse, will be assumed to be equal to 1unless otherwise mentioned.

Study Example 1 page 108.


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Problem 4.2Calculate the activity coefficient and activity of each ion in a solution

containing 75 mg/l Na+, 25 mg/l Ca2+, 10 mg/l Mg2+, 125 mg/l Cl-, 50

mg/l HCO3

-, and 48 mg/l SO4

2-.


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Solutions to Equilibrium Problems

Le Chateliers Principle states thatA chemicalsystem will respond to change with processes which

tend to reduce the effect of the change

For any chemical reaction Principle of conservation of mass must be obeyed

Electroneutrality must be maintained All positivelycharged species in solution must be balanced byequivalent numbers of negatively charged species

Proton condition Species with an excess of protonsmust be balanced by the species with a deficiency in

protons

All reactions involved must proceed towards a state ofequilibrium.


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Steps to Solve Equilibrium Problem Involving

Aqueous Phase only

Define the equilibrium problem

what chemical reactions are taking place

what is reacting with what

List all constituents of the system

all systems involving water include

H2O

H+

OH-

Include all ions, elements and neutral species present

initially


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Aqueous Phase only

For each element present initially list all forms or species which are likely to contain the

element and which are likely to present after equilibriumis attained

Identify concentrations of all species for eachelement so that mass and charge balances can bemade

List all appropriate equilibrium relationshipsbetween species of concern


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Aqueous Phase only

List associated equilibrium constants

List all mass and charge balance relationships forthe system

List proton conditions

Steps 5 to 8 will produce a number of equationsequal to the unknown species.

Solve the equations simultaneously


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Aqueous Phase only

If gaseous or solid phase are involved, thenequations expressing mass and charge

balances between and within each phase mustbe included.

Study Example 2


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Problem 4.4

A solution is prepared by diluting 10-2 mol of ammonia to 1 liter with

distilled water. Calculate the equilibrium concentration for each chemical

species in the water.


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Acids and Bases

strong acids and bases ionize completely in

dilute solutions (water)

weak acids and bases ionize partially in dilutesolutions (water)

acids increase H+ concentration

Bases increase OH- concentration


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[H+][OH-] = Kw

[ ] = activity or approximately molar concentration

[H+] is expressed in terms of pH. pH has an effect

on:

Equilibrium between most of the chemical species

Effectiveness of coagulation

Potential of water to be corrosive

Suitability of water to microorganisms Other quality characteristics of the water

Thus it is very important to understand the factors that

have an effect on pH of water.


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The pH and p(x) Concept

In pure water (no other materials): Activity = Molar concentration

[H+] = [OH-] Kw= 10

-14 @ 25 C [H+][OH-] = 10-14

[H+] = [OH-] = 10-7

pH = 7 (is the neutral pH)

][

1log

]log[

HpH

HpH


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pH Meters

scale 0 to 14

pH less than 7 indicates acidic condition;

[H+] > [OH-]pH more than 7 indicates basic condition;

[H+] < [OH-]

electrodes measure hydrogen-ion activitynot molar concentration


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The concept of expressing [H+] activity can be

used with other ions. So

x = concentration of a given chemical species or

equilibrium constant. Then

xxxp1

log)(log)( 1010

ww KpK

OHpOH

log

]log[


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Since,[H+][OH-] = Kw= 10-14 @ 25 C

Then,-log[H+] log[OH-] = -log Kw

pH + pOH = pKw

pKw= 14 @ 25 C

for weak acids and bases: pKA is the negative log of the ionization constant for

weak acids

pKB is the negative log of the ionization constant for

weak bases

Tables 4-1 and 4-2 show KA, pKA, KB, and pKB for

weak acids and bases.


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For weak acid and its conjugate base or weak

base and its conjugate acid:

pKA+ pKB = 14 @ 25 Cor

KAKB = 10-14 = Kw

Example

Boric acid has pKA = 9.24

Borate has pKB

= 4.76

pKA+ pKB = 14


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Solving Acid Base Equilibrium

Problems

Assumptions

equilibrium occurs very fast (thus neglect

kinetic considerations)

strong acids and bases are completely ionized in

water (except when the added concentration is 10-7)


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Tools to be used in Solving Problems equilibrium relationships

mass balance

charge balance

proton condition

How to Solve Problems identify unknowns

generate equations = unknowns

solve equations simultaneously

use graphical solutions use computers for complex problems

Study Examples 3, 4, 5, and 6


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Problem 4.5

Calculate the equilibrium pH of a solution containing (a) 10-3 M H2SO4; (b)

10-8 M H2SO4


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Logarithmic Concentration Diagram

Log C pH diagram represents mass balance

of each constituent at every pH value. Toconstruct the diagram, develop equations of C

as a function of pH, Kw, KA, CT


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These curves show the log of the concentration as a function of pH

How does a logarithmic concentration diagram

change when the concentration is changed? How does a logarithmic concentration diagramchange when the Ka is changed?

