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Tonight’s topicsSolubility rules

Table 1.2 in the text

Expressions for concentration

pp 9-20

Ionic strengthpp 21-22

Activity and activity coefficients

pp 22-40

Page numbers refer to the handout revised Chapter 1

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Our goal: Describe equilibrium composition of solutions

Some solutions not at equilibriumEquilibrium assessment still valuable

ApplicationsWater and wastewater treatmentNatural and disturbed systems

Techniques apply to analysis and design of treatment systems

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Common environmental systems are dilute systems

1 kg water 1 L water

1 mg/L 1 mg/kg = 1 part per million by mass

1 g/L 1 g/kg = 1 part per billion by mass

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Copper (Cu) in waterWhere does the Cu come from?What form(s) is (are) present?

Think about a material balance TOTCu = ?

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Possible forms of copper in a drinking water systemCu pipe (Cu0)Cuprous (Cu(I))

Cu+ and complexesCupric (Cu(II))

Cu2+ and complexesPrecipitated copper

CuCO3.Cu(OH)2 = malachite

Adsorbed copper S.O-Cu (S represents a solid surface)

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Why does speciation matter?

Speciation affects:MobilityToxicityReactivity

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Structure and charge distribution of a water molecule

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Hydrogen atomsThere is excess positive charge here

Oxygen atom

Unshared electron pairs create negative charge

in these regions

The H2O molecule approximates a tetrahedron

Four-sided moleculeOxygen in the centerHydrogen at two cornersPolar molecule

Oxygen is electrophilic Electron affinity toward oxygen

Hydrogen region is more positive

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Water is an unusual fluid

Unusual fluid properties result from:Molecular shapeHydrogen bonds

Unusual properties include:DensityHeat capacityHeat conductivitySolvent properties

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Properties of water videoProperties of water

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Solutes in waterWhat happens when a substance (A)

dissolves in water?A-A bonds breakH2O-H2O bonds break

A-H2O bonds form

Hydrophilic Dissolution in water favored

Hydrophobic Dissolution in water not favored

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Ions in solution

Cations = positive charge (Na+, Ca2+, Fe3+,…)

Anions = negative charge (Cl-, SO42-,

PO43-,…)

Charged solute surrounded by and stabilized by water molecules

What are examples of uncharged dissolved substances?

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Electroneutrality

(anion charge) = (cation charge)

Charge balanceImportant fundamental equation Used to solve equilibrium problems

For example…

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General solubility rulesSee Table 1.2, page 8 of our text

What is the difference among soluble, slightly soluble, and insoluble?

Are the following compounds soluble in water? NaNO3

CaCO3

PbS

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Ways of expressing concentrationsMass/volume and mass/mass unitsGas phase concentrations

Ideal gas lawConcentrations represented by a single

elementHardness and alkalinityEquivalentsComposite parameters

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What is a mole?Abbreviation = mol

Avogadro’s number6.0231023

That many items make up 1.0 mol

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Solubility calculationsHow many grams of each of the

following compounds should be added to 1.0-L of water to prepare a 0.01 M solution?

Assume these substances dissolve completely

NaCl

Ca(OH)2

FeS

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10-2 M solution of NaCl requires:

10-2 (mol/L)

(22.99 g Na/mol + 35.45 g Cl/mol)

= 0.584 g/L

For Ca(OH)2 = 0.761 g/ L

For FeS = 0.879 g/L

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Gas phase concentrationsWe apply gas phase concentrations in

Chapter 7

Ideal gas lawPi = (ni / Vtotal) R T = Ci R T

Vi = (ni / Ptotal) R T

R = 8.314 J/mol.K = 0.0821 L.atm/mol.K

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What is the concentration of oxygen in the atmosphere?

