SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실.

Presentation on theme: "SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실."— Presentation transcript:

1 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실

2 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

3 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Arrhenius theory  Acid : substance that liberates H +  Base : substance that supplies OH -

4 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

5 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

6 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.

7 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  HAc + H 2 O H 3 O + + Ac - Acid1 Base2 Acid2 Base1  R f = k 1  [ HAc ] 1  [ H 2 O ] 1  R r = k 2  [ H 3 O + ] 1  [ Ac - ] 1  k 1, k 2 = specific reaction rate  [ ] = concentration  Acid -base pair, conjugate pair = Acid1 and Base1, Acid2 and Base2 k 1 k2k2

8 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  R f = R r  K = k 1 / k 2 =  K a ( ionization constant / dissociation constant) K a = 55.3K =  Brönsted - Lowry theory : K a = acidity constant HAc + H 2 O H 3 O + + Ac - (c-x) x x K a = c >> x, c - x  c  K a   x 2 = K a C  x = [ H 3 O + ] = (7-16) [ H 3 O + ] [ Ac - ] [ HAc ] [ H 2 O ] [ H 3 O + ] [ Ac - ] [ HAc ] x2x2 c - x x2x2 c

9 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  B + H 2 O BH + + OH -  K b =  x = [ OH - ] = (7-24) [BH + ] [ OH - ] [ B ]

10 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실 [H 3 O + ] [ OH - ] [H 2 O ] 2

11 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  K a  K b =  = [ H 3 O + ] [ OH - ] = K w (7-12)(7-33)  K b = K w / K a  K a = K w / K b [ H 3 O + ] [ B - ] [ HB ] [BH + ] [ OH - ] [B - ]

12 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Triprotic (tribasic)acid, such as phosphoric acid, ionizes in three stages. H 3 PO 4 + H 2 O = H 3 O + + H 2 PO 4 - = K 1 = 7. 5  10 - 3 H 2 PO 4 - + H 2 O = H 3 O + + HPO 4 2 - = K 2 = 6.2  10 - 8 HPO 4 2- + H 2 O = H 3 O + + PO 4 3 - = K 3 = 6.2  10 - 13 [ H 3 O + ] [H 2 PO 4 - ] [H 3 PO 4 ] [ H 3 O + ] [H 2 PO 4 2- ] [H 2 PO 4 - ] [ H 3 O + ] [H 2 PO 4 2- ] [H 2 PO 4 - ]

13 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Triprotic (tribasic)acid, such as phosphoric acid, ionizes in three stages. PO 4 3 - + H 2 O HPO 4 2 - + OH - K b1 = = 4.8  10 - 2 HPO 4 2 - + H 2 O H 2 PO 4 - + OH - K b2 = = 1.6  10 - 7 H 2 PO 4 - + H 2 O H 3 PO 4 + OH - K b3 = = 1.3  10 - 12 [ H 3 PO 4 ] [OH - ] [ H 2 PO 4 2 ] [ HPO 4 2 - ] [OH - ] [ PO 4 3- ] [ H 2 PO 4 - ] [OH - ] [ PO 4 2 - ]

14 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  H n A(parent acid) : there are n+ 1 possible species in solution.  H n A + H n-j A -j + + H A - (n -1) + A n -  j represents the number of protons dissociated from the parent acid and goes from 0 to n.  C a = total C oncentration of a ll species  [H n A ] + [H n-j A -j ] + + [H A - (n -1) ] + [A n - ] = C a  Conjugate acid-base pair : K j K b(n+1- j) = K w ( K j : various acidity constant)  K 1 K b3 = K 2 K b2 = K 3 K b1 (Phosphoric acid system)

15 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  H n A(parent acid) : there are n+ 1 possible species in solution.  H n A + H n-j A -j + + H A - (n -1) + A n -  Amphoteric(ampholyte) : [H n-j A -j ],, [H A - (n -1) ], + NH 3 CH 2 COO -.  Zwitterion : + NH 3 CH 2 COO -.; electrically neutral.  Isoelectric point : The pH at which the Zwitterion concentration is a maximum

16 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.

17 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.

18 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.

