Harris: Quantitative Chemical Analysis, Eight Edition

CHAPTER 07: ACTIVITY &

THE SYSTEMATIC TREATMENT OF QUILIBRIUM

ACTIVITY

- Activity Coefficients are related to the hydrated radius.y y- The rate at which an ion diffuses through solution or migrates in an electric field

depends on the hydrated radius, not on the size of the bare ion.

Activity

7-1. The Effect of Ionic Strength on Solubility of Salts

< Color plate 2 >

Fe(SCN)2+ Fe3+ + SCN-

(red) (pale yellow) (colorless)

FeCl3 (1mM)KSCN (1.5mM)

FeCl3 (1mM)KSCN (1.5mM)

0 M(red)

0.4 M (by adding KNO3)(the red color fades)

•Adding any “inert” salt to any sparingly soluble salt (or complex)increases the solubility of the latterthe solubility of the latter. Ionic Atmosphere Model

Color Plate 2

Ionic Atmosphere Model

- An ionic atmosphere surrounds ions in solution.- The greater the ionic strength of a solution, the higher the charge in the ionic

hatmosphere. - Each ion-plus-atmosphere contains less net charge and there is less attraction

between any particular cation and anion.

7-1. The Effect of Ionic Strength on Solubility of Salts

Fe(SCN)2+ Fe3+ + SCN-

(red) (pale yellow) (colorless)(red) (pale yellow) (colorless)

•Adding any “inert” salt to any sparingly soluble salt (or complex)increases the solubility of the latter.

Solubility of potassium hydrogen tartrate (p.144)

What do we mean by “ Ionic Strength ” ?

Ionic Strength : a measure of the total concentration of ions in solution.() (moles/L) The more highly charged an ion, the more it is counted.

Lewis and Randall

1i

2iiC

21 Z Ci : concentration of ith species

Zi : charge of ith species

Langelier

2 5 10-5 TDS TDS : Total Dissolved Solids (mg/L) 2.5 10 TDS TDS : Total Dissolved Solids (mg/L)

Russel

1.6 10-5 Specific conductance (mho/cm)

7-2. Activity coefficients

Thermodynamic equilibrium constant (or Real equilibrium constant) (K)

bB

aA

dD

cC

bB

aA

ba

dD

cdc

ba

dc

Kc B][[A]D][[C]

B}{A}{D}{C}{ K

C

BABAB][[A]B}{A}{

K C i E ilib i CKc , Concentration Equilibrium Constant

* K does not predict that there should be any effect of ionic strength on ionic Kc does not predict that there should be any effect of ionic strength on ionicreaction. To account for the effect of ionic strength (), the concentration [ ]should be replaced by activity { }.

* Very commonly, tabulated equilibrium constants are not K but Kc, measuredunder a particular set of conditions.

* If 0, then Kc K

a) Activity coefficients for electrolytes

a-1) Debye-Hückel (Debye-Hückel Limiting Law

- logi = 0.51Zi21/2logi 0.51Zi

* when < 510-3 M, TDS < 200 mg/L

a-2) Extended Debye-Hückel

0 51Z 2/12

)305/(10.51Z

log- 2/1

2/12i

i

(at 25 C)

: hydrated radius (picometers = 10-12 m)

* when < 0 1 M TDS < 4 000 mg/L* when < 0.1 M, TDS < 4,000 mg/L

a) Activity coefficients for electrolytes

a-3) Günterlberg Approximation

From Extended Debye-Hückel Theory,(if temp = 15 C and = 300 10-12 m (pm)

2/12i

i0.5Z

log-

2/1i 1 log

* when < 0.1 M, TDS < 4,000 mg/L

* This approximation is used to calculate for ions of various chargeh h 15 Cat temp. other than 15 C .

a) Activity coefficients for electrolytes

a-4 ) Davies Equation

2/1

2.0

1ZA log- 2/1

2/12ii

A 0.5 (at 25 C)

