chapter 11

Analytical Chemistry Dr. Yehia Mechref

1

Chapter 11 Solving Equilibrium Problems for Complex Systems

POLYFUNCTIONAL ACIDS AND BASES

Phosphoric acid is a typical polyfunctional acid. In aqueous solution it undergoes the following three dissociation reactions:

+ ++ OHPOHOHPOH 342243 3

43

4231a

1011.7

]POH[]POH][OH[

+

=

=K

+ ++ OHHPOOHPOH 324242 8

42

243

2a

1032.6

]POH[]HPO][OH[

+

=

=K

+ ++ OHPOOHHPO 334224 13

24

343

3a

1052.4

]HPO[

]PO][OH[

+

=

=K

Generally, Ka1>Ka2>Ka3 Generally, Ka1>Ka2 by a factor of 104 to 105 because of electrostatic forces. That is, the first dissociation involves separating a single positively charged hydronium ion from a singly charged anion. In the second step, a hydronium ion is separated from doubly charged anion, a process that requires considerably more energy. A second reason that Ka1>Ka2 is a statistical one. In the firs step, a proton can be removed from two locations, whereas in the second step, only from one. Thus, the first dissociation is twice as probable as the second. COMBINING EQUILIBRIUM-CONSTAT EXPRESSIONS When two adjacent stepwise equilibria are added, the equilibrium constant for the resulting overall reaction is the product of the two constants. Thus for the first two dissociation equilibria for H3PO4 we may write,


2

+ ++ OHPOHOHPOH 342243 + ++ OHHPOOHPOH 324242 + ++ OH2HPOOH2POH 324243

and

108343

24

23

2a1a

1049.41032.61011.7

]POH[]HPO[]OH[

+

==

=KK

Similarly, for the rection

+ ++ OH3POOH3POH 334243 we may write

22108343

34

33

3a2a1a

1000.2105.41032.61011.7

]POH[]PO[]OH[

+

==

=KKK

Sodium carbonate is a polyfunctional base. Carbonate ion, the conjugate base of the hydrogen carbonate ion, is involved in the stepwise equilibria:

++ OHHCOOHCO 3223 4

23

31b

1013.2

]CO[]HCO][OH[

=

=K

++ OHCOHOHHCO 3223 83

322b

1025.2

]HCO[

]COH][OH[

=

=K


3

The overall basic dissociation reaction of sodium carbonate is described by the equations

++ OH2COHOH2CO 3222312

84

23

322

2b1b

1079.4

1025.21013.2

]CO[

]COH[]OH[

==

=KK

CALCULATION OF THE pH OF SOLUTIONS OF NaHA An Amphiprotic salt is a species that can act as an acid and as a base when dissolved in a suitable solvent. These salts are formed during neutralization titration of polyfunctional acids and bases Generally, when a 1 mol of NaOH is added to a solution containing 1 mol of the acid H2A, 1 mol of NaHA is formed. The pH of the solution is determined by two equilibria established between HA- and water.

+

++

++

OHAHOHHA

andOHAOHHA

22

32

2

Therefore, a solution of NaHA will be acidic or basic , depending on the relative magnitude of the equilibrium constants for these processes:

]HA[]AH][OH[

]HA[]A][OH[

2

1a

w2b

23

2a

+

==

=

KKK

K

where Ka1 and Ka2 are the acid dissociation constants for H2A. If Kb2 is greater than Ka2, the solution is basic otherwise, it is acidic. Calculating the pH of a solution of NaHA.


4

A solution of NaHA can be described in terms of mass balance: CNaHA = [H2A] + [HA-] + [A2-] And charge balance: [Na+] + [H3O+] = [HA-] + 2[A2-] + [OH-] Since the sodium ion concentration is equal to the molar analytical concentration of the salt, the last equation can be rewritten as CNaHA + [H3O+] = [HA-] + 2[A2-] + [OH-] Moreover, the dissociation constants for H2A are;

]HA[]A][OH[

]AH[]HA][OH[

23

2a

2

31a

+

+

=

=

K

K

Thus, we have four equations and need one more to solve for the five unknowns. Kw = [H3O+] [OH-] Subtracting the mass-balance equation from the charge balance equation. CNaHA + [H3O+] = [HA-] + 2[A2-] + [OH-] CNaHA = [H2A] + [HA-] + [A2-]

[H3O+] = [A2-] + [OH-] - [H2A] We then rearrange the acid-dissociation constant expression for H2A to obtain

1

32 K

]][[][a

HAOHAH+

= and for HA- to give,

]OH[]HA[]A[

3

2a2+ = K


5

Substituting these equations and the expression for Kw into the equation obtained from the subtraction of the mass-balance and charge-balance equations.

[H3O+] = ]OH[

]HA[

3

2a+K +

]OH[ 3w+

K - 1a

3 ]HA][OH[K

+

Multiplication by [H3O+] gives

[H3O+]2 = ]HA[2aK + wK -

1a

23 ]HA[]OH[

K

+

We rearrange to obtain,

w2a1a

23 ]HA[1

]HA[]OH[ KKK

+=

+

+

This equation can be rearranged to

1a

w2a3

/]HA[1]HA[]OH[

KKK

+

++=

Under most circumstances, we can assume that

NaHA]HA[ C Thus,

1aNaHA

wNaHA2a3 /1

]OH[KC

KC K+

+=+

Frequently, the ratio CNaHA/Ka1 is much larger than unity and Ka2 CNaHA is considerably greater then Kw, thus the above equation can be rewritten as

2a1a3 ]OH[ KK+


6

EXAMPLE Calculate the hydronium ion concentration of a 0.100 M NaHCO3 solution. First, we examine the assumptions. The dissociation constants for H2CO3 are Ka1 = 4.45 x 10-7 Ka2 = 4.69 x 10-11 CNaHA/Ka1 = 2.2 x 105 Ka2 CNaHA = 4.69 x 10-12, is almost 100 times larger than Kw.

Thus, 91173 106.41069.41045.4]OH[+ ==

EXAMPLE Find the hydronium ion concentration of 0.100 M NaH2PO4 solution Ka1 = 7.11 x 10-3 Ka2 = 6.32 x 10-8 CNaHA/Ka1 = 14 Ka2 CNaHA = 6.32 x 10-9, is substantially larger than Kw. Thus the second assumption is valid but not the first, Thus,

5

32

8

1aNaHA

NaHA2a3

1062.1

)1011.7/()1000.1(00.11.01032.6

/1]OH[

+

=+

=

+= KCC K

chapter 11

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