19 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

20 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  pH = log = – log [ H 3 O + ]  pH = - log a H+  hydronium ion concentration  activity coefficient = hydronium ion activity  pH = - log (    c )  p ; negative logarithm of the term. Ex) pOH = – log [OH - ], pK a = – log K a, pK w = – log K w  pH + pOH = pK w  pK a + pK b = pK w [ H 3 O + ] 1

21 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  H n A : there are n+ 1 possible species in solution.   0 =,  1 = C a = total acid,  = fraction  In general,  j = and  n =  0 +  j + +  n-1 +  n = 1 *  value K 1 = =,  1 = K 2 = = =,  2 = in general,  j = (7-69) [ A - n ] CaCa [H n A ] CaCa [H n-1 A -1 ] CaCa [H n-j A -j ] CaCa [H n-2 A 2- ] [H 3 O + ] 2 K 1 [H n A] (K 1 K 2 …K j )  0 [H 3 O + ] j [H n-1 A - ] [H 3 O + ] [H n A]  2 C a [H 3 O + ] 2  0 C a K1  1 C a [H 3 O + ]  0 C a [H n-2 A 2- ] [H 3 O + ] [H n-1 A - ] K10K10 [H 3 O + ] K1K20K1K20 [H 3 O + ] 2

22 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실   0 + + + = 1   0 = [H 3 O + ] n / { [H 3 O + ] n + K 1 [H 3 O + ] n-1 + K 1 K 2 [H 3 O + ] n-2 + K 1 K 2... K n } { D = [H 3 O + ] n + K 1 [H 3 O + ] n-1 + K 1 K 2 [H 3 O + ] n-2 + K 1 K 2... K n }   0 = [H 3 O + ] n / D  [H n A] = [H 3 O + ] n C a / D  [H n-j A -j ] = K 1 K j [H 3 O + ] n-j C a / D {C a = [H n A ] + [H n-j A -j ] + + [H A - (n -1) ] + [A n - ]} K10K10 [H 3 O + ] (K 1 K 2 …K j )  0 [H 3 O + ] j

23 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 [H 3 O + ] Solve the resulting expression for [H 3 O + ] Check all assumptions If all assumptions prove valid, convert [H 3 O + ] pH Eqs. (7-73) to (7-76)

24 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Proton Balance Equations (PBE) a) Always start with the species added to water ( Na 2 HPO 4 ) b) On the left side of the equation, place all species that can from when protons are consumed by the starting species. ( [H 2 PO 4 - ],[H 3 PO 4 ]) c) On the right side of the equation, place all species that can form when protons are released from the starting species.( [PO 4 3- ] ) 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 [H 3 O + ] to the left side of the equation, and [ OH - ] to the right side of the equation. Ex) [H 3 O + ] + [H 2 PO 4 - ] + 2[H 3 PO 4 ] = [ OH - ] + [PO 4 3- ]

25 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실 Proton Balance Equation What is the PBE when H 3 PO 4 is added to water? The species H 2 PO 4 - forms with the release of one proton The species HPO 4 -2 forms with the release of two protons The species PO 4 -3 forms with the release of three protons [H 3 O + ] = [OH - ]+[H 2 PO 4 - ]+2[HPO 4 -2 ]+3[PO 4 -3 ]

26 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) 2 양성자산만을 함유한 용액 (Soln. containing only a diprotic acid) (b) 양쪽전해질만을 함유한 용액 (Soln. containing only an ampholyte) (c) 2 산성염기만을 함유한 용액 (Soln. containing only a diacidic base) (4) 두개의 독립적인 산 - 염기쌍 (Two independent acid-base pairs) (a) 2 약산만을 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)

27 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Solutions of Strong Acids and Bases Strong Acids and Bases  10 -2 ( HCl ) PBE : [H 3 O + ] = [ OH - ] + [Cl - ] = + C a (7-84)  [H 3 O + ] 2 - C a [H 3 O + ] - K w = 0 (7-85) [H 3 O + ] = Or [OH - ] =  Concentration of Acid  10 -6 M : [H 3 O + ]  C a  Concentration of Base  10 -6 M : [OH - ]  C b [H 3 O + ] K w C b +  C b 2 + 4K w 2 C a +  C a 2 + 4K w 2

28 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Conjugate Acid - Base Pairs HB + H 2 O H 3 O + + B - B - + H 2 O OH - + HB H 2 O + H 2 O H 3 O + + OH - PBE : [H 3 O + ] + [ HB ] = [ OH - ] + [ B - ] [ HB ] = ( [H 3 O + ] C b ) / ( [H 3 O + ] + K a ) [ B - ] = (K a C a ) / ( [H 3 O + ] + K a ) Result :[H 3 O + ] = K a (C a - [H 3 O + ] + [ OH - ] ) / ( C b + [H 3 O + ] - [ OH - ] ) (7-99)

29 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Solutions Containing Only a Weak Acid C b =0, [H 3 O + ]  [ OH - ] [H 3 O + ] 2 + K a [H 3 O + ] - K a C a = 0 [H 3 O + ] = ( - K a +  K a 2 + 4K a C a ) / 2 C a  [H 3 O + ]  [H 3 O + ] =  K a C a (7-102)