* when < 0.5 M, TDS < 20,000 mg/L

Remark : There is no satisfactory theory that provides a good estimatef th ti it ffi i t f i i t th f t th 0 5 Mof the activity coefficient for ionic strengths of greater than 0.5 M

: The estimation of ion size places a limit on the accuracy of the p ycalculated activity coefficient.

a) Activity coefficients for electrolytes

Table 7-1

Table 7-1

Fig. 7-4. Activity coefficients for differently charged ions with a constant hydrated radius of 500 pm

b) Mean Activity Coefficients ( )

Theory : activity coefficient only for individual ions exists.Experiment : only the mean activity coefficient () can be derived

from measurements because ions are available only in pairsin pairs.

For electrolyte A BFor electrolyte AmBn

K [A]m [B]n m nKsp = [A]m[B]nAmB

n

= [A]m[B]nmn

= [A]m[B]nm+n

m+n = +m -

n

b) Mean Activity Coefficients

m+n = +m -

n

O

m+n = Am B

n

Or

C) Activity Coefficient of nonionic compounds (neutral molecules)

i) In aqueous solution

log = Ks (Empirical equation) = 10Ks > 1

Ks : salting out coefficient ( 0.01 ~ 0.15)

* If < 0.1 M, then 1. The activity of a neutral molecule is equal to itsconcentration if < 0.1

* Salting out effect : the effect of decreasing the solubility of molecule species (O2(aq)) by increasing salt concentrationspecies (O2(aq)) by increasing salt concentration

Example) {O2(aq)} = O2[O2(aq)] = KHPO2 = const.

At distilled water, O2 = 1.0At distilled water, O2 1.0But at salty water, [O2(aq)] decreases, and O2 increases to more that 1.0

C) Activity Coefficient of nonionic compounds (neutral molecules)

ii) In gases

{H2(g)} = H2 PH2

fugacity fugacity coefficientfugacity fugacity coefficient

* For most gases at or below 1 atm, 1.g

Deviation from ideal gas law results in deviation of fugacity coefficientfrom unit (1.0)

Example

Example) O2(g) O2(aq)

KH = 1.2910-3 at 25 CKs = 0.132P 23 8 H X 0 21 (V l f ti f O i d i )PH2O = 23.8 mmHg, XO2 = 0.21 (Volume fraction of O2 in dry air)

Determine the molar dissolved oxygen activity and concentrationat 25C for 1) distilled water

2) Sacramento River water (specific conductivity = 450 mho)2) Sacramento River water (specific conductivity = 450 mho)3) Pacific Ocean water ( = 0.7)

Example

i) distilled water

, 0

{O2(aq)} = O2[O2(aq)] = KHPO2 = const.

= 10Ks = 1

{O2(aq)} [O2(aq)]

22 760pressurevapor -sureAtmos.Pres

OO XP

203021023.8-760

203.021.0

760

{O2(aq)} [O2(aq)] = KHPO2{O2(aq)} [O2(aq)] KH PO2

= 1.29 10-3 PO2

= 1.29 10-3 0.203= 2.62 10-4 M= 8.4 mg/L

Example

ii) Sacramento River water

= 1.6 10-5 450= 0.007

Assume that the partial pressure of oxygen is the same over the pacific it i th S t Riocean as it is over the Sacramento River.

log = K log Ks = 0.132 0.007

= 1.002{O2(aq)} [O2(aq)] = KHPO

continued

Example

2(aq)][O(aq)}{O 22 OH PXKγ

mg/L48M1062.2203.01029.1)}({O 43

2

aq

mg/L4.8

= 1.002

M1062.2002.1

203.01029.1)]([O 43

2

aq

mg/L 4.8

Example

iii) Pacific Ocean water

= 0.7 , Ks = 0.132

log = 0.132 0.7 = 1.24

M1062.2203.01029.1)}({O 432

aq

mg/L 4.8

203010291 43 OH PK

M1011.224.1

203.01029.124.1

)]([O 42

2

OH PKaq

mg/L756 mg/L75.6

Example

Remark : i) The activity of oxygen, {O2(aq)} = 8.4 mg/L

ii) Wet chemical methods (Winkler Method) will produce a result of8.4 mg/L for distilled and Sacramento water but 6.45 mg/L for the

ifiPacific ocean water.

