Polyfunctional Acids and Bases [PDF]

Homework

Fundamentals of Analytical Chemistry


7-10, 12, 14, 18,

20, 27

Chapter 15 Polyfunctional Acids and Bases

Polyprotic Acids
Acids that can donate

more than 1 proton

per molecule  

Strong acid – H2SO4 Several weak acids


‘Well behaved’ behaved’ dissociation 

For most cases we can safely assume that protons are removed sequentially

Polyprotic Acids


Consider the dissociation of the weak polyfunctional acid H2A 

First step: H2A  H+ + HA• K1 = [H+][HA-] / [H2A] • Subscript 1 indicates the first proton ‘removed’ removed’



Second Step: HA-  H+ + A2• K2 = [H+][A2-] / [HA-] • Subscript 2 indicates the second proton ‘removed’ removed’



• All molecules will have their first proton removed before any will have the second removed, etc… etc…

For higher order polyprotics, polyprotics, subscript indicated the number of the proton removed from the ‘parent’ parent’ molecule

Polyprotic Acids
Possible 

combinations



H2A /



A2• Conjugate base of the weak acid HA• Kb = Kw / K2

HA-

• Buffer • Henderson Hasselbalch using pK1

HA- / A2• Also a buffer (conjugate weak acid/base pair) • HendersonHenderson-Hasselbalch using pK2!

H2A • Since the protons dissociate sequentially, this is like any other weak acid • Calculate [H+] like any monoprotic acid using K1.



Polyprotic Acids



H2A / A2• • •

Will react H2A + A2-  2HAMay have either of the two buffers OR HA-

1

Acid Salt
Partially neutralized 

 

polyprotic acid

Only ‘new’ new’ situation relative to monoprotic acids Can act as either an acid or a base [H+] a function of Ka (acting as an acid) and Kb (acting as a base)

Acid Salt
Deriving [H + ] =



• Let Kn+1 = Ka as an acid, then Kb = Kw / Kn • For a diprotic acid, Kn = K1, and Kn+1 = K2

K n +1C HA− + K w C −  1 +  HA K n  

If CHA- / K1 is much greater than 1, and also Kn+1CHA- is much greater than Kw, then

[ H + ] ≈ K n K n +1

Acid Salt

Titration of Polyprotic Acids


Final equation

shows that under the assumed conditions, [H+] (and therefore pH) is independent of the concentration of the acid salt

Legend
Point  

A

 

B (blank)

First buffer region pH from pK1


Point 


Region

Only a weak acid Treat like any monoprotic acid using K1


Region

Legend

C

Acid Salt

 


Point  

E

Conjugate base of ‘acid salt’ salt’ Kb = Kw / K2


Region 

D

Second buffer region Use pK2 to calculate pH

F

Strong base

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Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH

Calculations
Consider the

titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH 

Assume the formula of the acid is H2A


Before any titrant 



[H ] = − K + +

1

K12 + 4KaCHA 2

is added =

mL of titrant is added H2A + OH-  HA- + H2O I 2.50 0.50 0 ~~ ∆ -0.50 -0.50 +0.50 ~~ F 2.00 ~0 0.50 ~~

−1.12x10−3 + (1.12x10−3 )2 + 4 ∗1.12x10−3 ∗ 0.10 2

pH = -log (0.0100) = 2.00

Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH [H + ] =


After 5.00

− (C A− + K a ) + (C A− + K a ) 2 + 4 K a C HA

Buffer, but with K1> 10-3, we have to use the quadratic form for [H+]

Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH
After 25.0

2

 0.5   0.5  −3 −3 2 −3  2.0  − (  + 1.12 x10 ) + (  + 1.12 x10 ) + 4 ∗1.12 x10 ∗    30   30   30  [H ] = 2 [ H + ] = 3.51x10− 3 ; pH = 2.46 +

mL of titrant is added H2A + OH-  HA- + H2O I 2.50 2.50 0 ~~ ∆ -2.50 -2.50 +2.50 ~~ F ~0 ~0 2.50 ~~

