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Acid-Base Reactions and Equilibrium

By Brnsted-Lowry

1) Acid is proton (H+) donor or (OH-) accepter

- substance which can donate a hydrogen ion (proton), H+.

2) Base is proton (H+) accepter or (OH-) donor- substance which can accept a hydrogen ion (proton), H

+.

Note:

By Lewis

1) Acid is an electron pair acceptor- A Lewis acid is a substance which can accept an electron pair from a Lewise base to form a covalent bond.

2) Base is an electron pair donor

For example, in the formation of an ammonium ion, the proton (H+) is the Lewis acid accepting the unshared electron pair on the ammonia.

NH3 + H2O NH4+ + OH-

H H

.. ..

H : N : + H : O : .. ..

H

Ammonium ion

Lewis base Lewis acid

H + H -.. ..

H : N : H + : O :

.. ..H

The structure above is incomplete!

----------------------------NSEN

Acid-Base and Salts(Deutsch and Whitney, 1972)

- Acids, bases and salts are among the most important of all chemical compounds because they are electrolytes and they dissociate intoions when dissolved in water.

- Acids, bases, and salts are all electrolytes. Their solutions conduct electricity.

Acids

- An acid is a compound which produces hydronium ion H3O+(hydrogen ions H+) in water.

- In water, all ions are surrounded by the polar water molecules. Thus it is common to refer to the hydronium ion H3O+concentration,

rather than the hydrogen ion H+concentration in water.

- The hydronium ion H3O+is simply a hydrogen ion H+attached to a water molecule.

- e.g., sulfuric acid H2SO4dissociates to form hydrogen and sulfate ions according to the following equation:

H2SO4 H+ + SO4

2-

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Bases

- A base is a compound which produces hydroxide ions or hydroxyl ions OH-in water.- Some bases contain hydroxyl ions (OH-) but others do not. Bases that do not contain hydroxide ions will react with water molecules to

form hydroxyl ions (OH-).

Ammonia NH3 + H2O NH4+ + OH

Methylamine CH3NH2 + H2O CH3NH3+ + OH

Aniline C6H5NH2 + H2O C6H5NH3+ + OH

- e.g., sodium hydroxide, NaOH dissociates to form sodium ions and hydroxyl ions according to the following equations:

NaOHNa + + OH -

- If the concentration of hydrogen ions H+ (or hydronium ions) exceeds the concentration of hydroxyl ions OH -, the solution is acidic.

- If the concentration of hydroxyl ions exceeds the concentration of hydrogen ions (or hydronium ions), the solution is basic or alkaline.

- If the concentrations of two ions are equal, the solution is neutral.

----------------------------

pH

- Since pH has a significant effect on the chemical reactions which are of concern in the operations of nuclear power plants, control ofpH is very important in the operations of the plants.

The pH of a solution is defined as:

pH = - log {H+} - log [H+]

{H+} [H

+] = 10

pH

The pOH of a solution is defined as:

pOH = - log {OH-} - log [OH

-]

[OH-] {OH

-} = 10

-pOH

Since

{H+}{OH

-} = 1 x 10

-14

pH + pOH = 14

Pure water is neutral (pH = 7) at 25C.

pH + pOH = 7 + 7 = 14

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Example: What is the hydrogen ion concentration [H+] and the hydroxyl ion concentration

[OH-] in a solution with a pH of 5.5? Ignore the ionic strength effect.

a) [H+] {H+} = 10-pH= 10-5.5 = 3.16 x 10-6M (moles/L)

b) KW= {H+}{OH

-} [H

+] [OH

-] = 10

-14

[OH-] = 10-14/[H+] = 10-14/ (3.16 x 10-6) = 3.16 x 10-9

or

pH + pOH = 14; thus pOH = 14 pH = 14 5.5 = 8.5

[OH-] {OH-} = 10-pOH = 108.5 = 3.16 x 10-9M (moles/L)

Fig 1-9 shows the pH of various common substances.

Neutralization

- the reaction of the hydrogen ions H+(or hydronium ion, H3O+) of an acid and the hydroxyl ions OH- of a base to form undissociated

molecules of water.- e.g., when dilute solutions of hydrochloric acid HCl and sodium hydroxide NaOH are mixed, the following reaction occurs:

H3O+ + Cl - + Na+ + OH

2H2O + Na+ + Cl-

--------------------------

Acid dissociation equilibrium constant, Ka(greater is Ka, greater is acidity)

HCl H+

+ Cl-

{H+}{Cl- }

Keq= --------------- = Ka Ka= acid dissociation equilibrium constant

{HCl }Ka= 10

3

pKa= 3 (pKa = - log Ka)

H2SO4 H+ + HSO4

- pKa1= 3

HSO4- H

+ + SO4- pKa2 = +1.96

{H+

}{HSO4-

}Ka1= ----------------- Ka1= 10

3= 1stacid dissociation equilibrium constant

{H2SO4} pKa1= 3

{H+}{SO4-}

Ka2= ---------------- Ka2= 10-1.96

= 2nd

acid dissociation equilibrium constant

{HSO4-} pKa2= +1.96

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Base dissociation equilibrium constant, Kb

NaOH Na+ + OH

-(pKa = 14.77)

{Na+}{OH-}

Keq= ----------------- = Kb Kb= base dissociation equilibrium constant

{NaOH } Kb= 1 since pKb= 0

The large Kavalues indicate that the acid has a strong tendency to donate a proton to water; i.e.,

it is a strong acid.

The large Kbvalues indicate that the substance has a strong tendency to accept a proton from

water; i.e., it is a strong base. (SJ, p. 90-91)

1. Strong Acid vs. Strong Base

a. Strong Acids- usually means it dissociate (or ionize) completely (almost completely)

e.g., Table 4-1

Perchloric acid HClO4 pKa = 7

HCl pKa = 3H2SO4 pKa = 3

HNO3 pKa = 0

Hydronium ion H3O+

pKa = 0

b. Strong Base

- usually means it dissociate (or ionize) completely- e.g.,

Sodium hydroxide NaOH pKa = 14.77

Potassium hydroxide KOH pKa = 16

(Note: pKa + pKb = 14)

2. Weak Acids - partial dissociation

CH3COOH acetic acid (HAc) pKa =HCOOH formic acid pKa =H2CO3 carbonic acid pKa =H2S hydrogen sulfide pKa =

HCN hydrocyanic acid pKa =NH4

+ ammonium pKa =

* Strong acid to weak acid series is really a continuum (see pages 90 & 91 in SJ)

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Example: If you put 10

-3moles HCl in 1 L of water, HCl ionizes completely in water. What

will be pH of the water? Neglecting the ionic strength effect. pKa of HCl = 6.1

HCl H+ + Cl-

10-3M 10-3M 10-3M

H2O H+ + OH-10-7 10-7

pH = - log {H+} - log [H

+] = - log [10

-3] = 3 Thus, 10

-3M HCl gives pH 3.

