Harris: Quantitative Chemical Analysis, Eight Edition

CHAPTER 07: ACTIVITY &

THE SYSTEMATIC TREATMENT OF QUILIBRIUM

- Activity Coefficients are related to the hydrated radius. - 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

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)

µ ≅ 0 M (red)

µ ≅ 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 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 atmosphere. - 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)

•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. (µ) (moles/L) The more highly charged an ion, the more it is counted.

Lewis and Randall

∑=i

2iiC

21 Zµ Ci : concentration of ith species

Zi : charge of ith species

Langelier

µ ≅ 2.5 × 10-5 × TDS TDS : Total Dissolved Solids (mg/L)

Russel

µ ≅ 1.6 × 10-5 × Specific conductance (µmho/cm)

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

γγγγ

γγγγ

⋅⋅

⋅=⋅⋅

== C

Kc , Concentration Equilibrium Constant

* Kc does not predict that there should be any effect of ionic strength on ionic reaction. To account for the effect of ionic strength (µ), the concentration [ ] should be replaced by activity { }. * Very commonly, tabulated equilibrium constants are not K but Kc, measured under a particular set of conditions. * If µ → 0, then Kc → K

7-2. Activity coefficients

a) Activity coefficients for electrolytes

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

- logγi = 0.51Zi2⋅µ1/2

* when µ < 5×10-3 M, TDS < 200 mg/L

a-2) Extended Debye-Hückel

)305/(10.51Z log- 2/1

2/12i

i µαµ

γ⋅+

= (at 25 °C)

α : hydrated radius (picometers = 10-12 m)

* when µ < 0.1 M, TDS < 4,000 mg/L

a-3) Günterlberg Approximation

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

2/1

2/12i

i 10.5Z log-

µµ

γ+

=

* when µ < 0.1 M, TDS < 4,000 mg/L

* This approximation is used to calculate γ for ions of various charge at temp. other than 15 °C .

a) Activity coefficients for electrolytes

a-4 ) Davies Equation

−

+⋅= µ

µµγ 2.0

1ZA log- 2/1

2/12ii

A ≅ 0.5 (at 25 °C)

* when µ < 0.5 M, TDS < 20,000 mg/L

Remark : There is no satisfactory theory that provides a good estimate of 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 calculated activity coefficient.

a) Activity coefficients for electrolytes

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 (γ± )

Theory : activity coefficient only for individual ions exists. Experiment : only the mean activity coefficient (γ±) can be derived from measurements because ions are available only in pairs.

γ±m+n = γ+m ⋅ γ-

n

For electrolyte Am⋅Bn

Ksp = [A]m⋅[B]n⋅γAm⋅γB

n

= [A]m⋅[B]n⋅γ±m⋅γ±n

= [A]m⋅[B]n⋅γ±m+n

b) Mean Activity Coefficients γ±m+n = γ+

m ⋅ γ-n

γ±m+n = γAm ⋅ γB

n

Or

C) Activity Coefficient of nonionic compounds (neutral molecules)

i) In aqueous solution

logγ = Ks⋅µ (Empirical equation) γ = 10Ks⋅µ > 1

Ks : salting out coefficient (≅ 0.01 ~ 0.15)

* If µ < 0.1 M, then γ ≅ 1. The activity of a neutral molecule is equal to its concentration if µ < 0.1

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

Example) {O2(aq)} = γO2[O2(aq)] = KH⋅PO2 = const.

At distilled water, γO2 = 1.0 But at salty water, [O2(aq)] decreases, and γO2 increases to more that 1.0

ii) In gases

{H2(g)} = γH2 ⋅PH2

fugacity fugacity coefficient

* For most gases at or below 1 atm, γ ≅ 1. Deviation from ideal gas law results in deviation of fugacity coefficient from unit (1.0)

C) Activity Coefficient of nonionic compounds (neutral molecules)

Example) O2(g) O2(aq) ⇌

KH = 1.29×10-3 at 25 °C Ks = 0.132 PH2O = 23.8 mmHg, XO2 = 0.21 (Volume fraction of O2 in dry air)

Determine the molar dissolved oxygen activity and concentration at 25°C for 1) distilled water 2) Sacramento River water (specific conductivity = 450 µmho) 3) Pacific Ocean water (µ = 0.7)

Example

i) distilled water

, 0 ∴=µ

22 760pressurevapor -sureAtmos.Pres

OO XP ×

=

( ) 203.021.0760

23.8-760=

=

{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

{O2(aq)} = γO2[O2(aq)] = KH⋅PO2 = const.

