Equilibrium Equilibrium ChemistryChemistry

CE 541CE 541

Important to:Important to: Determine the relationship between constituents in waterDetermine the relationship between constituents in water Understand the effect of alterations of water on the different Understand the effect of alterations of water on the different

chemical species presentchemical species present Limitations of Equilibrium CalculationsLimitations of Equilibrium Calculations

Dynamic changes (in wastewater and surface water)Dynamic changes (in wastewater and surface water) Due to exposure to sunDue to exposure to sun Due to exposure to pollution (organic and inorganic)Due to exposure to pollution (organic and inorganic)

Rapid reactions (reaction between acids and bases)Rapid reactions (reaction between acids and bases) Very slow reactions (oxidation-reduction in natural waters)Very slow reactions (oxidation-reduction in natural waters) Precipitation reactionsPrecipitation reactions Lack of information on accurate equilibrium constants for Lack of information on accurate equilibrium constants for

many of the reactions taking place in natural watersmany of the reactions taking place in natural waters

Ion Activity CoefficientsIon Activity Coefficients

““The activity of ion or molecule can be The activity of ion or molecule can be found by multiplying its molar found by multiplying its molar concentration by an activity coefficient, concentration by an activity coefficient, ””

{A} = {A} = [A] [A]{A} = activity{A} = activity[A] = concentration[A] = concentration = activity coefficient= activity coefficient

For practical reasons and rough For practical reasons and rough calculations [A] is used in place of {A}calculations [A] is used in place of {A}

Some investigators found that “Some investigators found that “the the activity coefficients for ions in an activity coefficients for ions in an electrolyte were related to the electrolyte were related to the concentration of charged particles in the concentration of charged particles in the solutionsolution”. They introduced the ”. They introduced the ionic ionic strengthstrength as an empirical measure of the as an empirical measure of the interactions among all the ions in a interactions among all the ions in a solution.solution.

= ionic strength= ionic strength Ci = molar concentration of the iCi = molar concentration of the ithth ion ion Zi = charge of the iZi = charge of the ithth ion ion

Langelier estimated Langelier estimated as (TDS as (TDS 2.5 2.5 10 10-5-5))

i

iiZC2

2

1

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

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

This relationship is used for solutions with This relationship is used for solutions with up to 0.5 M. up to 0.5 M.

)(1

5.0log 2 GuntelbergZ

)(2.01

5.0log 2 DavisZ

ConclusionConclusion

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

Study Example 1 page 108.

Problem 4.2Problem 4.2Calculate the activity coefficient and activity of each ion Calculate the activity coefficient and activity of each ion in a solution containing 75 mg/l Nain a solution containing 75 mg/l Na++, 25 mg/l Ca, 25 mg/l Ca2+2+, 10 , 10 mg/l Mgmg/l Mg2+2+, 125 mg/l Cl, 125 mg/l Cl--, 50 mg/l HCO, 50 mg/l HCO33

--, and 48 mg/l SO, and 48 mg/l SO442-2-. .

Solutions to Equilibrium Solutions to Equilibrium ProblemsProblems

Le Chatelier’s Principle states thatLe Chatelier’s Principle states that “ “A A chemical system will respond to change chemical system will respond to change with processes which tend to reduce the with processes which tend to reduce the effect of the changeeffect of the change””

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

obeyedobeyed Electroneutrality must be maintainedElectroneutrality must be maintained “ “All All

positively charged species in solution must be positively charged species in solution must be balanced by equivalent numbers of negatively balanced by equivalent numbers of negatively charged speciescharged species””

Proton conditionProton condition “ “Species with an excess of Species with an excess of protons must be balanced by the species with protons must be balanced by the species with a deficiency in protonsa deficiency in protons””

All reactions involved must proceed towards a All reactions involved must proceed towards a state of equilibrium.state of equilibrium.