If you put H2SO4in water, H2SO4ionizes in water (pKa1= -3, pKa2 = +1.96)

H2SO4 H++ HSO4

- pKa1= -3

HSO4- H++ SO4

- pKa2 = +1.96

only the first H+is strong

Example: If you put 10-3mole of NaOH in 1 L of water, NaOH dissociates completely. What

will be pH of the water? Neglecting the ionic strength effect.

NaOH Na+ + OH-

10-3 10-3 10-3

H2O H+ + OH

-

10-7 10-7

pOH = - log {OH-} - log [OH

-] = - log [10

-3] = 3

pOH + pH = 14

pH = 14 pOH

pH = 14 - 3 = 11 Thus, 10-3

M NaOH gives pH 11.

Example: Calculate the pH of the following solutions (advanced, do it later).

Perchloric acid, HClO4(pKa = 7)

a) 1 x 10-3M HClO4b) 1 x 10-8M HClO4

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(Solution)

a) For 1 x 10-3M HClO4

HClO4 H+ + ClO4

H2O H+ + OH

-

[H+][ClO4]

Ka = ------------------ = 1 x 107 (1)

[HClO4]

Kw= [H+][OH

-] = 10

-14 (2)

Mass Balance

CT, ClO4 = [HClO4] + [ClO4

] = 10-3 (3)

Charge Balance

[H+] = [ClO4] + [OH-] (4)

From the mass balance (3), CT, ClO4 = [HClO4] + [ClO4

] = 10

-3

Since HClO4 is a very strong acid, it completely dissociates, [HClO4] is negligible

CT, ClO4 = [ClO4

] = 10-3

From charge balance (4), [H+] = [ClO4

] + [OH

-]

Kw

[H+

] = [ClO4

] + ----------[H

+]

10-14

[H+] = 10

-3 + ---------- = 10

-3 + 10

-11= 10

-3

10-3

pH = 3 error < 5%

b) 1 x 10-8

M - HClO4

Since CT

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[H+][ClO4

]

Ka= ------------------ = 1 x 107 (1)

[HClO4]

Kw= [H+][OH-] = 10-14 (2)

Mass Balance

CT, ClO4 = [HClO4] + [ClO4] = 10-8 (3)

Charge Balance

[H+] = [ClO4

] + [OH

-] (4)

From the mass balance (3), CT, ClO4 = [HClO4] + [ClO4

] = 10

-8

Since HClO4 is a very strong acid, it completely dissociates, [HClO4] is negligible

CT, ClO4 = [ClO4

] = 10-8

From charge balance, [H+] = [ClO4

] + [OH

-]

Kw

[H+] = [ClO4

] + ----------

[H+]

10-14

[H+] = 10

-8 + ---------

[H+]

[H+

]2

- 10-8

[H+

] - 10-14

= 0

10-8

+ [(10-8

)2 + (4 x 10

-14)]

1/2

[H+] = --------------------------------------- = 1.05 x 10

-7pH = 6.98

2

Example: Calculate the pH of the following solutions (advanced, do it later).

Sodium hydroxide, NaOH (totally dissociates)

a) 1x 10-5

M NaOH

NaOH Na+ + OH

-

10-5

H2O H+ + OH

-

Kw= [H+][OH-] = 10-14 (1)

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Mass Balance

CT, Na = [Na+] = 10-5 (2)

Charge Balance

[Na+] + [H

+] = [OH

-] (3)

(10-5) + [H+] = [OH-]

Substitute [OH-] = [H

+] + (10

-5) into (1)

[H] ([H+] + 10-5) = 10-14

[H+]

2 + 10

-5 [H] - 10

-14 = 0

( ) ( ) ( )( )2

5 5 14

1010 10 4 1 10

9.999 102(1)

H x

+

= =

pH = 9.00

b) 1x 10-9 M NaOH

NaOH Na+ + OH-

10-9

H2O H+ + OH-

Kw= [H+][OH

-] = 10

-14 (1)

Mass Balance

CT, Na = [Na+] = 10

-9 (2)

Charge Balance

[Na+] + [H

+] = [OH

-] (3)

(10-9) + [H+] = [OH-]

Substitute [OH-] = [H+] + (10-9) into (1)

[H] ([H+] + 10

-9) = 10

-14

[H+]

2 + 10

-9 [H] - 10

-14 = 0

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( ) ( ) ( )( )2

9 9 14

10 810 10 4 1 10

9.999 10 9.95 102(1)

H x x

+

= = =

pH = 7.00

3. Conjugate acid vs. Conjugate base

Bronsted-Lowry said:"Removal of a proton from any acid (weak acid) produces its conjugate base"

a. When hydrocyanic acid HCN is added in water, HCN partially dissociates

Hydrpgen cyanide Cyanide ionHCN CN

- + H+

weak acid conjugate base

At equilibrium,

{CN-}{H

+}

(1) Ka= ---------------- = acid dissociation equilibrium constant{HCN}

Ka= 10-9.3

( PKa= 9.3 )

b. Addition of a proton (H+) to conjugate base produces conjugate acid.

- remove H+from H2O

- when cyanide ion (CN-) is added in water as NaCN, NaCN dissociates completely to

form HCN

NaCN(aq) Na+ + CN-

CN- + H2O HCN + OH

-

conjugate acid

At equilibrium,

{HCN}{OH-} {HCN}{OH

-}

Kb= ------------------ = ------------------ = base dissociation equilibrium constant{CN-}{H2O} {CN

-}

Kb= 10-4.7

(pKb= 4.7) Note pKb= - log Kb

See Table 4-1 Acidity and Basicity Constants for substances in Aqueous Solution. (SJ, p. 446)

[10-3M NaCN in solution is identical to 10-3M NaOH (strong base) plus 10-3M HCN (weak acid) ]

