Equilibria in Solutions of Weak Acids and Bases

Weak Acids  

Most acidic substances are weak acids, that is acids that are only partially ionized in aqueous solution.  We can express the extent to which a weak acid ionizes by using the equilibrium constant of the ionization reaction. If we represent a general weak acid as HA, we can write the equation for its ionization reaction as follows

HA(aq)  +  H2O(l)  <==>  H3O+(aq)  +  A-(aq)

Since [H2O] is omitted from equilibrium expressions in aqueous solutions the equilibrium constant expression is

     [H3O+][A-]
Ka = ----------
        [HA]

Tbe subscript a on the K denotes that it is an equilibrium for the ionization of an acid, and Ka is called the acid-dissociation constant.   The magnitude of Ka indicates the tendency of the hydrogen atom to ionize.

  Example 1

A-0.10 M solution of formic acid, HCHO2, has a pH of 2.38.  Calculate Ka for formic acid and determine its percentage ionization.  The ionization equilibrium for formic acid can be written as follows:

HCH2O2(aq)  +  H2O(l)  <==>    H3O+(aq)  +  CHO2-(aq)

-log[ H3O+] = 2.38

log[ H3O+] = -2.38

[ H3O+] = antilog (-2.38) = 4.2 x 10-3 M

Next we construct a table of initial, change, and equilibrium concentrations.

Concentrations(M) HCHO2(aq)
+
 H2O(l) <==> H3O+(aq)
+
 CHO2-(aq)
Starting 0.10 0 0
Change -4.2 x 10-3 +4.2 x 10-3 +4.2 x 10-3
Equilibrium 0.10 - 4.2 x 10-3 4.2 x 10-3 4.2 x 10-3

We now substitute these values into the equilibrium constant expression.  Notice that to the number of significant figures we are using that 0.10 - 4.2 x 10-3  is 0.10, so we will use 0.10 M as the equilibrium concentration of HCHO2.
     [H3O+][CHO2-]   (4.2 x 10-3)(4.2 x 10-3)
Ka = ------------- = ------------------------ = 1.8 x 10-4
        [HCHO2]                0.10

The percentage of acid that ionized is given by the concentration of H3O+ or CHO2- at equilibrium , divided by the initial acid concentration times 100%.

                     [H3O+]            4.2 x 10-3
Percent ionization = ------- x 100% = ----------- x 100% = 4.2%
                     [HCHO2]             0.10

  Example 2 

What is the pH of a 0.30 M solution of acetic acid?  The Ka for acetic acid is 1.8 x 10-5.   

The ionization equilibrium for acetic acid can be written as follows:

HC2H3O2(aq)  +  H2O(l)  <==>    H3O+(aq)  +  C2H3O2-(aq)

     [H3O+][C2H3O2-] 
Ka = -------------- = 1.8 x 10-5
        [HC2H3O2]

Next we construct our usual table.

 

Concentrations(M) HC2H3O2(aq) 
+
 H2O(l) <==> H3O+(aq)
+
 C2H3O2-(aq)
Starting 0.30 0 0
Change -x +x +x
Equilibrium 0.30 - x x x

The next step is to substitute the equilibrium concentrations into the equilibrium constant expression.

        [H3O+][C2H3O2-]                         (x)(x)
Ka = ----------------------- = 1.8 x 10-5 = ---------------
            [HC2H3O2]                               0.30 - x

 

This expression leads to a quadratic in x, which we could solve by using the quadratic formula.  We can simplify the problem and avoid the quadratic, if we assume that 0.30 - x is about 0.30, in other words x is negligible compared to 0.30.  After we get a value for x we will look back to see if this assumption was valid.  By using this assumption we have:

              x 2             
Ka = ----------  = 1.8 x 10-5 
           0.30                               

Solving for x, we have

x2 = (0.30)(1.8 x 10-5) = 5.4 x 10-6 

x = (5.4 x 10-6)1/2 = 2.3 x 10-3    1/2 power is square root ( I don't have a square root symbol)

We now look back to check the validity of our simplifying assumption that 0.30 - x ~ 0.30.  We see that x is indeed negligible compared to 0.30.