Steps to Solve Equilibrium Steps to Solve Equilibrium Problem Involving Aqueous Phase Problem Involving Aqueous Phase

onlyonly Define the equilibrium problemDefine the equilibrium problem

what chemical reactions are taking placewhat chemical reactions are taking place what is reacting with whatwhat is reacting with what

List all constituents of the systemList all constituents of the system all systems involving water includeall systems involving water include

HH22OO HH++

OHOH--

Include all ions, elements and neutral Include all ions, elements and neutral species present initiallyspecies present initially

Steps to Solve Equilibrium Steps to Solve Equilibrium Problem Involving Aqueous Phase Problem Involving Aqueous Phase

onlyonly For each element present initiallyFor each element present initially

list all forms or species which are likely to list all forms or species which are likely to contain the element and which are likely contain the element and which are likely to present after equilibrium is attainedto present after equilibrium is attained

Identify concentrations of all species Identify concentrations of all species for each element so that mass and for each element so that mass and charge balances can be madecharge balances can be made

List all appropriate equilibrium List all appropriate equilibrium relationships between species of relationships between species of concernconcern

Steps to Solve Equilibrium Steps to Solve Equilibrium Problem Involving Aqueous Phase Problem Involving Aqueous Phase

onlyonly List associated equilibrium constantsList associated equilibrium constants

List all mass and charge balance List all mass and charge balance relationships for the systemrelationships for the system

List proton conditionsList proton conditions

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

Solve the equations simultaneouslySolve the equations simultaneously

Steps to Solve Equilibrium Steps to Solve Equilibrium Problem Involving Aqueous Phase Problem Involving Aqueous Phase

onlyonly

If gaseous or solid phase are If gaseous or solid phase are involved, then equations involved, then equations expressing mass and charge expressing mass and charge balances between and within each balances between and within each phase must be included.phase must be included.

Study Example 2Study Example 2

Problem 4.4Problem 4.4

A solution is prepared by diluting 10A solution is prepared by diluting 10-2-2 mol of ammonia to mol of ammonia to 1 liter with distilled water. Calculate the equilibrium 1 liter with distilled water. Calculate the equilibrium concentration for each chemical species in the water.concentration for each chemical species in the water.

Acids and BasesAcids and Bases

strong acids and bases ionize strong acids and bases ionize completely in dilute solutions (water)completely in dilute solutions (water)

weak acids and bases ionize partially weak acids and bases ionize partially in dilute solutions (water)in dilute solutions (water)

acids increase Hacids increase H++ concentration concentration Bases increase OHBases increase OH-- concentration concentration

[H[H++][OH][OH--] = K] = Kww

[ ] = activity or approximately molar [ ] = activity or approximately molar concentrationconcentration

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

Equilibrium between most of the chemical Equilibrium between most of the chemical speciesspecies

Effectiveness of coagulationEffectiveness of coagulation Potential of water to be corrosivePotential of water to be corrosive Suitability of water to microorganismsSuitability of water to microorganisms Other quality characteristics of the waterOther quality characteristics of the water Thus it is very important to understand the Thus it is very important to understand the

factors that have an effect on pH of water.factors that have an effect on pH of water.

The pH and p(x) ConceptThe pH and p(x) Concept

In pure water (no other materials):In pure water (no other materials): Activity = Molar concentrationActivity = Molar concentration [H[H++] = [OH] = [OH--]] KKww = 10 = 10-14-14 @ 25 @ 25 C C [H[H++][OH][OH--] = 10] = 10-14-14

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

pH = 7 (is the neutral pH)pH = 7 (is the neutral pH)

][

1log

]log[

HpH

HpH

pH MeterspH Meters

scale 0 to 14scale 0 to 14 pH less than 7 indicates acidic pH less than 7 indicates acidic

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

condition; [Hcondition; [H++] < [OH] < [OH--]] electrodes measure hydrogen-ion electrodes measure hydrogen-ion

activity not molar concentrationactivity not molar concentration

The concept of expressing [HThe concept of expressing [H++] ] activity can be used with other ions. activity can be used with other ions. SoSo

x = concentration of a given x = concentration of a given chemical species or equilibrium chemical species or equilibrium constant. Thenconstant. Then

xxxp

1log)(log)( 1010

ww KpK

OHpOH

log

]log[

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

Then,Then, -log[H-log[H++] – log[OH] – log[OH--] = -log K] = -log Kww

pH + pOH = pKpH + pOH = pKww

pKpKww = 14 @ 25 = 14 @ 25 C C

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

constant for weak acidsconstant for weak acids pKpKBB is the negative log of the ionization is the negative log of the ionization

constant for weak basesconstant for weak bases

Tables 4-1 and 4-2 show KTables 4-1 and 4-2 show KAA, pK, pKAA, K, KBB, and , and pKpKBB for weak acids and bases. for weak acids and bases.