C = P / RT

= 0.2095 / (0.0821 × 298)

{atm / (L.atm/mol.K) × K}

= 8.56 × 10-3 mol/L

32 g/mol = 0.27 g O2 / L air

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Mass or moles per volumeTypical units are:

mg/L, g/L, mol/L, mmol/L

Atomic weights listed inside the front cover of our text

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A lake contains 10-4 M NO3- (nitrate)

Express the concentration as mg NO3- / L

NO3- = 14 g N/mol + (316) g O/mol

= 62 g NO3-/mol

10-4 mol NO3- / L 62 g NO3

- / mol NO3-

= 6.2 10-3 g/L = 6.2 mg NO3

- / L

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Species expressed as a single elementFor example, nitrogen (N)

An important nutrient for growing corn

An important nutrient in eutrophication

Bacteria convert N compoundsDifferent species promote different

reactions NO3

- (nitrate)

NO2- (nitrite)

NH4+ (ammonia)

org-N (organic nitrogen)

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Concentrations from groundwater analyses:

NO3- = 12 mg/L

NO2- = 7 mg/L

NH+4 = 9 mg/L

What is the total N?

12 (mg NO3-/L) (1/62) (mmol NO3

-/mg NO3-) 1

(mmol N/mmol NO3-) 14 (mg N/mmol N)

= 2.7 mg NO3- as N/L (or mg NO3

--N/L)

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7 (mg NO2-/L) (1/46) (mmol NO2

-/mg NO2-) 1

(mmol N/mmol NO2-) 14 (mg N/mmol N)

= 2.1 mg NO2- as N/L (mg NO2

--N/L)

9 (mg NH4+/L) (1/18 mmol NH4

+/mg NH4+) 1

(mmol N/mmol NH4+) 14 (mg N/mmol N)

= 7 mg NH4+ as N/L (mg NH4

+-N/L)

Total N = 11.8 mg N/L

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Species expressed as a different compoundHardness = multivalent metal ions

Ca2+, Mg2+, Fe2+, Mn2+

Alkalinity = acid neutralizing capacity

CO32-, HCO3

-, OH-

Often reported as CaCO3 mg/L as CaCO3 (common expression)

mg as CaCO3 / L (logical expression)

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If water contains 35 mg Ca2+/L and 15 mg Mg2+/L, what is the hardness expressed as mg CaCO3 /L?

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3

135 1 8.73 10

40.08 10

mg Ca mol Ca mol hardness mol hardness

L mg Ca mol Ca L

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3

115 1 6.17 10

24.305 10

mg Mg mol Mg mol hardness mol hardness

L mg Mg mol Mg L

Total hardness = 1.49 x 10-3 mol/L

3 53 3 3

3

1.49 10 1 10 149mol CaCO mg CaCO mg as CaCOmol hardness

L mol hardness mol CaCO L

What is an equivalent concentration? It depends on the reaction

1 eq = 1 mole charge

1 eq = 1 mole H+

1 eq = 1 mole e-

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What is the normality (eq/L) of a 10-4 M

solution of Fe(NO3)3?

It depends on the reaction.

For precipitation of Fe(OH)3(s) the reaction is:

Fe3+ + 3 H2O = Fe(OH)3(s) + 3 H+

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4 4 410 3 3 10 3 10mol Fe eq eq

NL mol Fe L

For reduction of Fe3+ to Fe2+

Fe3+ + e- = Fe2+

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4 4 410 1 10 10mol Fe eq eq

NL mol Fe L

Pause for introductions…Interview a neighbor, report back to

class:NameBirthplaceAcademic backgroundWork experienceFamous person you want to have dinner

with

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Chemical reactivity or activityOverall tendency to participate in

reactionsNotation used in our text

Concentration: [i] = mole of i per L ci = mole of i per L

Activity: {i} = activity of i Dimensionless

{i} i ci

i activity coefficient

“represented by”

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How do solution properties affect reactivity?

Reactivity = f(solution properties)Dissolved solids can be:

Neutral (examples?)Positively charged (cation)Negatively charged (anion) Charges can be

Monovalent ( 1) Divalent ( 2) Trivalent ( 3) Greater (z)

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Ionic strengthA measure of ionic charge

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2 i iI c z ci = mole of i per L

zi = valence (charge) associated with species i

Solution includes 10-3 M NaNO3 and 10-4 M CaCl2. Assuming everything dissolves, what is the ionic strength?