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{CN-}{H+} {HCN}{OH-}

KaKb = ---------------- ------------------- = {H+}{OH

-} = Kw = 10

-14 at 20C

{HCN} {CN-}

Ka Kb= Kw

(10-9.3)(10-4.7) = 10-14 or (5.01 x 10-10) (2.02 x 10-5) = 10-14

Kw= water dissociation constant = molar ion product of water

pKa + pKb = PKw

9.3 + 4.7 = 14

c. Dissociation of water

H2O H+ + OH-

{H+}{OH

-}

Keq = ---------------- = {H+}{OH

-} = Kw = 10

-14

{H2O}

{H+}{OH

-} = Kw = 10

-14at 20C

a. NH4+is added in water as NH4Cl

NH4Cl NH4+ + Cl-

NH4+ NH3(aq) + H

+

weak acid conjugate

base

(NH3consumes H+)

{NH3}{H+}

Ka = --------------- = 10-9.3

{NH4+}

- Removal of a proton from acid produces its conjugate base.

b. NH3is added in water

NH3 + H2O NH4+ + OH

-

weak base conjugateacid

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{NH4+}{OH

-}

Kb= ------------------- = 10-4.7

{NH3}

- Addition of a proton to conjugate base produces conjugate acid.

c. KaKb= (10-9.3) (10-4.7) = 10-14 = Kw

--------------------------------------------------------

Q: How do you add a conjugate base like CN-to a system?

A: Add as the salt of a weak acid - a conjugate base is the salt of the weak acid.

HCN CN- + H+

weak acid conjugate base

{H+}{CN-}

Keq= -------------- = Ka = 10-9.3 PKa = 9.3 at 25C

{HCN}

Ka = acid dissociation constant (greater is Ka, greater is acidity)

Add CN- as NaCN

NaCN Na+ + CN

CN + H2O HCN + OH

weak base conjugate acid

{HCN}{OH}

Keq = -------------------- = Kb = 104.7

{CN}

PKa = 4.7 at 25 C

Kb= base dissociation constant (greater is Kb, greater is the weak base)

{H+}{CN} {HCN}{OH}

KaKb= --------------- --------------------{HCN} {CN}

= { H+}{OH} = Kow

= (10-9.3) (104.7) = 1014

Kw = water dissociation constant = molal ion product of water

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-------------------------

NSEN

3-3.1 Acid-Base Equilibria of Dilute Solutions(Cohen, p. 44)

Since free protons cannot be expected to exist in solution to any significant extent, the acidic or basic properties of a solute cannot be realized

unless the solvent itself possesses acid or basic properties; i.e., can accept or donate protons. Thus, the general case for significant acid or basicsolutions must be written

HA H+ + A-

A- + H2O HA + OH-

--------------------------------------------------------------HA + H2O HA + H

+ + OH-

- Water can act as both a proton acceptor and a proton donor (amphiprotic) and therefore its solution can show both acidic and basic properties.

Water dissociates to protons (hydronium) and hydroxyl ions

2H2O H3O+ + OH-

hydronium

with similar reaction for heavy water.

The equilibrium for this reaction is written

{H3O+

}{OH-

} H3O+

[H3O+

] OH-

[OH-

]Kw= -------------------- = ---------------------------------

{H2O}2 H2O

2[H2O]2

a H3O+ C a OH

-

= ------------------------ (3.12)

a2H2O

where or a refer to activity coefficient

For dilute solutions, up to 10-4M, the activities are nearly equal to the molal concemtrations, m, and the water activity is essentially constant.Thus,

Kw= [H3O+] [OH-] or mH+CmOH-= Kw (3.13)

Table 3.1 and Fig. 3.2 present three sets of data for the molal ion product of pure water.

Table 3.1 Molar Ion Product of Water (Cohen, 1980)

Jones (1958) Noyes (1910) Harned and Owen (1950)

Temperature Temperature TemperatureoF

oC Kw x 10

14 oF

oC Kw x 10

14 oF

oC Kw x 10

14

75 24 0.9 64.4 18 0.64 64.4 18 0.57

200 93 43 212 100 55 212 100 54.6

300 149 208 313 156 268 313 156 222

400 204 500 424 218 645 424 218 403

480 249 645 583 306 340 583 306 308

540 282 628

* Some possibility exists that the experimental results of Noyes et al. (1910) may be in error (Cohen, 1980).

log.

. . ( )KwT

T= + 447099

60875 00170607 (3.14)

where T = temperature (F, R ?)

????? The computed results using the above Eqn.(3.14) do not match with the data presented in Table 3.1 ?????

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Figure 3.2. Molal ion product of water.

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Solving Acid-Base Equilibrium Problems

Steps to solve equilibrium problems.

Write:

1.Equilibrium ReactionsandEquations

2.Mass balance: CT= concentration of all species (of the compounds involved)

3. Charge balance: (positive charge) = (negative charge)

4. Proton balance(proton condition): (proton gainer) = (proton loser)

Summation of all species Summation of all species

that gain a proton = that lose a proton

in solution in solution

Example: Soda ash (Na2CO3) is added to pure water.

Na2CO3 2 Na+ + CO3

2-

1. a. Equilibrium reactions:

H2CO3 H+ + HCO3

-

HCO3-

H+

+ CO32-

H2O H

+ + OH-

b. Equilibrium equations:

{H+}{HCO3-}

Ka1= ------------------ (1){H2CO3}

{H+}{CO3

2-}

Ka2= ----------------- (2)

{HCO3-

}

Kw= {H+} {OH

-} (3)

2. Mass balance:

CT,CO3 = [H2CO3] + [HCO3-] + [CO3

2-] (4)

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3. Charge balance:

[Na+] + [H+] = 2[CO32-] + [HCO3

-] + [OH-] (5)

4. Proton condition: zero level / reference species; CO3 2-, H2O

2[H2CO3] + [HCO3-] + [H

+] = [OH

-] (6)

Note H+is same as H3O

+

6 unknowns and 6 equations

Note:

HCO3 + H2O H2CO3 + OH

(1)

CO32 + H2O HCO3

- + OH (2)

[H2CO3][OH ]

Kb1= --------------------- (1)

[HCO3] [H2O]

[HCO3

][OH

]Kb2= --------------------- (2)

[CO32 ] [H2O]

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Example: NaAsO2(herbicide) is added to distilled water.