For weak acid and its conjugate For weak acid and its conjugate base or weak base and its base or weak base and its conjugate acid:conjugate acid:

pKpKAA + pK + pKBB = 14 @ 25 = 14 @ 25 C C

oror

KKAAKKBB = 10 = 10-14-14 = K = Kww

Example Example

Boric acid hasBoric acid has pKpKAA = 9.24= 9.24

Borate hasBorate has pKpKBB = 4.76= 4.76

pKpKAA + pK + pKBB = 14= 14

Solving Acid – Base Solving Acid – Base Equilibrium ProblemsEquilibrium Problems

AssumptionsAssumptions equilibrium occurs very fast (thus equilibrium occurs very fast (thus

neglect kinetic considerations)neglect kinetic considerations) strong acids and bases are strong acids and bases are

completely ionized in water (except completely ionized in water (except when the added concentration is ≈ when the added concentration is ≈ 1010-7-7))

Tools to be used in Solving ProblemsTools to be used in Solving Problems equilibrium relationshipsequilibrium relationships mass balancemass balance charge balancecharge balance proton conditionproton condition

How to Solve ProblemsHow to Solve Problems identify unknownsidentify unknowns generate equations = unknownsgenerate equations = unknowns solve equations simultaneouslysolve equations simultaneously use graphical solutionsuse graphical solutions use computers for complex problemsuse computers for complex problems

Study Examples 3, 4, 5, and 6Study Examples 3, 4, 5, and 6

Problem 4.5Problem 4.5

Calculate the equilibrium pH of a solution containing (a) Calculate the equilibrium pH of a solution containing (a) 1010-3-3 M H M H22SOSO44; (b) 10; (b) 10-8-8 M H M H22SOSO44

Logarithmic Concentration Logarithmic Concentration DiagramDiagram

Log C – pH diagram represents Log C – pH diagram represents mass balance of each constituent mass balance of each constituent at every pH value. To construct at every pH value. To construct the diagram, develop equations of the diagram, develop equations of C as a function of pH, KC as a function of pH, Kww, K, KAA, C, CTT

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 diagram change when the Ka is changed?

How does a logarithmic concentration diagram change when the concentration is changed?

How does a logarithmic concentration diagram change when the Ka2 is changed?

Logarithmic Concentration Logarithmic Concentration Diagram for Monoprotic Diagram for Monoprotic

Acids and BasesAcids and BasesDiagram of Monoprotic AcidDiagram of Monoprotic Acid0.02 M acetic acid solution. Use:0.02 M acetic acid solution. Use:

the above acetic acid equations can be used the above acetic acid equations can be used for all monoprotic acids. Monoprotic acid is for all monoprotic acids. Monoprotic acid is an acid which yields one proton when ionized.an acid which yields one proton when ionized.

][][][

][][

25@108.1][

]][[

25@10]][[

5

14

AcOHH

CAcHAc

CKHAc

AcH

CKOHH

T

A

w

Line of [HLine of [H++] is ] is obtained from:obtained from:

log [Hlog [H++] = -pH] = -pH

Line of [OHLine of [OH--] is ] is obtained from:obtained from:

log [OHlog [OH--] = pH – ] = pH – pKpKww

Concentration of Concentration of Acetic Acid and Acetic Acid and Acetate can be Acetate can be obtained from:obtained from: ][

][

][][

HK

KCAc

and

HK

KCHAc

A

AT

A

AT

Intersection of [HAc] Intersection of [HAc] and [Acand [Ac--] lines is ] lines is called the System called the System Point and is located Point and is located at:at:

pH = pKpH = pKAA

To left of the system To left of the system point, [Hpoint, [H++] > K] > KAA, so, so

Both lines pass Both lines pass through the system through the system pointpoint

)1....(log]log[

)..(log]log[

slopewithdigonalpHpKCAc

linehorizonalCHAc

AT

T

To the right of the To the right of the system point, [Hsystem point, [H++] ] < K< KAA, so, so

Both lines pass Both lines pass through the system through the system pointpoint

)..(log]log[

)1....(log]log[

linehorizontaCAc

slopewithdiagonalpHpKCHAc

T

AT

Just below the system point and fromJust below the system point and from [HAc] + [HAc] + [Ac-] = C[Ac-] = CTT

[HAc] = [Ac[HAc] = [Ac--] = (1/2)C] = (1/2)CTT

log [(1/2)Clog [(1/2)CTT] = log C] = log CTT + log (1/2) = log C + log (1/2) = log CTT – 0.3 – 0.3

