SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
Modern Theories of Acids, Bases, and Salts
 Acid-Base Equlibria
 Sörensen’s pH Scale
 Species Concentration as a Function of pH
 Calculation of pH
 Acidity Constants

SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Arrhenius theory
 Acid : substance that liberates H+
 Base : substance that supplies OH -
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
Acid : a substance , charged or uncharged, that
is capable of donating a proton
Base : a substance , charged or uncharged, that
is capable of accepting a proton from acid
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Protophilic : Capable of accepting protons from the
solute : acetone, ether
“염기성용매”

Protogenic : proton -donating compound : acetic acid

Amphiprotic : Both proton accptors and proton
donors : water, alcohols

Aprotic : neither accept nor donate protons :
hydrocarbons
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
 Acid : a molecule or ion that accepts an
electron pair to form a covalent bond.
 Base : a substance that provides the pair of
unshared electrons by which the base
coordinates with an acid.
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
k1

HAc + H2O
Acid1


Base2
k2
H3O+ + Ac Acid2
Base1
Rf = k1  [ HAc ]1  [ H2O ]1
Rr = k2  [ H3O + ] 1  [ Ac -] 1
 k1, k2 = specific reaction rate
 [ ] = concentration
 Acid -base pair , conjugate pair = Acid1 and Base1, Acid2 and Base2
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Rf = Rr
K = k1 / k2 =

Ka( ionization constant / dissociation constant)

[ H3O
Ka = 55.3K =

+
] [ Ac -]
[ HAc ] [ H2O ]
[ H3O
+
] [ Ac -]
[ HAc ]
Brönsted - Lowry theory : Ka = acidity constant
HAc + H2O
(c-x)
H3O+ + Ac x
x
x2
Ka =
c-x
 x2= KaC
c >> x , c - x  c  Ka 
 x = [ H3O + ] = 𝑲𝒂𝑪
x2
c
(7-16)
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

BH+ + OH -
B + H2O
[BH
+
] [ OH -]

Kb =

x = [ OH - ] = 𝑲𝒃𝑪
[B]
(7-24)
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

H2O + H2O

K =
[H3O
+
H3O + + OH ] [ OH -]
[H2O ]2
(7-29)
Kw( autoprotolysis constant / ion product of water)
 Kw = K  [H2O ]2
[H3O + ]  [ OH -] = Kw  1  10 -14 at 25 oC
* In pure water
[H3O + ] = [ OH - ]  𝟏 × 𝟏𝟎−𝟏𝟒  1  10 -7

SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Ka  Kb =
[ H3O
+
] [ B -]
[ HB ]

[BH
= [ H3O + ] [ OH -] = Kw

Kb = Kw / Ka

Ka = Kw / Kb
+
] [ OH -]
[B -]
(7-12)(7-33)
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Triprotic (tribasic)acid, such as phosphoric acid, ionizes in three stages.
H3PO4 + H2O = H3O+ + H2PO4[ H3O
+
] [H2PO4-]
[H3PO4 ]
= K1 = 7. 5  10 - 3
H2PO4- + H2O = H3O+ + HPO42 [ H3O
+
] [H2PO42-]
[H2PO4
-]
= K2 = 6.2  10 - 8
HPO42- + H2O = H3O+ + PO43 [ H3O
+
] [H2PO42-]
[H2PO4-]
= K3 = 6.2  10 - 13
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Triprotic (tribasic)acid, such as phosphoric acid, ionizes in three stages.
PO4 3 - + H2O
Kb1 =
HPO4 2 - + OH -
[ HPO42 - ] [OH - ]
[ PO4 3- ]
HPO42 - + H2O
Kb2 =
Kb3 =
H2PO4 - + OH -
[ H2PO4 - ] [OH - ]
[ PO4
H2PO4 - + H2O
= 4.8  10 - 2
2-
]
= 1.6  10 - 7
H3PO4 + OH -
[ H3PO4 ] [OH - ]
[ H2PO4 2]
= 1.3  10 - 12
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

HnA(parent acid) : there are n+ 1 possible species in solution.