NaAsO2 Na+ + AsO2

1) Equilibrium reactions

AsO2 + H2O HAsO2 + OH

H2O H+

+ OH

[HAsO2] [OH] Kw

Equilibrium eqn: Kb= ------------------- = ------ (1)

[AsO2] [H2O] Ka

(since KaKb= Kw)

Kw = [H+] [OH] (2)

2) Mass balance: CT,As = [AsO2] + [HAsO2] (3)

3) Charge balance: [H+] + [Na+] = [OH] + [AsO2] (4)

4) Proton condition: zero level (reference level); AsO2-, H2O

[HAsO2] + [H+] = [OH

] (5)

5 unknowns and 5 equations

Note:HAsO2 H

++ AsO2

[ AsO2] [H

+]

Ka= -------------------[HAsO2]

AsO2 + H2O HAsO2 + OH

[HAsO2] [OH]

Kb= ---------------------[AsO2

-] [H2O]

[H+][ AsO2

] [HAsO2] [OH

]

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

[HAsO2] [AsO2-] [H2O]

1

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Three methods for solving acid-base equilibrium problems:

1) Exact solution (solve a set of algebraic equations simultaneously).2) Approximation solutions (make assumptions, then solve algebraic equations within 5% error)3) Graphical solution (approximation method)

General Equation for the pH of a weak acid - exact solution (Butler p. 120)

Based on Ka

HA H+ + A

-

H2O H+ + OH

-

1) Equilibrium- assume effect of ionic strength is negligible; i.e., = 1 for simplicity

Acid dissociation constant

[H+][A

-]

Ka = ----------- (1)

[HA]

Molal ion product of water

Kw = [H+][OH

-] (2)

2) Mass balance on A

C T, A = [A-

] + [HA] (3)

3) Charge balance:

[H+] = [A

-] + [OH

-] (4)

4) Proton balance (Proton condition):

Reference species (zero level species): HA, H2O

[H+] = [A-] + [OH-] (5)

From the mass balance eqn (3):

[HA] = C T, A - [A-] (3)

Substitute (3) into (1) and solve for [A-]:

[H+] [A

-]

Ka= -----------------

(C T, A) - [A-]

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Solve for [A-]:

Ka(C T, A) - Ka[A-] = [H+] [A-]

[A-] ([H

+] + Ka) = Ka(C T, A)

Ka (C T, A)

[A-] = ----------------- (1)

[H+] + Ka

Substitute (1) into the charge balance (4)

Ka (C T, A)

[H+] = -------------- + [OH-] (7)

[H+] + Ka

Solve for C T, A

( [H+] - [OH-] ) ([H+] + Ka)

C T, A= ---------------------------------

Ka

[H+] + Ka

= ( [H+] - [OH

-] ) (-----------------)

Ka

Kw [H+

]= ([H

+] - -------) ( ------ + 1 ) (8)

[H+] Ka

[H+] Kw

C T, A= (1 + ------- ) ([H+] - ------- ) (9)

Ka [H+]

This is a general (exact) equation for pH of a weak acid.

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Based on Kb

HA H+ + A

-

A - + H2O HA + OH-

H2O H+ + OH-

Note:[H+][A-] [HA][OH-]

Ka Kb = ------------ ---------------- = [H+] [OH

-] = Kw

[HA] [A-]

Kw

kb= ---------Ka


[HA] [OH-] Kw

Kb = --------------- = ------- (1)[A-] Ka

Kw = [H+][OH

-] (2)


C T, A = [A-] + [HA] (3)

3) Charge balance:

[H+

] = [A-

] + [OH-

] (4)


Reference species (zero level species): HA, H2O

[H+] = [A-] + [OH-] (5)


[HA] = C T, A - [A-] (3)

Substitute (3) into (1) and solve for [HA]:

(C T, A- [A-] ) [OH-]

Kb= ---------------------------- (6)

[A-]

Solve for [Ac-]:

Kb[A-] = (C T, A- [A

-] ) [OH

-] = C T, A[OH

-]- [A

-] [OH

-]

Kb[A-] + [A-] [OH-] = [A-] (Kb + [OH

-] ) = C T, A[OH-]

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C T, A[OH

-]

[A-] = ----------------- (1)

Kb + [OH-]


C T, A[OH-][H+] = ----------------- + [OH-] (7)

Kb + [OH-]

Solve for C T, A

C T, A[OH-]

[H+] - [OH

-] = -----------------

Kb + [OH-]

C T, A[OH-] = ( [H

+] - [OH

-] ) (Kb + [OH

-])

KbC T, A= ( [H

+] - [OH

-] ) (--------- + 1)

[OH-]

Kw Kb[H+]

C T, A= ([H+] - -------) ( ----------- + 1 ) (8)

[H+] Kw

Or from the previous En:

[H+] Kw

C T, A= (1 + ------- ) ([H+

] - ------- )Ka [H

+]

[H+] Kw

= (1 + ----------- ) ([H+] - ------- )

Kw/ kb [H+]

kb [H+] Kw

= (1 + ----------- ) ([H+] - ------- ) (9)

Kw [H+]

This is a general (exact) equation for pH of a weak acid.

For weak base (Sato)

Based on Kb

AOH A+ + OH-

H2O H+ + OH

-

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21

Note:


[A+] [OH

-]

Kb = --------------- (1)[AOH]

Kw = [H+][OH

-] (2)


C T, A = [A+] + [AOH] (3)

3) Charge balance:

[A+] + [H

+] = [OH

-] (4)


[AOH] = C T, A - [A+] (3)

Substitute (3) into (1) and solve for [HA]:

[A+] [OH

-]

Kb= ------------------

C T, A- [A+]

Solve for [A+]:

Kb(C T, A- [A+]) = [A+] + [OH-]

KbC T, A- Kb[A+

] = [A+

] + [OH-

]KbC T, A = [A

+] [OH

-] + Kb[A

+]

= [A+] ([OH-] + Kb)

KbC T, A[A+] = ----------------- (1)

Kb + [OH-]


KbC T, A

[H+

] + ----------------- = [OH-

] (5)Kb + [OH

-]

Solve for C T, AKbC T, A

[OH-] [H

+] = -----------------

Kb + [OH-]

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[OH-] + Kb

C T, A= ( [OH-] [H

+] ) (------------------)

Kb

[OH-]

= ( [OH-] [H+] ) (----------- + 1 )Kb

Kw Kw / [H+]

= ( ------ [H+] ) (-------------- + 1 )

[H+] Kb

Kw Kw

= (------- [H+] ) ( ----------- + 1 ) (6)

[H+] Kb[H

+]

This is a general (exact) equation for pH of a weak base.

Cohen (1980, p. 50) gives the following eqn:

Kw Kw

C T, A= (1 + ------------ ) (-------- [H+] ) (7)

kb [H+] [H+]

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1. Exact solution

Example: Find the pH and chemical species concentrations of 0.01M CH3COOH (acetic acid,HAc):

CH3COOH CH3COO + H

+

HAc Ac- + H +

Neglect ionic strength effect (= 1 for simplicity).