30 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Solutions Containing Only a Weak Base C a = 0, [ OH - ]  [H 3 O + ] [H 3 O + ] = K a [ OH - ] / ( C b - [ OH - ] ) = K a K w / [H 3 O + ] C b - K w C b [H 3 O + ] 2 - K w [H 3 O + ] - K a K w = 0 [H 3 O + ] = ( K w +  K w 2 + 4C b K a K w ) / 2C b K a  [H 3 O + ]  [H 3 O + ] =  K a K w / C b C b  [ OH - ]  [ OH - ] =  K b C b

31 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Solutions Containing a Single Conjugate Acid - Base pair C a, C b  [H 3 O + ] or [ OH - ]  [H 3 O + ] = K a C a / C b Example: acetic acid and sodium acetate

32 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Two Conjugate Acid - Base pair ( Polyprotic) PBE : [H 3 O + ] + [H 2 A] ab + [ HA - ] b + 2[ H 2 A] b = [OH-] + [ HA - ] a + 2[ A -2 ] b + [ A -2 ] ab  [H 3 O + ] 4 + [H 3 O + ] 3 ( K 1 + 2C b + C ab ) + [H 3 O + ] 2 [K 1 (C b - C a ) + K 1 K 2 -K w ] - [H 3 O + ] [K 1 K 2 (2C a + C ab ) + K 1 K w ] - K 1 K 2 K w = 0

33 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Solutions Containing Only a Diprotic Acid C ab and C b = 0 C a  K 2 : [H 3 O + ] 3 + [H 3 O + ] 2 K 1 - [H 3 O + ] K 1 C a -2 K 1 K 2 C a = 0 [H 3 O + ]  2 K 2,2 K 1 K 2 C a is drop : [H 3 O + ] 2 + [H 3 O + ] K 1 - K 1 C a = 0 C a  K 2 & [ H 3 O + ]  2 K 2, K 2 <<K 1 : [H 3 O + ] = - K a +  K a 2 + 4K a C a 2

34 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Solutions Containing Only an Ampholyte. C a, C b = 0 [H 3 O + ] =  ( K 1 K 2 C ab + K 1 K w ) / ( K 1 + C ab ) K 2 C ab  K w, [H 3 O + ] =  ( K 1 K 2 C ab ) / ( K 1 + C ab ) C ab  K 1, [H 3 O + ] =  K 1 K 2

35 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Solutions Containing Only a Diacidic Base. C a, C ab = 0, C b >>K b2 & [ OH - ] >> 2K b2  K w is drop [ OH - ] 2 + [ OH - ]K b1 - K b1 C b = 0 C b  [ OH - ], [ OH - ] =  K b1 C b

36 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실 Two independent Acid - Base pair  HB 1 + H 2 O H 3 O + + B 1 - K 1 = [H 3 O + ][B 1 - ] / [HB 1 ]  HB 2 + H 2 O H 3 O + + B 2 - K 2 = [H 3 O + ][B 2 - ] / [HB 2 ]  PBE : [H 3 O + ] + [HB 1 ] B1 + [HB 2 ] B2 = [ OH - ] + [B 1 - ] A1 + [B 2 - ] A2

37 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  Solutions Containing Two Weak Acids. C b1, C b2 = 0 [H 3 O + ] 2 + [H 3 O + ] (K 1 + K 2 ) - ( K 1 C a1 +K 2 C a2 ) = 0 C a1, C a2  [H 3 O + ], [H 3 O + ] =  K 1 C a1 + K 2 C a2.

38 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실 Solutions Containing a Salt of a Weak Acid and a Weak Base.  In case ammonium acetate, C a1, C b2 = 0 and K w = negligibly small  NH 4 + + Ac - HAc + NH 3  acid 1 Base 2 Acid 2 Base 1  C a1 =C b2 =C s,( C s : salt concentration.)  C s >> K 1 or K 2 [H 3 O + ] 2 C s - [H 3 O + ] K 1 K 2 - K 1 K 2 C s = 0 ( C s >> [H 3 O + ] ) * [H 3 O + ] =  K 1 K 2  H n A [H 3 O + ] 2 - [H 3 O + ] K 1 (n - 1) - n K 1 K 2 = 0 [H 3 O + ] =  n K 1 K 2

39 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실 Effect of Ionic Strength upon Acidity Constants.  HB + H 2 O H 3 O + + B  K = =  pK` = pK + (Debye-Hückel equation) Z: charge  : ionic strength a H3O+ a B a HB [H 3 O + ] [B] [HB]  H3O+  B  HB 0.51(2Z - 1)   1 +  

40 SKKU Physical Pharmacy Laboratory 성균관대학교 물리약학연구실  약품 및 생체 내에서의 완충제  ‘ 제약용 완충용액의 제조 ’ 반드시 포함  삼투성과 pH 를 조절하는 방법 다음 주제를 최대한 자세히 조사 / 정리하여 제출하시오. (Due date: 2015. 10. 2 )