HnA + Hn-j A-j + • • • + H A- (n -1) + An  j represents the number of protons dissociated from the parent acid and goes from 0 to n.
 Ca = total Concentration of all species

[HnA ] + [Hn-j A-j ] + • • • + [H A- (n -1) ] + [An -] = Ca

Conjugate acid-base pair : Kj Kb(n+1- j) = Kw ( Kj : various acidity constant)

K1Kb3 = K2Kb2 = K3Kb1 (Phosphoric acid system)
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

HnA(parent acid)
: there are n+ 1 possible species in solution.

HnA + Hn-j A-j + • • • + H A- (n
-1)
+ An -

Amphoteric(ampholyte) : [Hn-j A-j ] , • • , [H A- (n -1) ] , + NH3CH2COO -.

Zwitterion : + NH3CH2COO -.; electrically neutral.

Isoelectric point : The pH at which the Zwitterion concentration is a maximum
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

‘Zwitter’ in German means 'between'.

A zwitterion is a molecule that contains both
a negatively and a positively charged group.

These are bonded through intermediate groups.

Amino acids and proteins behave as zwitterions.
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
Depending on the pH of a solution, macromolecules such as proteins which contain many
charged groups, will carry substantial net charge, either positive or negative.
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

-
Polyelectrolytes
molecules that contain multiple same charges, e.g.DNA and
RNA

Polyampholytes
- molecules that contain many acidic and basic groups
- the close association allows these molecules to interact
through opposing charged groups.
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
 Depending on the pH of a solution, macromolecules such as proteins
which contain many charged groups, will carry substantial net charge,
either positive or negative.

Cells of the body and blood contain many polyelectrolytes (molecules
that contain multiple same charges, e.g.DNA and RNA) and
polyampholytes (many acidic and basic groups) that are in close
proximity.

The close association allows these molecules to interact through
opposing charged groups
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
1





pH = log [ H O + =] – log [ H3O + ]
3
pH = - log aH+
hydronium ion concentration  activity coefficient = hydronium ion activity
pH = - log (  c )
p ; negative logarithm of the term.
Ex) pOH = – log [OH-], pKa = – log Ka, pKw = – log Kw


pH + pOH = pKw
pKa + pKb = pKw
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

HnA : there are n+ 1 possible species in solution.
[HnA ]

0 =

In general,
j =
, 1 =
Ca
[Hn-jA-j ]
[Hn-1A-1 ]
Ca
and n =
Ca
Ca = total acid,  = fraction
0 + j + • • • + n-1 + n = 1
[A
-n
]
Ca
*  value
K1 =
K2 =
[H
n-1
[H
[H
n-2
A-] [H3O+]
A]
A2-] [H3O+]
[H
n
n-1
A-]
in general , j =
=
=
1 Ca [H3O+]
0 C a
[H n-2 A2-] [H3O+]2
K1[H
(K1K2 …K j)0
[H3O+]j
n
A]
K10
, 1 =
=
[H3O+]
2 Ca [H3O+]2
0 CaK1
, 2 =
K10
[H3O+]
(7-69)
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
K10
(K1K2 …K j)0

0 +

0 = [H3O+]n / { [H3O+]n + K1[H3O+]n-1 + K1 K2 [H3O+]n-2 + • • • K1 K2 . . . Kn }
[H3O+]
+• • • +
[H3O+]j
=1
{ D = [H3O+]n + K1[H3O+]n-1 + K1 K2 [H3O+]n-2 + • • • K1 K2 . . . Kn }