Calculating with the ' short ' form :

Acid salt

pH = pK a + log(b / a ) = 2.95 + log(0.5 / 2.0) pH = 2.35 ( significant difference)

Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH [H + ] =

K n+1C HA− + K w 2.5 ; C HA− = = 0.050 50 C −  1 +  HA  Kn  

(3.91x10 −6 ∗ 0.050) + 1.0 x10 −14 1 + 0.050 1.12 x10−3 + [ H ] = 6.54 x10 −5 ; pH = 4.18

[H + ] =

(

or with the ' short ' form : [ H + ] = (1.12 x10−3 )(3.91x10 −6 ) [ H + ] = 6.62 x10 −5 ; pH = 4.18

)

Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH
After 35.0

mL of titrant is added

H2A + OH-  I 2.50 3.50 ∆ -2.50 -2.50 F ~0 1.00 THEN HA- + OH-  I 2.50 1.00 ∆ -1.00 -1.00 F 1.50 ~0

HA- + H2O 0 ~~ +2.50 ~~ 2.50 ~~ A2- + H2O 0 ~~ +1.00 ~~ 1.00 ~~

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Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH
Buffer with   


After 50.0

HA- / A2-

pH = pKa + log (b/a) pH = 5.41 + log (1.00 / 1.50) pH = 5.23

Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH
Only species affecting the  

Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH

pH is A2-

Kb = KW / K2 CB = 2.5 / 75  1.0 x10 −14   2.5  ∗ [OH − ] =   −6    3.91x10   75  [OH − ] = 9.23x10−6 ; pOH = 5.03 pH = 14.00 − pOH = 8.97

  

[OH-] = 2.5/100 pOH = 1.60 pH = 12.40

different than a monoprotic titration past the equivalence point


After 75.0

For polyprotic you must be past the last equivalence point.

A2- + H2O 0 ~~ +2.50 ~~ 2.50 ~~

mL of titrant is added

H2A + OH-  I 2.50 7.50 ∆ -2.50 -2.50 F ~0 5.00 THEN HA- + OH-  I 2.50 5.00 ∆ -2.50 -2.50 F ~0 2.50

HA- + H2O 0 ~~ +2.50 ~~ 2.50 ~~ A2- + H2O 0 ~~ +2.50 ~~ 2.50 ~~

Sulfuric Acid pH
Unique 




No 

HA- + H2O 0 ~~ +2.50 ~~ 2.50 ~~

Titration of 25.0 mL of 0.10 M o-phthalic acid with 0.10 M NaOH

Titration of 25.0 mL of 0.10 M ophthalic acid with 0.10 M NaOH
Strong base

mL of titrant is added

H2A + OH-  I 2.50 5.00 ∆ -2.50 -2.50 F ~0 2.50 THEN HA- + OH-  I 2.50 2.50 ∆ -2.50 -2.50 F ~0 ~0



situation

First proton is completely dissociated (strong acid) Second proton comes from a weak acid (partially dissociated) Dissociation of the first proton affects the amount of dissociation for the second proton


What

is the pH for a 0.010 M solution of sulfuric acid?

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Sulfuric Acid pH Assume 1 mL of solution, then M = mmol acid H2SO4  HSO4- + H+ I 0.010 0 0 ∆ -0.010 +0.010 +0.010 F ~0 0.010 0.010 THEN HSO4-  SO42- + H+ I 0.010 0 0.010 ∆ -x +x +x F 0.010x 0.010+x 0.010-x

K2 = [H+][SO42-] / [HSO4-]; K2 = 0.0102 = (0.010+x)(x) / (0.010 - x) (0.0102)(0.010 - x) = (0.010+x)(x) 1.02x10-4 - 0.0102x = x2 + 0.010x Rearranging: x2 + 0.010x + 0.0102x – 1.02x10-4 = 0 Quadratic – solve for x x = 0.0042, [H+] = 0.010+0.0042 = 0.0142 pH = 1.85 Assume one proton dissociates, pH = 2.00 Assume both protons dissociate, pH = 1.70

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