Note: We know

HAc is acid, thus pH < 7HAc is weak acid, thus pH > 2 (noting that strong acid such as 0.01 M HCl has pH ~ 2)

Thus, we expect pH of HAc is 2 < pH < 7

(Solution)

1) Equilibrium

HAc H+ + Ac

pKa = 4.7

H2O H+ + OH

pKw = 14

a) Acid dissociation equilibrium eqn:

- assume effect of ionic strength is negligible; i.e., = 1 for simplicity

[H+] [Ac]

Ka = --------------- = 104.7

= 2.00 x 105

(1)[HAc]

b) Water equilibrium eqn:

Kw = [H+][OH] = 1 x 10 14 (2)

2) Mass balance on Ac, CT, Ac

CT, Ac = [Ac] + [HAc] = 0.01 (3)

3) Charge balance: positive charge species = negative charge species

[H+] = [Ac] + [OH] (4)

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24


all species that gain a proton = all species that lose a proton(proton gainers = proton losers)

Reference species (zero level species): HAc, H2O

[H+] = [Ac] + [OH] (5)

Note H+is same as H3O

+

We have 4 unknowns and 4 equations. Now solve the set of algebraic equations (1)-(4) or (5),

simultaneously.

From Eq. (2) [ ][ ]

OHx

H

+=

1 10 14

Then substitute [OH-] into the proton condition Eq. (5)

[ ] [ ][ ]

H Acx

H

+

+= +

1 10 14

and solve for [Ac-]

[ ] [ ][ ]

Ac Hx

H

+

+=

1 10 14

(6)

Substitute Eq (6) into the mass balance En (3):

[ ] [ ][ ]

0 011 10

14

. = +

+

+HAc H

x

H

and solve for [HAc]

[ ]

[ ] [ ]HAc H

x

H=

+

+

0 011 10

14

. (7)

Substituting Eqs. (6) and (7) into Eq. (1) yields

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25

[ ] [ ][ ]

[ ][ ]

K

H Hx

H

Hx

H

xa =

=

+ +

+

+

+

1 10

0 011 10

2 10

14

14

5

.

2x10-7

- 2 x 10-5

[H+] + 2 x 10

-19/[H

+] = [H

+]2 1 x 10

-14

[H+]2 + 2 x 10-5[H+] 2x10-7 2 x 10-19/[H+] = 0

[H+]

3 + 2 x 10

-5[H

+]2 2x10

-7[H

+] 2 x 10

-19= 0

This is a third degree polynomial equation. Use methods such as trial and error solution ornumerical method (e.g., Newton-Raphson, Jacobian iteration, and Birge-Vietta).

[H+] = 10 3.36

[OH-] = Kw / [H+] = 10-14/ 103.36 = 2.291 x 1011

Using Eq. (6), solve for Ac-

[ ] [ ][ ]

Ac Hx

H

+

+=

1 10 14

[Ac-] = 10 3.36 - (10-14/ 103.36) = 4.365 x 10-4

From mass balance eq. (3)

[HAc] = 0.01 [Ac-] = 9.563 x 10

-3

Summary

[H+] = 103.36 thus, pH = log [H+] = 3.36

[OH-] = 2.291 x 1011[Ac-] = 4.365 x 10-4

[HAc] = 9.563 x 10-3

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26

2. Approximate solution

Example: Find the pH and chemical species concentrations of 0.01M CH3COOH (acetic acid,

HAc):

CH3COOH CH3COO + H

+

HAc Ac- + H +

Neglect ionic strength effect (= 1).

(Solution)

1) Equilibrium

HAc H+ + Ac (pKa = 4.7)

H2O H+ + OH (pKw = 14)

[H+] [Ac

]

Ka = --------------- = 104.7= 2.00 x 105 (1)

[HAc]

Kw = [H+][OH] = 1 x 10 14 (2)

2) Mass balance on Ac

C T, Ac = 0.01= [Ac] + [HAc] (3)

3) Charge balance:

positive charge species = negative charge species

[H+] = [Ac] + [OH] (4)

4) Proton balance (Proton condition): proton gainers = proton losers

Reference species (zero level species): HAc, H2O

[H+] = [Ac] + [OH] (5)

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27

Assumptions:

i) [H+] >> [ OH-] because HAc is acid (approximately 100 x greater )

ii) [HAc] >> [Ac-] because HAc is weak acid (Ka is very small)

a. Solution using assumption (i)

From proton condition Eq. (5)

[H+] = [Ac

] + [OH

]

from assumption (i), [OH] ~ 0;

[H+] = [Ac

]

Let [H+] = [Ac-] = x (6)

From mass balance Eq. (3),C T, Ac = 0.01 = [Ac

-] + [HAc]

0.01 = x + [HAc]

thus [HAc] = 0.01 x (7)

Substituting Eqs (6) and (7) into the equilibrium eqn (1):

[H+

] [Ac-

] x2

Ka = --------------- = ------------ = 2.00 x 10

-5

[HAc] 0.01 x

Then, solve for x

2x10-7

2 x 10-5

x = x2

x2 + 2 x 10-5x 2 x 10-7 = 0

This is a quadratic equation, and can be solved.

Note: If a general quadratic equation is a x2+ bx + c = 0, a general solution is

x b b ac

a=

2

4

2

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28

x x x x

x x

=

=

2 10 2 10 4 1 2 10

2 1

2 10 8 9465 10

2

5 5 2 7

5 4

( ) ( )( )

( )

.

x = - 4.57 x 10-4 and 4.37 x 10-4 only one answer is correct

x = [H+] = 4.37 x 10

-4

[OH-] = 10-14/ 4.37 x 10-4 = 2.29 x 1011

[Ac-] = 4.37 x 10

-4

[HAc] = 0.01 - 4.37 x 10-4

= 9.56 x 10-3

* Check the assumption using charge balance eqn (4)

[H+] = [Ac

-] + [OH

-] (4)

(4.37 x 10-4) = (4.37 x 10-4) + (2.29 x 1011)

%errorright hand side left hand side

left hand sidex=

100

(4.37 x 10-4

) + (2.29 x 1011

) - (4.37 x 10-4

)% error = -------------------------------------------------------- x 100

(4.37 x 10-4)

= 5.2 x 10-6

< 5 % thus the assumption is acceptable.