How does a logarithmic concentration diagramchange when the concentration is changed? How does a logarithmic concentration diagramchange when the Ka2 is changed?
http://chemistry.beloit.edu/Rain/moviepages/logCKa2.htmhttp://chemistry.beloit.edu/Rain/moviepages/logCKa2.htmhttp://chemistry.beloit.edu/Rain/moviepages/logCKa2.htmhttp://chemistry.beloit.edu/Rain/moviepages/logConc3.htmhttp://chemistry.beloit.edu/Rain/moviepages/logCKa1.htmhttp://chemistry.beloit.edu/Rain/moviepages/logConc1.htmhttp://chemistry.beloit.edu/Rain/moviepages/logCKa2.htmhttp://chemistry.beloit.edu/Rain/moviepages/logConc3.htmhttp://chemistry.beloit.edu/Rain/moviepages/logCKa1.htmhttp://chemistry.beloit.edu/Rain/moviepages/logConc1.htm


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for Monoprotic Acids and BasesDiagram of Monoprotic Acid

0.02 M acetic acid solution. Use:

the above acetic acid equations can be used for all monoproticacids. Monoprotic acid is an acid which yields one proton

when ionized.

][][][

][][

[email protected]][

]][[

25@10]][[

5

14

AcOHH

CAcHAc

CKHAc

AcH

CKOHH

T

A

w


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Line of [H+] is obtained

from:

log [H+

] = -pH

Line of [OH-] is obtained

from:

log [OH-] = pH pKw

Concentration of Acetic

Acid and Acetate can beobtained from:

][][

][][

HK

KCAc

and

HK

KCHAc

A

AT

A

AT


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Intersection of [HAc]

and [Ac-] lines is called

the System Point and is

located at:

pH = pKA

To left of the systempoint, [H+] > KA, so

Both lines pass through

the system point

)1....(log]log[

)..(log]log[

slopewithdigonalpHpKCAc

linehorizonalCHAc

AT

T


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To the right of the

system point, [H+] < KA,

so

Both lines pass throughthe system point

)..(log]log[

)1....(log]log[

linehorizontaCAc

slopewithdiagonalpHpKCHAc

T

AT


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Just below the system point and from[HAc] + [Ac-]

= CT

[HAc] = [Ac-

] = (1/2)CTlog [(1/2)CT] = log CT + log (1/2) = log CT0.3

So, curves of horizontal and diagonal lines intersectat a point 0.3 below the system point.


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General Procedure for Construction of logC-

pH Diagrams for Monoprotic Acids and Bases

1. Draw horizontal line representing log CT2. Locate the system point at pH = pKA3. Draw 45 lines sloping to left and right of the

system point4. Locate a point of 0.3 logarithmic units below the

system point

5. Connect horizontal and 45 lines with shortcurves passing through the points

6. [H+] and [OH-] lines are drawn as 45 lineswhich intersect at pH = 7 and log C = -7

7. Change in concentration will only shift the[HAc] and [Ac-] curves up or down


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Diagram of Monoprotic Base

Similar procedure System point at pH = pKw

pKB

0.01 M NH3 Solution

For ammonia,

pKB = 4.74

so, system point is located at:

14 4.74 = 9.26


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for a Weak Acid and a Weak Base

The logC-pH diagram is

made by superimposingthe curves for each

material on a single

diagram.

Study Examples 11 to 15

0.1 M acetic acid and 0.1 M ammonia


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for Polyprotic Acids and Bases

Take a solution containing 0.01 M carbonic

acid (H2CO3) as an example.

1410]][[

10][][][

33.10107.4][

]][[

37.6103.4][

]][[

14

22

33

*

32

2

11

2

3

2

3

1

7

1*

32

3

ww

T

AA

AA

pKKOHH

CCOHCOCOH

pKKHCO

COH

pKKCOH

HCOH


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Solving the equation for individual carbonicspecies gives

212

223

21

13

2

211

0

*

32

/][/][1][

]/[/][1][

]/[]/[1

][

AAA

T

T

AA

TT

AAA

TT

KKHKH

C

CCO

HKKH

CCHCO

HKKHK

CCCOH


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The diagram was

constructed in the same

manner as in the case ofmonoprotic acids and bases

except that:

Slope of the line for [CO32-]

changes from +1 to +2when pH drops below pKA1

([H+] >> KA1)

Slope of the line for

[H2CO3*

] changes from -1to -2 when pH becomes

greater than pKA2 ([H+]


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General Procedure for Construction of logC-

pH Diagrams for Diprotic Acids and Bases

Draw horizontal line representing log CT Locate the system point at pH values equal to

pKA1 and pKA2 (pKw-pKB1 and pKw-pKB2 for abase)

Draw 45 lines sloping to left and right of eachsystem point to the adjacent system point

The slope of lines changes from -1 to -2 and from+1 to +2

The procedure for construction of logC-pHdiagram for polyprotic acids and bases aresimilar except that the slope of the diagonalchanges from +1 to +2 and then from +2 to +3 asit reaches the adjacent system points.


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Problem 4.25

Draw a log C-pH diagram for a 10-2 M solutionof hydrogen sulfide. Assume a closed system.

From the diagram, determine the pH for

solutions that contain the following: 10-2 M H2S

10-2 M Na2S

0.5 10-2 M HS- and 0.5 10-2 M S2- 0.5 10-2 M H2S and 0.5 10-2 M HS-


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ce 541 - equilibrium chemistry

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