Buffers

A buffer solution is a solution of a weak acid and its salt (conjugate base) or a weak base and its salt (conjugate acid).  Both components must be present.  A buffer solution has the ability to resist change in pH upon the addition of small amounts of either acid or base.  A buffer solution must have a relatively large concentration of acid to react with any OH- ions that are added to it, and it must contain similar concentration of base to react with any added H3O+ ions.  The acid and base components of the buffer must not neutralize each other.  These requirements are met by an acid-base conjugate pair.

To understand how a buffer works consider the acetic acid equilibrium

CH3COOH(aq)  +  H2O(l)  <==>    H3O+(aq)  +   CH3COO-(aq)

Suppose a strong acid is added to a solution contains CH3COO- ions and CH3COOH molecules in about equal concentrations.  The newly added H3O+ ions transfer protons to the to a relatively large reservoir of CH3COO- ions to form CH3COOH molecules.  Only a small fraction of these newly formed CH3COOH molecules ionize, so the pH is virtually unchanged.  If a small amount of strong base is added instead, the OH- ions remove protons from the CH3COOH molecules.  As a result, the concentration of  CH3COO increases, but the OH- concentration and thus the pH remains nearly unchanged.  So the weak acid acts as a reservoir of hydrogen ions to react with OH- ions, whereas its conjugate base acts as a sink for H3O+.

The Henderson-Hasselbalch Equation

Calculating the pH of a buffer solution involves the equilibrium constant expression for the weak acid or weak base used in the buffer.  We can rearrange the equilibrium constant expression into a form more convenient to use however.   Using a weak acid for example and rearranging the expression for Ka

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

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

 take the negative logarithm of both sides:

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

 Which gives,

               [A-]        
pH = pKa + log ----   This is known as the Henderson-Hasselbalch equation
               [HA]

We are using equilibrium concentrations of acid and conjugate base in solution for the expression above.  However only a tiny fraction of the acid donates protons to the conjugate base.  Thus, a good approximation for the equilibrium concentrations is to just use the initial concentrations of the acid and salt.

 Example 1 

(a) What is the pH of a 0.30 M solution of ammonia, NH3, which is also 0.36 M in ammonium nitrate, NH4NO3? (b) What is the pH after the addition of 20.0 mL of 0.050 M HCl to 80.0 mL of the buffer solution?

(a)  Use the Henderson-Hasselbalch equation (for a weak base).

               [NH4+]        
pH = pKb + log ------ = 4.74 + log (0.36/0.30) = 4.82  
               [NH3]

(b)  First we do some stoichiometric calculations:

mol NH4+ present initially 0.0800 L x 0.36 mol/L = 0.029 mol

mol NH3 present initially 0.0800 L x 0.30 mol/L = 0.024 mol

mol HCl added 0.0200 L x 0.40 mol/L = 0.0080 mol (this reacts with NH3 to become NH4+)

mol NH4+ present after HCl added 0.029 mol + 0.0080 mol = 0.030 mol

mol NH3 present after HCl added  0.024 mol - 0.008 mol = 0.023 mol

Since volume will cancel from the ratio we can use moles in the Henderson-Hassellbalch equation.

               [NH4+]        
pH = pKb + log ------ = 4.74 + log (0.030/0.023) = 4.86  
               [NH3]

 Example 2 

 What ratio of acetic acid to acetate must be used to prepare a buffer solution having a pH of 5.00?

               [CH3COO-]              [CH3COO-]
pH = pKa + log --------- = 4.74 + log --------- = 5.00  
               [CH3COOH]              [CH3COOH]
    [CH3COO-] 
log --------- = 5.00 - 4.74 = 0.26 
    [CH3COOH]

 Taking the antilog we obtain

[CH3COO-] 
--------- = 1.8 
[CH3COOH]

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