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

General Procedure for General Procedure for Construction of logC-pH Diagrams Construction of logC-pH Diagrams

for Monoprotic Acids and Basesfor Monoprotic Acids and Bases

1.1. Draw horizontal line representing log CDraw horizontal line representing log CTT

2.2. Locate the system point at pH = pKLocate the system point at pH = pKAA

3.3. Draw 45Draw 45 lines sloping to left and right lines sloping to left and right of the system pointof the system point

4.4. Locate a point of 0.3 logarithmic units Locate a point of 0.3 logarithmic units below the system pointbelow the system point

5.5. Connect horizontal and 45Connect horizontal and 45 lines with lines with short curves passing through the pointsshort curves passing through the points

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

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

Diagram of Monoprotic Diagram of Monoprotic BaseBase

Similar procedureSimilar procedure System point at pH System point at pH

= pK= pKww – pK – pKBB

0.01 M NH0.01 M NH33 Solution Solution

For ammonia, For ammonia,

pKpKBB = 4.74 = 4.74

so, system point is so, system point is located at:located at:

14 – 4.74 = 9.2614 – 4.74 = 9.26

Logarithmic Concentration Logarithmic Concentration Diagram for a Weak Acid Diagram for a Weak Acid

and a Weak Baseand a Weak Base

The logC-pH diagram is made by superimposing the curves for each material on a single diagram.

Study Examples 11 to 15

0.1 M acetic acid and 0.1 M ammonia

Logarithmic Concentration Logarithmic Concentration Diagram for Polyprotic Diagram for Polyprotic

Acids and BasesAcids and BasesTake a solution containing 0.01 M Take a solution containing 0.01 M carbonic acid (Hcarbonic acid (H22COCO33) as an ) as an example. example.

1410]][[

10][][][

33.10107.4][

]][[

37.6103.4][

]][[

14

2233

*32

211

23

23

17

1*32

3

ww

T

AA

AA

pKKOHH

CCOHCOCOH

pKKHCO

COH

pKKCOH

HCOH

Solving the equation for individual Solving the equation for individual carbonic species gives carbonic species gives

212

223

2113

2211

0*32

/][/][1][

]/[/][1][

]/[]/[1][

AAA

TT

AA

TT

AAA

TT

KKHKH

CCCO

HKKH

CCHCO

HKKHK

CCCOH

The diagram was The diagram was constructed in the constructed in the same manner as in the same manner as in the case of monoprotic case of monoprotic acids and bases except acids and bases except that:that:

Slope of the line for Slope of the line for [CO[CO33

2-2-] changes from ] changes from +1 to +2 when pH +1 to +2 when pH drops below pKdrops below pKA1A1 ([H ([H++] ] >> K>> KA1A1))

Slope of the line for Slope of the line for [H[H22COCO33

**] changes from ] changes from -1 to -2 when pH -1 to -2 when pH becomes greater than becomes greater than pKpKA2A2 ([H ([H++] << K] << KA2A2))

General Procedure for General Procedure for Construction of logC-pH Diagrams Construction of logC-pH Diagrams

for Diprotic Acids and Basesfor Diprotic Acids and Bases Draw horizontal line representing log CTDraw horizontal line representing log CT Locate the system point at pH values equal Locate the system point at pH values equal

to pKto pKA1A1 and pK and pKA2A2 (pK (pKww-pK-pKB1B1 and pK and pKww-pK-pKB2B2 for a base)for a base)

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

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

The procedure for construction of logC-pH The procedure for construction of logC-pH diagram for polyprotic acids and bases are diagram for polyprotic acids and bases are similar except that the slope of the similar except that the slope of the diagonal changes from +1 to +2 and then diagonal changes from +1 to +2 and then from +2 to +3 as it reaches the adjacent from +2 to +3 as it reaches the adjacent system points. system points.

Problem 4.25Problem 4.25

Draw a log C-pH diagram for a 10Draw a log C-pH diagram for a 10-2-2 M solution of hydrogen sulfide. M solution of hydrogen sulfide. Assume a closed system. From the Assume a closed system. From the diagram, determine the pH for diagram, determine the pH for solutions that contain the following:solutions that contain the following:

1010-2-2 M H M H22SS 1010-2-2 M Na M Na22SS 0.5 0.5 10 10-2-2 M HS M HS-- and 0.5 and 0.5 10 10-2-2 M S M S2-2-

0.5 0.5 10 10-2-2 M H M H22S and 0.5 S and 0.5 10 10-2-2 M HS M HS--