0 = [H3O+]n / D

[H n A] = [H3O+]n Ca / D

[H n-j A-j] = K1 • • • K j[H3O+]n-j Ca / D
{Ca = [HnA ] + [Hn-j A-j ] + • • • + [H A- (n
-1)
] + [An -]}
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
Calculation of pH
Proton Balance Equations (PBE)
Express the concentration of all species as
a function of equilibrium constants and [H3O+]
Eqs. (7-73) to (7-76)
Solve the resulting expression for [H3O+]
Check all assumptions
If all assumptions prove valid,
convert [H3O+]
pH
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Proton Balance Equations (PBE)
a) Always start with the species added to water ( Na2HPO4)
b) On the left side of the equation, place all species that can from when
protons are consumed by the starting species. ( [H2PO4-],[H3PO4])
c) On the right side of the equation, place all species that can form when
protons are released from the starting species.( [PO43-] )
d) Each species in the PBE should be multiplied by the number of
protons lost or gained when it is formed from the starting species
e) Add [H3O+] to the left side of the equation, and [ OH -] to the right
side of the equation.
Ex) [H3O+] + [H2PO4-] + 2[H3PO4] = [ OH -] + [PO43-]
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
Proton Balance Equation
What is the PBE when H3PO4 is added to water?
The species H2PO4- forms with the release of one proton
The species HPO4-2 forms with the release of two protons
The species PO4-3 forms with the release of three protons
[H3O+] = [OH-]+[H2PO4-]+2[HPO4-2]+3[PO4-3]
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
용액의 분류
(1)강산과 강염기의 용액(Solutions of strong acids and bases)
(2) 짝산-염기쌍(Conjugated acid-base pairs)
(a) 약산만을 함유한 용액(Soln. containing only a weak acid)
(b) 약염기만을 함유한 용액(Soln. containing only a weak base)
(c) 단일 짝산-염기쌍을 함유한 용액(Soln. containing a single
conjugated acid-base pair)
(3)두개의 짝산-염기쌍 (Two conjugate acid-base pairs)
(a) 2양성자산만을 함유한 용액(Soln. containing only a diprotic acid)
(b) 양쪽전해질만을 함유한 용액(Soln. containing only an ampholyte)
(c) 2산성염기만을 함유한 용액(Soln. containing only a diacidic base)
(4)두개의 독립적인 산-염기쌍 (Two independent acid-base pairs)
(a) 2약산만을 함유한 용액(Soln. containing two weak acids)
(b) 약산과 약염기의 염을 함유한 용액(Soln. containing a salt of a
weak acid and a weak base)
(c) 약산과 약염기를 함유한 용액(Soln. containing a weak acid and
a weak base)
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Solutions of Strong Acids and Bases
Strong Acids and Bases  10 -2 ( HCl )
PBE :
[H3O+]
= [ OH
-]
+ [Cl
-]
=
 [H3O+] 2 - Ca [H3O+] - Kw = 0
Kw
[H3O+]
+ Ca
(7-84)
(7-85)
Ca +  Ca2 + 4Kw
[H3O+] =
Or
[OH - ] =
2
Cb +  Cb2 + 4Kw
2
 Concentration of Acid  10 -6 M : [H3O+]  Ca
 Concentration of Base  10 -6 M : [OH - ]  Cb
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Conjugate Acid - Base Pairs
HB + H2O
B - + H2O
H2O + H2O
H3O+ + B OH - + HB
H3O+ + OH -
PBE : [H3O+] + [ HB ] = [ OH -] + [ B -]
[ HB ] = ( [H3O+] Cb) / ( [H3O+] + Ka)
[ B -] = (KaCa) / ( [H3O+] + Ka)
Result :[H3O+] = Ka (Ca - [H3O+] + [ OH -] ) / ( Cb + [H3O+] - [ OH -] ) (7-99)
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Solutions Containing Only a Weak Acid
Cb=0, [H3O+]  [ OH -]
[H3O+]2 + Ka [H3O+] - KaCa = 0
[H3O+] = ( - Ka +  Ka2 + 4KaCa ) / 2
Ca  [H3O+] 
[H3O+] =  KaCa
(7-102)
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Solutions Containing Only a Weak Base
Ca = 0, [ OH -]  [H3O+]
[H3O+] = Ka [ OH -] / ( Cb - [ OH -] ) = KaKw / [H3O+] Cb - Kw
Cb [H3O+]2 - Kw [H3O+] - KaKw = 0
[H3O+] = ( Kw +  Kw2 + 4CbKaKw ) / 2Cb
Ka  [H3O+] 
[H3O+] =  KaKw / Cb
Cb  [ OH -]  [ OH -] =  Kb Cb
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Solutions Containing a Single Conjugate
Acid - Base pair
Ca, Cb  [H3O+] or [ OH -]