γ = 10Ks⋅µ = 1

{O2(aq)} ≅ [O2(aq)]

ii) Sacramento River water

continued

µ = 1.6 × 10-5 × 450 = 0.007 Assume that the partial pressure of oxygen is the same over the pacific ocean as it is over the Sacramento River. logγ = Ks × µ = 0.132 × 0.007 ∴ γ = 1.002

Example

{O2(aq)} ≅ [O2(aq)] = KH⋅PO

mg/L 4.8

M1062.2002.1

203.01029.1)]([O 43

2

=

×=××

=∴ −−

aq

mg/L 4.8 M1062.2203.01029.1)}({O 43

2

=×=××= −−aq

Example

2(aq)][O(aq)}{O 22 OH PXKγ ==

γ = 1.002

iii) Pacific Ocean water

µ = 0.7 , Ks = 0.132 logγ = 0.132 × 0.7 ∴ γ = 1.24

M1062.2203.01029.1)}({O 432

−− ×=××=aq

mg/L 4.8=

M1011.224.1

203.01029.124.1

)]([O 43

22 −

−

×=××

××

=∴ OH PKaq

mg/L 75.6=

Example

Remark : i) The activity of oxygen, {O2(aq)} = 8.4 mg/L ii) Wet chemical methods (Winkler Method) will produce a result of 8.4 mg/L for distilled and Sacramento water but 6.45 mg/L for the Pacific ocean water. iii) A membrane-covered, specific oxygen electrode (D.O meter) will 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 dissolved oxygen concentration?

Example

1) Charge Balance

Solutions must have zero total charge ∑ 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.

The equilibrium problem can be solved by working n equations and n unknowns.

n equations n-2 : chemical equilibrium conditions 2 : Charge balance Mass balance

7-4 Systematic Treatment of Equilibrium

2) Mass Balance

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

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

7-4 Systematic Treatment of Equilibrium

Example) Analysis of Vittel (France) : Mineral Water

mg/L M.W. mM

Na+ 3 23.0 0.13 K+ 2 39.1 0.05 Mg2+ 36 24.3 2×1.48 Ca+ 202 40.1 2×5.04

HCO3- 402 61.0 6.59

Cl- 7 35.5 0.20 NO3

- 6 62.0 0.10 SO4

2- 306 96.0 2×3.19

13.22 mM

13.27 mM

* Water Analysis : ± 2% Wastewater or Sea water Analysis : ± 5% generally acceptable

Error = (0.05/13.27)×100 = 0.38%

Systematic Treatment of Equilibrium

General Prescription

Step 1 : Write all the pertinent chemical equations. * Because of not knowing all the chemical reactions, we undoubtedly oversimplify many equilibrium problems.

Step 2 : Write all the charge balance Write 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 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 = 3.9×10-11 ⇌

ii) F- + H2O HF(aq) + OH- ⇌

iii) H2O H+ + OH- Kw = [H+][OH-] = 1.0×10-14 ⇌

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

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

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

11--

b 101.5 F][

][OH [HF] K ×==

* 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 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 ?

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

Assumption ; 2[Ca2+] >> [H+] [F-] >> [OH-] ∴ 2[Ca2+] = [F-] log2 + log[Ca2+] = log[F-] ∴ log[F-] = log[Ca2+] + 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.

From Charge Balance,

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

[CO2(aq)] ≅ [H2CO3*] = KH⋅PCO2 = 0.0344×10-3.5 = 1.0×10-5 M

7-*32

-3

a1 104.45 ]COH[

]HCO][[H K ×==

+

11-

3

-23

a2 104.69 ]HCO[

]CO][[H K ×== −

+

-14-w 101.0 ]][OH[H K ×== +

charge balance [H+] = [HCO3

-] + 2[CO32-] + [OH-]

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

2-], [OH-] equation : 4

Solving the equations , , , , then pH = 5.7

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

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

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

Ka1 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/Ka2 ⇌

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

a2sp

-3

2-3

2

K/K][HCO ][Ca

K][HCO ][Ca ][H

+++ ==eq

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

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

- and Ca2 [H+] : actual [H+] in water

Langelier Saturation Index, I = pH - pHeq

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

∑=i

ii ZC 2

21 µ C = molar conc. of ion i

Z = valence of ion i

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

pHeq : 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+]

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

Erosion of Marble by Acidic Rain

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