iii) A membrane-covered, specific oxygen electrode (D.O meter) will) , p yg ( )produce an identical reading for three solutions.

iv) How a fish fees with respect to the dissolved oxygen content of these waters ? Does it care about dissolved oxygen activity or dissolvedDoes it care about dissolved oxygen activity or dissolvedoxygen concentration?

7-4 Systematic Treatment of EquilibriumThe equilibrium problem can be solved by working

n equations and n unknowns.

n equationsn-2 : chemical equilibrium conditions

2 Ch b l

1) Charge Balance

n equations 2 : Charge balanceMass balance

1) Charge Balance

Solutions must have zero total charge i i h i h positive charges = negative charges

n1[C1] + n2[C2] + ···· = m1[A1] + m2[A2] + ····

[Ci], ni = concentration, charge of i th cation[Ai], mi = concentration, charge of i th anion

* Activity coefficients do not appear in the charge balance.

7-4 Systematic Treatment of Equilibrium

2) Mass Balance

Example) When MA and HA both are added to a solution

C = [HA] + [A-]CT,A = [HA] + [A ]

CT,A : sum of the moles of HA and MA added per liter of solution

* Activity coefficients do not appear in the mass balance* It really refers to conservation of atoms, not to mass

Example) Analysis of Vittel (France) : Mineral Water

mg/L M W mMmg/L M.W. mM

Na+ 3 23.0 0.13K+ 2 39.1 0.05K 2 39.1 0.05Mg2+ 36 24.3 21.48Ca+ 202 40.1 25.04 Error = (0.05/13.27)100

0 38%

HCO3- 402 61.0 6.59

13.22 mM = 0.38%

Cl- 7 35.5 0.20NO3

- 6 62.0 0.10SO 2 306 96 0 2 3 19SO4

2- 306 96.0 23.1913.27 mM

* W t A l i 2%* Water Analysis : 2%Wastewater or Sea water Analysis : 5% generally acceptable

Systematic Treatment of Equilibrium

General Prescription

S 1 i ll h i h i l iStep 1 : Write all the pertinent chemical equations.* Because of not knowing all the chemical reactions,

we undoubtedly oversimplify many equilibrium problemswe undoubtedly oversimplify many equilibrium problems.

Step 2 : Write all the charge balanceW i h b lWrite the mass balance

Step 3 : Write the equilibrium constant for each chemical reaction.- This is one step in which activity coefficients appear.- Although it is proper to write all equilibrium constants in terms of

activities, the algebraic complexity of manipulating the activity coefficients often obscures the chemistry of a problemcoefficients often obscures the chemistry of a problem

7-5 Applying the systematic Treatment of Equilibrium

The Dependence of Solubility on pH.(Coupled Equilibria ; the product of one reaction is a reactant in the next reaction)

i) CaF2(s) Ca2+ + 2F- Ksp = [Ca2+][F-]2 = 3.910-11

ii) ( ) 11--

101 5][OH[HF]Kii) F- + H2O HF(aq) + OH-

iii) H2O H+ + OH- Kw = [H+][OH-] = 1.010-14

11-b 101.5

F][][[ ]K

) 2 w [ ][ ]

iv) Mass Balance ; [F-] + [HF] = 2[Ca2+]

v) Charge Balance ; [H+] + 2[Ca2+] = [OH-] + [F-]v) Charge Balance ; [H ] + 2[Ca2 ] = [OH ] + [F ]

Five unknowns [H+], [OH-], [Ca2+], [F-], [HF]

* To solve the problems is no simple matter for these 5 equations.- If we fix the pH, it becomes simpler.