[H3O+] = KaCa / Cb
Example: acetic acid and sodium acetate
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Two Conjugate Acid - Base pair ( Polyprotic)
PBE : [H3O+]
+ [H2A]ab + [ HA-]b + 2[ H2A]b=
[OH-] + [ HA-]a + 2[ A-2]b+ [ A-2]ab
 [H3O+] 4 + [H3O+] 3 ( K1 + 2Cb + Cab) + [H3O+] 2[K1(Cb - Ca) + K1K2 -Kw] [H3O+] [K1K2(2Ca+ Cab) + K1Kw] - K1K2 Kw = 0
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Solutions Containing Only a Diprotic Acid
Cab and Cb = 0
Ca  K2 : [H3O+] 3 + [H3O+] 2K1- [H3O+] K1Ca -2 K1K2 Ca = 0
[H3O+]  2 K2 ,2 K1K2 Ca is drop : [H3O+] 2 + [H3O+] K1 - K1 Ca = 0
Ca  K2 & [ H3O+]  2 K2 , K2 <<K1 : [H3O+] =
- Ka +  Ka2 + 4KaCa
2
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Solutions Containing Only an Ampholyte.
Ca , Cb = 0
[H3O+] =  ( K1K2Cab + K1Kw ) / ( K1 + Cab)
K2Cab  Kw, [H3O+] =  ( K1K2Cab ) / ( K1 + Cab)
Cab  K1 ,
[H3O+] =  K1K2
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Solutions Containing Only a Diacidic Base.
Ca, Cab = 0, Cb >>Kb2 & [ OH -] >> 2Kb2  Kw is drop
[ OH -]2 + [ OH -]Kb1 - Kb1Cb = 0
Cb  [ OH -] , [ OH -] =  Kb1Cb
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
Two independent Acid - Base pair

HB1 + H2O
H3O+ + B1-
K1 = [H3O+ ][B1-] / [HB1]

HB2 + H2O
H3O+ + B2-
K2 = [H3O+ ][B2-] / [HB2]

PBE :
[H3O+ ] + [HB1]B1 + [HB2]B2 = [ OH -] + [B1-]A1 + [B2-]A2
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실

Solutions Containing Two Weak Acids.
Cb1, Cb2 = 0
[H3O+]2 + [H3O+] (K1 + K2) - ( K1Ca1 +K2Ca2) = 0
Ca1 , Ca2  [H3O+] , [H3O+] =  K1 Ca1 + K2Ca2.
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
Solutions Containing a Salt of a Weak Acid and a Weak Base.



In case ammonium acetate, Ca1, Cb2 = 0 and Kw = negligibly small
NH4+ + AcHAc + NH3
acid1

Base2
Acid2
Base1
Ca1 =Cb2 =Cs ,( Cs : salt concentration.)
 Cs >> K1 or K2
[H3O+]2Cs - [H3O+] K1 K2 - K1K2Cs = 0 ( Cs >> [H3O+] )
* [H3O+] =  K1 K2
 HnA
[H3O+]2 - [H3O+] K1 (n - 1) - n K1K2 = 0
[H3O+] = n K1 K2
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
Effect of Ionic Strength upon Acidity Constants.

H3O+ + B
HB + H2O
a
H3O+

K=

pK` = pK +
Z: charge
aHB
aB
=
[H3O+] [B]
[HB]
0.51(2Z - 1)  
1+

H3O+
B
HB
(Debye-Hückel equation)
 : ionic strength
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실
다음 주제를 최대한 자세히 조사/정리하여 제출하시오.
(Due date: 2014. 10. 2 )

약품 및 생체 내에서의 완충제
 ‘제약용 완충용액의 제조’ 반드시 포함

삼투성과 pH를 조절하는 방법
SKKU Physical Pharmacy Laboratory
성균관대학교 물리약학연구실