Summary

[H+] = 4.37 x 10-4 thus, pH = log [4.37 x 10-4] = 3.36

[OH] = 2.29 x 10

11

[Ac

] = 4.37 x 10-4

[HAc] = 9.56 x 10-3

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b) Solution using assumptions (i) and (ii)

From proton condition eqn (5),

[H+] = [Ac+] + [OH-] and

from assumption (i), [OH-] ~ 0;

[H+] = [Ac

-]

Let [H+] = [Ac-]= x (a)

From mass balance Eq. (3),

CT, Ac= 0.01 = [Ac-] + [HAc] and

assumption (ii), [HAc] >> [Ac-] ;

0.01 = [HAc] (b)

Substituting (a) and (b) into the acid dissociation equilibrium eqn (1) yields

[H+] [Ac

-] x

2

Ka = ------------- = ------- = 2.00 x 10-5

[HAc] 0.01

then solve for x: x = [H+] = [Ac-] = 4.47 x 10-4

[OH-

] = 10-14

/ 4.47 x 10-4

= 2.24 x 1011

[HAc] = 0.01

* Check the assumption using charge balance eqn (4)

[H+] = [Ac

-] + [OH

-] (4)

(4.47 x 10-4) = (4.47 x 10-4) + (2.24 x 1011)

%errorright hand side left hand side

left hand sidex=

100

(4.47 x 10-4

) + (2.24 x 1011

) - (4.47 x 10-4

)

% error = -------------------------------------------------------- x 100(4.47 x 10-4)

= 2.3 x 10-9

< 5 % thus the assumption is acceptable.

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30

Summary

[H+] = 4.47 x 10-4 thus, pH = log [4.47 x 10-4] = 3.35

[OH-] = 2.24 x 10

11

[Ac-] = 4.47 x 10-4[HAc] = 0.01

Overall Summary:

1) Exact solution, pH = 3.362) Based on assumption (i), pH = 3.363) Based on assumptions (i) and (ii), pH = 3.35

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31

A guide for making approximations

1. Look at the size of the concentration (CT). If it is small compared to 10-7

, only [H+] and

[OH-] will be predominant and the pH will be near 7.2. If the concentration is large compared to 10-7, look at Ka.3. If Ka is small compared to 10-7, the acid will never become completely dissociated, and a

reasonable approximation is that [HA] is large compared to [A-]. Try this. If it doesnt

work, solve the exact equations.

4. If Ka is large compared to 10-7, look at the size of the concentration compared to Ka.5. If the concentration is small compared to Ka but large compared to 10 -7, complete

dissociation may be a good approximation. Try this. If it doesnt work, solve the exact

equations.

6. If the concentration is large compared to Ka, [H+] will certainly be large compared to[OH-]. In addition, a reasonable approximation is slight dissociation, [HA] predominant.

Try this. If it doesnt work, use the exact mass balance.

Flow Chart for Weak Acid Approximations

CT> 10-

Ka > A-

Ka >> 10-7

CTKa

Assume H+>> OH

-

and/or HA >> A-

CTcompared to 10-7

Ka compared to 10-7

CTcompared to Ka

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Example: (SJ, p.192, 4.14.3) Arsenious acid is a weak acid:

HAsO2 H+ + AsO

2- PKa = 9.92 Ka = 10

-9.22

If 10-3

moles of the herbicide NaAsO2, are added to 1 liter of distilled water at 25C, what is the

resulting pH? Determine by solving the appropriate set of equations analytically (usingapproximate method).

NaAsO2 Na+ + AsO

2-

AsO2- + H2O HAsO2 + OH

-

H2O H+ + OH

-

Equilibrium:

[HAsO2][ OH

-

] Kw 10

-14

Kb= ----------------------- = ------ = ----------

[AsO2-] Ka 10

-9.22

= 10-4.78

(1)

Kw = [H+][OH-] = 10-14 (2)

Mass Balance:

CT, AsO2= [HAsO2] + [AsO2-] = 10-3 (3)

Charge Balance:

[H+] + [Na

+] = [OH

-] + [AsO

2-] (4)

Proton Condition: (zero level sp. Na+, AsO

2-, H2O )

[HAsO2] + [H

+] = [ OH-] (5)

Assumptions:

Because the solution is basic, [H+]

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33

Substituting (3) and (5) into (1) gives

[HAsO2][ OH-] x2

Kb= ---------------------- = ----------- = 10-4.78

(1)

[AsO2-] 10-3 - x

x = 1.37 x 10-4

[HAsO2] = [OH

-] = 1.37 x 10

-4

[AsO2-] = 10

-3 - [HAsO

2]

= 10-3 - (1.37 x 10-4) = 8.63 x 10-4

Kw 10-14

[H

+

] = -------- = ---------------- = 7.30 x 10

-11

[OH-] 1.37 x 10-4

pH = 10.14

Check the assumptions using Charge Balance Eq. (4)

[H+] + [Na

+] = [OH

-] + [AsO

2-] (4)

(7.30 x 10-11

) + (1 x 10-3

) = (1.37 x 10-4

) + (8.63 x 10-4

)

1 x 10-3

= 1 x 10-3

Error < 5%

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34

Example(SJ, p.193, 4.14.7) What pH results when 10-4mole of NaH2PO4are added to 1 liter of

distilled water? The temperature = 25C; assume that = 10-2after the addition. Compare thevalue obtained when ionic strength effects are taken into account with that calculated when it is

neglected. (See Fig. 3-4 for activity coefficients.)

NaH2PO4 Na+ + H2PO4-10-4 10-4

H3PO4 H2PO4 + H

+

H2PO4 HPO4

2 + H

+

HPO42

PO43

+ H+

H2O H+ + OH

Equilibrium:

{H2PO4} {H+

}Ka1= -------------------- = 5.9 x 10

-3 (1)

{H3PO4}

{HPO42}{H

+}

Ka2= -------------------- = 6.2 x 10-8

(2)

{H2PO4}

{PO43

} {H+}

Ka3

= ------------------- = 4.8 x 10-13 (3)

{HPO42}

Kw = {H+}{OH

} = 10

14(4)

Mass balance:CT= [H3PO4] + [H2PO4

] + [HPO42] + [PO4

3] = 1 x 10-4

(5)

Charge balance:

[Na+] + [H

+] = [H2PO4

] + 2[HPO4

2] + 3[PO4

3] + [OH

-]

(6)

Proton condition:Reference level or zero level : H2O, H2PO4

[H3PO4] + [H+] = [HPO4

2] + 2[PO4

3] + [OH

] (7)

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Assumptions:

1) the solution is acidic because NaH2PO4forms multi proton compounds.2) The dominant reaction will probably be (2) because the compound is added as H2PO4

.