In this case we don’t use charge balance equations (v) because we don’t knowIn this case, we don t use charge balance equations (v) because we don t know charge balance exactly due to the addition of chemicals to get, for example, pH=3.

Fig. pH dependence of the concentration of Ca 2+, F- and HF in a saturated solution of CaF2

-

What is the pH of saturated solution with CaF2 ?

From Charge Balance

[H+] + 2[Ca2+] = [OH-] + [F-]

From Charge Balance,

Assumption ; 2[Ca2+] >> [H+][F-] >> [OH-]

2[Ca2+] [F-] 2[Ca2+] = [F-]log2 + log[Ca2+] = log[F-]

log[F-] = log[Ca2+] + 0.3 log[F ] log[Ca ] + 0.3

Find a point on pC-pH diagram (Fig. 9-3) where log[Ca2+] is located 0.3 log unit vertically below log[F-]. pH 7.11, Check the assumptions.

What is the pH of distilled water in equilibrium with the air ?

[CO (aq)] [H CO *] = K P = 0 034410-3 5 = 1 010-5 M[CO2(aq)] [H2CO3*] = KHPCO2 = 0.034410 3.5 = 1.010 5 M

7--3

1 104 45]HCO][[H

K

*32

a1 104.45]COH[

K

11--2

3 104 69]CO][[HK

3a2 104.69

]HCO[ K

-14-w 101.0 ]][OH[H K w ]][[

charge balance[H+] = [HCO -] + 2[CO 2-] + [OH-] [H+] = [HCO3 ] + 2[CO3

2 ] + [OH ]

unknowns : [H+], [HCO3-], [CO3

2-], [OH-]equation : 4

equation : 4

Solving the equations , , , , then pH = 5.7

* What is the pH of distilled water in equilibrium with the air ?

* About 100 years ago, Friedrich Kohlrausch measured conductivity of water.In removing ionic impurities from water, they found it necessary to distillthe water 42 consecutive times under vacuum to reduce conductivity to

li i i la limiting value.

Box 6-4. Carbonic Acid (p. 134)

K f H CO i b t 102 t 10 4 ti ll th K f th b li idKa1 for H2CO3 is about 102 to 10 4 times smaller than Ka for other carboxylic acid.

Erosion of Marble by Acidic Rain

Erosion of Marble by Acidic Rain

Erosion of Marble by Acidic Rain

Water Stabilization

CaCO3(s) Ca2+ + CO32- Ksp

H+ + CO32- HCO3

- 1/Ka23 3

CaCO3(s) + H+ Ca2+ + HCO3- K= Ksp /Ka2

-3

2-3

2

K/K][HCO ][Ca

K][HCO ][Ca ][H

eq

a2sp K/KKq

[H+]eq: [H+] in water if it were in equilibrium with CaCO3(s) at the existing solution concentration of HCO3

- and Ca2

[H+] > [H+] d t t d (di l ti )

so ut o co ce t at o o CO3 a d Ca[H+] : actual [H+] in water

[H+] > [H+]eq undersaturated (dissolution)[H+] < [H+]eq supersaturated (precipitation)

Erosion of Marble by Acidic Rain

Langelier Saturation Index, I = pH - pHeq

I < 0 undersaturated (dissolution)I < 0 undersaturated (dissolution)I > 0 supersaturated (precipitation)

pHeq = log Ksp/Ka2 – Log [Ca2+] – log [HCO3-]

pH : pH of water if it were in equilibrium with CaCO3(s) at the existingpHeq : pH of water if it were in equilibrium with CaCO3(s) at the existing solution concentration of HCO3

- and Ca2+, pHeq = -log [H+]eq.pH : actual pH of water, pH = -log [H+]p p p g

Ksp, Ka2 must be adjusted for temperature and ionic strength to Ksp’ and Ka2’

i

ii ZC 2

21 C = molar conc. of ion i

Z = valence of ion i

pHeq = (pKa2’ – pKsp’) + pCa2+ + pHCO3-