[H3O+] = x =  2.3 x 10-3  M

pH = -log( 2.3 x 10-3 ) = 2.64

  Example 3 

What is the pH of a 0.10 M solution of hydrofluoric acid?  The Ka for hydrofluoric acid is 6.8 x 10-4.   

 The ionization equilibrium for hydrofluoric acid can be written as follows:

HF(aq)  +  H2O(l)  <==>    H3O+(aq)  +  F-(aq)

     [H3O+][F-] 
Ka = ---------- = 6.8 x 10-4
        [HF]

Next we construct our usual table.

 

Concentrations(M) HF(aq) 
+
 H2O(l) <==> H3O+(aq)
+
 F-(aq)
Starting 0.10 0 0
Change -x +x +x
Equilibrium 0.10 - x x x

The next step is to substitute the equilibrium concentrations into the equilibrium constant expression.

         [H3O+][F-]                          (x)(x)
Ka = --------------- = 6.8 x 10-4 = ------------
            [HF]                               0.10 - x

 

When we try solving this equation using the approximation 0.10 - x ~ 0.10  (assuming that x is negligible compared to 0.10) we obtain x = 8.2 x 10-3 M.   Since x is greater than 5 percent of 0.10, we must work the problem using the quadratic formula.

x2 = (0.10 - x)(6.8 x 10-4) = 6.8 x 10-5 - (6.8 x 10-4)x

x2 + (6.8 x 10-4)x - 6.8 x 10-5 = 0

     -6.8 x 10-4 ± [(6.8 x 10-4)2 + 4(6.8 x 10-5)]1/2
x = ------------------------------------------------ 
                             2  
     -6.8 x 10-4 ± 1.6 x 10-2
x = ------------------------- = 7.9 x 10-3 or -8.3 x 10-3
                2

Of these two solutions we choose 7.9 x 10-3 as the only one which is physically reasonable.

pH = -log( 7.9 x 10-3 ) = 2.10

Polyprotic Acids 

 Acids which have more than one ionizable H atom are known as polyprotic acids.  For example, each of the H atoms in sulfurous acid, H2SO3, can ionize in successive steps:

H2SO3(aq)  + H2O(l)  <==>  H3O+(aq)  +  HSO3-(aq)      Ka1 = 1.7 x 10-2   

HSO3-(aq)  + H2O(l)  <==>  H3O+(aq)  +  SO3-2(aq)      Ka2 = 6.4 x 10-8   

In the preceding example  Ka2 is much smaller than Ka1.  This observation is general: It is always easier to remove the first proton froma polyprotic acid than the second.  Similarly, for an acid with three ionizable protons, it is easier to remove the second proton than the third.  Thus the Ka values become successively smaller as successive protons are removed.

 Weak Bases 

A number of substances behave as weak bases in water.  Such substances react with water, removing protons from H2O, thereby forming the conjugate acid of the base and OH- ions:

Weak base +  H2O  <==>  conjugate acid  +   OH- 

The most commonly encountered weak base is ammonia or one of its derivatives.

NH3(aq)  +   H2O(l)  <==>  NH4+(aq)  +   OH-(aq)   

     [NH4+][OH-] 
Kb = ---------- 
        [NH3]

The constant Kb is called the base-dissociation constant, by analogy with the acid-dissociation constant, Ka, for weak acids.

  Example 4 

What is the pH of a 0.15 M solution of NH3?  The Kb for ammonia is 1.8 x 10-5 

 Build the usual table.

 

Concentrations(M) NH3(aq) 
+
 H2O(l) <==> NH4+(aq)
+
 OH-(aq)
Starting 0.15 0 0
Change -x +x +x
Equilibrium 0.15 - x x x 
     [NH4+][OH-]    (x)(x)
Kb = ---------- = ---------
        [NH3]      0.15 - x

Assume that we neglect x in comparison to 0.15. Then we have

  x2 
----- = 1.8 x 10-5
 0.15

x2 = (0.15)(1.8 x 10-5) = 2.7 x 10-6

x = [NH4+] = [OH-] = 1.6 x 10-3 M

pOH = -log(1.6 x 10-3) = 2.80

pH = 14.00 - pOH = 14.00 - 2.80 = 11.20

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