3) [HPO42] >> [PO4

3] because Ka3 > [H3PO4] because Ka2

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36

[H2PO4] = (1 x 10-4 ) x = (1 x 10-4 ) (2.459 x 10-6)

= 9.754 x 10-5

From (1)

[H2PO4] [H

+] (9.754 x 10

-5) (2.459 x 10

-6)

[H3PO4] = --------------------- = ------------------------------------

Ka1 5.9 x 10-3

= 4.065 x 10-8

From (3)

Ka3[HPO42] (4.8 x 10-13) (2.459 x 10-6)[PO4

3] = ----------------- = ------------------------------------

[H+] (2.459 x 10

-6)

= 4.8 x 10-13

Check:

Using charge balance (6)

[Na

+

] + [H

+

] = [H2PO4

] + 2[HPO42

] + 3[PO43

] + [OH

-

]

(1 x 10-4

) + (2.459 x 10-6

) = (9.754 x 10-5

) + 2(2.459 x 10-6

)

+ 3 (4.8 x 10-13) + (4.067 x 10-9)

(1.0246 x 10-4

) = (1.02462 x 10-4

) ---- checked OK

i) when = 10-2

,

HPO42[HPO42

] H+[H

+

]Ka2= ----------------------------------- = 6.2 x 10

-8 (2)

H2PO4 [H2PO4]

The activity coefficients are (see note below)

HPO42= 0.66 (a)

H+= H2PO4= 0.90 (b)

Substituting (a) and (b) into (2)

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0.66 [HPO42

] 0.9 [H+]

Ka2= ----------------------------------- (2)0.9[H2PO4

]

Let [H+] = [HPO42] = x (c)

[H2PO4] = 1 x 10-4 [HPO4

2] = 1 x 10-4 x (d)

0.66 x2

Ka2= ------------ = 6.2 x 10-8

10-4 x

x2 + 9.39 x 108 x 9.39 x 1012 = 0

6

12288

100177.32

)1039.9(4)1039.9(1039.9

=+= xxxxx

[H+] = [HPO4

2] = x = 3.0177 x 10

-6

pH = - log (3.0177 x 10-6) = 5.52 and pOH = 8.39

[OH] = Kw/ [H

+] = (1 x 10

-14)/ (3.0177 x 10

-6) = 4.067 x 10

-9

[H2PO4] = (1 x 10

-4 ) x = (1 x 10

-4 ) (3.0177 x 10

-6)

= 9.698 x 10-5

From (1)

[H2PO4] [H

+] (9.898 x 10

-5) (3.0177 x 10

-6)

[H3PO4] = --------------------- = ------------------------------------Ka1 5.9 x 10

-3

= 4.960 x 10-8

From (3)

Ka3[HPO42] (4.8 x 10-13) (9.698 x 10-5)

[PO43

] = ----------------- = ---------------------------------

[H+] (3.0177 x 10

-6)

= 1.543 x 10-11

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38

----------Note

Ba

zA

i

i

i+

=

1

)(log

2

10

At 25C,

A = 0.5085,

B = 0.3281 x 108

for H2PO4,

ai= 4.00 x 10-8

and z = 1

for HPO42,

ai= 4.00 x 10-8

and z = 2

for H+, ai= 9.00 x 10

-8 and z = 1

17980.010)103281.0)(1000.4(1

10)2(5085.0log

288

22

4210 =

+

=

xxHPO

HPO42= 100.17980

= 0.66

04495.010)103281.0)(1000.4(1

10)1(5085.0log

288

22

4210 =

+

=

xxPOH

H2PO4= 10

0.045

= 0.91

04495.010)103281.0)(1000.4(1

10)1(5085.0log

288

22

10 =

+

=

+

xxH

H+= 100.045

= 0.91

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------------------------NSEN

Bases of interest in water reactor

KOH, LiOH, NH4OH

- The bases of interest in water reactor technology areKOH, LiOH, and NH4OH.

- NaOH is not ordinarily utilized because of problems resulting from activation of 23Na to 24Na.

- LiOH has been shown experimentally to be associated at high concentrations, and the same is probably true for KOH at high temperatures.

- NH4OH is a weak base.- At the normal concentrations of interest (10-4M), however, only the dissociation of NH4OH need to be taken into account in computing

solution pH.

Dissociation constants for NH4OH and LiOH are presented in Table 3.2and Fig. 3.3.

Table 3.2 Dissociation Constants: NH4OH, LiOH, and HSO4-(Cohen, 1980)

Jones (1958) Write (1961) Write (1961)

(a) Marshall &

Johns (1966)NH4OH NH4OH LiOH HSO4

-

(oF) (

oC) Kbx 10

6Kbx 10

6Kb Kax 10

6

75 24 17.9

120 49 21.9 0.135

160 71 19.1 0.0731

200 93 15 16.1 0.0804

240 116 12.4 0.0502 880

280 138 10.2 0.0348

300 149 7.80 300

320 160 7.37 0.0500

360 182 5.18 0.0467

400 204 2.83 3.42 0.0502 42.0

440 227 2.19 0.0320

480 249 1.13 1.30 0.0305 10.0

520 271 0.84 0.0234

540 282 0.43560 293 0.35 2.00

(a), Data from W.L. Marshall and E. V. Jones, " Reactor Chemistry Division Annual Progress

Report For Period Ending Jan. 31, 1965", ORNL-3789, p. 148, Oak Ridge National Laboratory;

also Phys. Chem. 70, 12 (December 1966).

Temperature

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40

Figure 3.3. Dissociation constants KB, for NH4OH and LiOH (Cohen 1980).

Fig. 3.4

The calculated pH at high temperature of pure water, dilute solutions of ammonia, and strong base, are plotted in Fig. 3.4, where SB is

molal concentration of added base.

In Fig. 3.4, the point for K2SO4, at 560 F, indicates that a 0.5 x 10-4M solution of K2SO4is more alkaline at that temperature than a solution of

NH3which as a pH at 75F of 10.0 (11.1 ppm NH3).

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Figure 3.4. pH of solutions of strong base (SB), NH3, and pure water (Cohen 1980).

The dissociation equilibria are calculated from the following equations:

Dissociation of base:

[ ][ ]

[ ]

KaM OH

MOH

=

+

(3.18)

Dissociation of water:

[ ][ ]K H OH w = + (3.16)Mass balance for base:

[ ] [ ] [ ]M MOH B+ + = (3.17)Charge balance (Condition of electroneutrality):

[ ] [ ] [ ]M H OH+ + + = (3.15)For univalent weak bases,

[ ][ ]

[ ][ ]

B K

HH

K

K H

w w

B

=

+

+

+

+

1 (3.19)

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42

Salts

- When an acid reacts with a base, the positive ions from the base and negative ions from the acid form an ionic compound called a salt,such as NaCl.