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i ci zi ci zi2

Na+ 10-3 1 10-3

NO3- 10-3 -1 10-3

Ca2+ 10-4 2 4 10-4

Cl- 210-4 -1 210-4

I = 1.3 × 10-3

See Figure 1.6 on page 32 of our textAt this ionic strength:

1 0.96 monovalent

2 0.88 divalent

3 0.75 trivalent

Ionic strength effects are greater for higher charge ions…

Minimal effect on neutral solutes

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Activity depends onConcentrationEnvironmental components

Other constituents’ concentrationsTemperaturePressure (usually 1 bar)

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Chemical activity measured relative to a standard state

Standard state includes:ConcentrationReference environmental conditions

Also known as standard environmental conditions

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Standard state common definitions

StateStandard Conc.

T (C)P (bar)

Other

Solid Pure solid 25 1 -

Liquid

Pure liquid 25 1 -

Gas Pure gas 25 1

Ideal gas behavior

Solute

1.0 M 25 1

Infinite dilution

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Activity coefficients

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realsystem reactivity per moleof i

standardstate reactivity per moleof ii

Real

StandardState

1.01.0

ii i i

i

i ii iC

i CCi

i

Dimensions from these terms are not visible

= f(ionic strength)For example, the Davies equation:

A = 0.51 for water at T = 25CApplicable at I < 0.5 M

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0.52

0.5log 0.2

1i

IA z I

I

That symbol is 1.0, not I

Compare activity coefficients of Na+ and

Ca2+ at I = 10-1 and I = 10-4.

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I

Na+ Ca2+

Log Log

10-4 5.04 10-3 0.988

2.02 10-2 0.955

10-1 0.112 0.772 0.448 0.356

Chemical equilibriumEquilibrium occurs when:

No driving force to change composition

Consider generic reaction:

a A + b B c C + d D

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The forward reactionRate of conversion of A and B to C and D

can be described by:= kf {A}a {B}b

kf is the rate constant for the forward reaction

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Reverse reactionRate of conversion of C and D to A and B

can be described by:= kr {C}c {D}d

kr is the rate constant for the reverse reaction

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Net reaction rate is: rnet = kf {A}a {B}b – kr {C}c {D}d

Net rate is zero at equilibrium

kf {A}a {B}b = kr {C}c {D}d

Rearranging terms yields:

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c d

feqa b

req

k C DK

k A B

In general:

Q = activity quotientAt equilibrium Q = Keq Dynamic equilibrium

Forward and reverse reactions continueRates are equal

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c d

a b

C DQ

A B

What is pH? pH = - log{H+} - log[H+]

pH = 7 -log[H+] = 7Log[H+] = -7

[H+] = 10-7 mol/L

What is the range of pH values?

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Temperature effects Keq depends on the change in enthalpy

(Hr) associated with the reaction

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21,,

11lnln

12 TTR

HKK r

TeqTeq

How does water temperature affect pH?

Equilibrium expression:{H+} {OH-} = Kw = 10-14.0 (T = 25 C)

Charge balance in pure water:[H+] = [OH-]

What is the pH at T = 1 C?

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Hr = Hproducts – Hreactants

Reaction is: H2O = H+ + OH-

Hr = Hproducts – Hreactants

= 0 – 230.0 – (-285.83) = 55.83 kJ/mol.K

2

1

3,

, 1 2

1 1 55.83 10 1 1ln

8.314 298 274eq T r

eq T

K H

K R T T

2

2 1

1

,, ,

,

ln 1.97 exp( 1.97)eq Teq T eq T

eq T

KK K

K

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KT = 274 = 1.39x10-15

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1581.39 10

{ } { } 3.73 10{ }

H HH

pH = - log{H+} = 7.4

Combining chemical reactions

H2CO3 HCO3- + H+ Ka1 = 10-6.35

HCO3- CO3

2- + H+ Ka2 = 10-10.33

H2CO3 CO32- + 2 H+ Ka = 10-16.68

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Key conceptsExpressions for concentrations

Mass/volumeMol/volumeEquivalentsAs another compound

Activity and ionic strength effects

Equilibrium

Temperature effects

Combining chemical reactions

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Team surveyIf you are here…

Turn results in before you leave

If you are not here…Send results by Monday

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