- Saltis an ionic compounds that forms when a metal atom or a positive radical replaces the hydrogen of an acid.

(Benjamin, p 7)

- Some ions are so stable when surrounded by water molecules that, upon contact with water, they are very likely break off of any compoundthey are associated with and become surrounded by water molecules, and are then very unlikely to participate in any subsequent reactions. Suchcompounds are called salts.

Examples are sodium (Na+) and chloride (Cl-). Compounds of the form NaaX(s)or Z Cl b (s)tend to dissociate completely to Na plus Xa-or Zb+plus

Cl-, respectively, whenever they are exposed to water.

NaaX a Na+ + Xa-

Z Cl bZb+ + bCl-

The Xa-and Zb+ ions might react with other dissolved species, but Na+and Cl-rarely do.

- A salt is a compound which dissociates in solution to form positive ions other than the hydrogen or hydronium ions and negative ions otherthan the hydroxyl ion.

- e.g., sodium chloride, NaCl, dissociates to form sodium ions and chloride ions according to the following equation:

NaCl Na+ + Cl -

- the reaction between a salt and water to form a weakly acidic or weakly basic solution.- For example,

o The salt NaCl is said to be derived from HCl and NaOH.o The sodium carbonate Na2CO3 is said to be derived from carbonic acid H2CO3and NaOH.

- The strength of the acids and bases from which a salt is derived determines whether the solution of the salt is acidic or basic.

- Fig. 1-10illustrates the relationships among acids and bases.

- For example,o A salt derived from a strong acid and a strong base will not undergo hydrolysis. Its solution will be neutral.

e.g., NaCl.

o A salt derived from a strong acid and a weak acid and a weak base will undergo hydrolysis. Its solution will be acidic. e.g., Ammonium chloride NH4Cl

o A salt derived from a weak acid and a strong base will undergo hydrolysis. Its solution will be basic. e.g., Sodium carbonate, Na2CO3

o A salt derived from a weak acid and a weak base will undergo hydrolysis. Its solution can be acidic or basic, depending on the relativestrength of the acid and base from which it is derived.

e.g., Ammonium carbonate, (NH4)2CO3forms a basic solution.

High Volatility Materials(Cohen, p. 68)

3-5.1.1 NH3

Because of its volatility,NH3is attractive as a source of alkalinity in reactor and boiler technology.

In a simple (single-stage) evaporation process the limiting concentration ratio will not exceed the reciprocal of the distribution constant, vapor to

liquid.

NH3is quite stable, thermally, at the temperature of concern in reactor technology, but is subject to radiolysis.

Morpholine and cyclohexylamine are also used in conventional boiler technology as a source of basicity in the condensing portion of the steam

cycle.

The equilibrium is between un-ionized ammonia in the liquid and NH3in the vapor.

[NH3]l [NH3]v (3.25)

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[ ]

[ ]K

NH

NHD

v

l

=3

3

Subscripts land vdenote the liquid and vapor phase, respectively.

--------------------------------

write [NH4+] = NH3and

[NH4OH] = (1 ) NH3

[NH4+]

= ----------

NH3

[NH4OH]

1 = -------------NH3

Mass balance: CT, NH3= NH3 = [NH4OH] + [NH4+]

[NH4OH] = NH3 [NH4+] = NH3 NH3 = (1 ) NH3

--------------------------------

(Cohen 1980) - derived by Sato

Equilibrium Reactions:

NH4OHNH4+ + OH

NH4+ NH3 + H

+

NH3 + H2O NH4+ + OH

H2O H+ + OH

Equilibrium Equations:

[ ][ ][ ]

KNH OH

NHb =

+

4

3

(1)

[ ][ ]Kw H OH = + (2)

[ ][ ]

[ ]K

H NH

NHa =

+

+

3

4

not used

Mass balance:

CT, NH3= NH3 = [NH3] + [NH4+] (3)

Charge balance:

[NH4+] + [H+] = [OH-] (4)

From mass balance(3),

[NH4OH] = CT, NH3= [NH4+] (3)

From charge balance (4),

[OH-] = [NH4+] + [H+] (4)

Substituting (3) and (4) into (1) yields

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[ ] [ ][ ]

[ ] [ ] [ ]{ }[ ]

KNH OH

NH

NH NH H

C NHb

T NH

= =

+

+ + + +

+

4

3

4 4

43

,

[ ][ ] [ ]

1 3

4 4

4

4 4Kb

NH

NH NH H

NH

NH NH H

=

+ ++

+

+

+ ++

+

Let = [NH4+] / NH3 =

[ ]1 1

4

1

4

1 1 1

Kb NH H NH H

Kb OH OH

=

++

+

++

+

=

1 1 1

OH K

OH

OH K

K OHb

b

b

= +

=

+

OH

K OH

OH K

b

b

=

+

=

+

K

OH Kb

b

[ ]

1 1

1

1

1

1

=

+

=

+

+

=

+

=

+

=

++

K

OH K

OH K K

OH K

OH

OH K K

OH

K

K H

b

b

b b

b

b b b

w /

[ ]

1 1

1

=

+

+

Kb H

Kw

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[ ]

[ ][ ]

[ ][ ]

[ ]

[ ] [ ]

1 1

1

1

14

4

1

14

4

4

4 4

=

+

+

=

+

+

+

+

=

+

+ =+

+

Kb H

Kw

NH OH

NH OH

H

H OH

NH

NH OH

NH OH

NH OH NH

[ ]1

4

3

=

NH OH

NH

or

[ ] [ ]

[ ] [ ]

[ ] [ ] [ ][ ] [ ]

[ ]

[ ] [ ]

1 14

14

4 4

4 4 4

4 4

4

4 4

3

=

+

=

+

+

+

=

+ +

+

+ +

=

+ +

NH

NH

NH

NH OH NH

NH OH NH NH

NH OH NH

NH OH

NH OH NH

[ ]1

4

3

=

NH OH

NH

----------------------

The relationship is

[ ]1

1

1

=

+ +

K

KHB

W

(3.26)

where [H+] is obtained from Eqn (3.19)

Various values of are assumed from which NH3is calculated, and a curve of vs NH3 constructed. KD is then given by

[ ]

( )

NH

NHKv D

3

31 =

(3.27)

The apparent distribution coefficient is, therefore

[ ]( )( )

NH

NHK Kv D D

3

3

1

= = '

Fi 3 12 h K f NH f i f f i

7-1 Acid and Base_S11

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