Buffer Solution

Nie salah satu dari syllabus A-level. Tapi rasenye matrikulasi pun blajar juga.

Buffer Solution 

“Solutions which resist changes in pH when small quantities of acid or 

alkali are added.”

Acidic Buffer (pH < 7)    

•          a weak acid   +   its sodium or potassium salt                                         

ethanoic acid              sodium ethanoate

Alkaline Buffer (pH > 7)  

•          weak base     +        its chloride                                                         

 ammonia          ammonium chloride

Acidic Buffer Solution

•          It is essential to have a weak acid for an equilibrium to be present so that ions 

        can be remove and produced. 

•          The dissociation is small and there are few ions.

                             CH₃COOH(aq)       ⇋         CH₃COO¯(aq)    +    H⁺(aq)

relative concs.            HIGH                              LOW                   LOW

• A strong acid can’t be used as it is fully dissociated and cannot remove 

            H⁺(aq):     HCl(aq)   →   Cl¯(aq)    +    H⁺(aq)        

Adding acid

•          Any H⁺ is removed by reacting with CH₃COO¯ ions to form CH₃COOH via the equilibrium.  

        Unfortunately, the concentration of CH₃COO¯ is small and only a few H⁺ can be “mopped up”.

•          A much larger concentration of CH₃COO¯ is required.

•          To build up the concentration of CH₃COO¯ ions, sodium ethanoate is added.

Adding alkali

•          This adds OH¯ ions which react with H⁺ ions   

                                         H⁺(aq)+OH¯(aq) ⇋ H₂O(l)

•          Removal of H⁺ from the weak acid equilibrium means that, according to Le Chatelier’s Principle, 

more CH₃COOH will dissociate to form ions to replace those being removed.

                        CH₃COOH(aq)        ⇋        CH₃COO¯(aq)    +    H⁺(aq)

•          As the added OH¯ ions remove the H⁺ from the weak acid system, the equilibrium moves to the 

        right to produce more H⁺ ions.

•          Obviously, there must be a large concentration of undissociated acid molecules to be available.

Alkaline buffer solution

•          Very similar but is based on the equilibrium surrounding a weak base;

                                          NH₃(aq)  +   H₂O(l)       ⇋       OH¯(aq)   +   NH₄⁺(aq)

            relative concs.              HIGH                              LOW              LOW

•          but one needs        a large conc. of  OH¯(aq)  to react with any  H⁺(aq) added and a large conc of

       NH₄⁺(aq)  to react with any OH¯(aq) added.

•          There is enough NH₃ to act as a source of OH¯ but one needs to increase the concentration of 

       ammonium ions by adding an ammonium salt.

•          Use ammonia (a weak base)  +  ammonium chloride (one of its salts)

Buffer solutions –Ideal concentration

•          The concentration of a buffer solution is also important

•          If the concentration is too low, there won’t be enough CH₃COOH and CH₃COO¯ 

       to cope with the ions added.

      

Summary

•          For an acidic buffer solution one needs:

v  large [CH₃COOH(aq)]          -   for dissociating into  H⁺(aq) when alkali is added

v  large [CH₃COO¯(aq)]           -   for removing  H⁺(aq) as it is added

•          This situation can’t exist if only acid is present; a mixture of the acid and salt is used.

•          The weak acid provides the equilibrium and the large CH₃COOH(aq) concentration.

•          The sodium salt provides the large CH₃COO¯(aq) concentration.

•          One uses a  weak acid  +  its sodium or potassium salt

Calculating the pH of an acidic buffer solution

•          Consider an acid buffer, for example that formed by the mixing of ethanoic acid and 

       sodium ethanoate.

•          The expression for K_a of ethanoic acid is:

                                   K_a      =      [H₃O⁺(aq)] [CH₃COO¯(aq)]

                                                         [CH₃COOH(aq)]

             re-arrange        [H₃O⁺(aq)]   =   [CH₃COOH(aq)] x K_a

                                                           [CH₃COO¯(aq)]

•          As a buffer contains a weak acid not all the acid molecules dissociate and the equilibrium lies

        to the left.

CH₃COOH(aq)   +   H₂O(l)   ⇋   CH₃COO¯(aq)   +   H₃O⁺(aq)

•          At equilibrium, [CH₃COOH(aq)] is almost the same as the initial [CH₃COOH(aq)],

•          [CH₃COO¯(aq)] in the buffer will be effectively be the initial [CH₃COONa] as this 

      dissociates completely.    

•          Taking account of these assumptions, the expression for a general buffer becomes            

                             [H₃O⁺(aq)]  =  K_a  x  [acid]        

                                                             [salt]    

•          Taking – log of each side gives: ― log [H₃O⁺(aq)]   =   ― log K_a  + log   [salt]

                                                                                                             [acid]

            Or        pH   =   pK_a  + log   [salt]

                                                     [acid]

•          pH of a buffer solution depends on the ratio of [salt]:[acid]

•          If [salt]=[acid] then the relationship simplifies to pH = pK_a (or [H₃O⁺] = K_a)

                       

           

      

Calculate the pH of a buffer whose [HA] is 0.1 mol dm^-3 and [A¯] of 0.1 mol dm^-3.

                                                                       K_a                 =      [H⁺(aq)] [A¯(aq)]

                                                                                                    [HA(aq)]

re-arrange                                                     [H⁺(aq)]            =   [HA(aq)] x K_a

                                                                                                  [A¯(aq)]

from information given  [A¯] = 0.1 mol dm^-3 ; [HA] = 0.1 mol dm^-3

If the K_a of the weak acid HA is 2 x 10^-4 mol dm^-3.

[H⁺(aq)] =   0.1 x 2 x 10^-4                       =  2 x 10^-4 mol dm^-3

                         0.1

pH        =  - log₁₀ [H⁺(aq)]          =   3.699

Calculate the pH of the solution formed when 500cm³ of 0.1 mol dm^-3 of weak acid HX is mixed

 with 500cm³ of a 0.2 mol dm^-3 solution of its salt NaX.   K_a =  4 x 10^-5 mol dm^-3.

                               K_a      =      [H⁺(aq)] [X¯(aq)]

                                                      [HX(aq)]                    

re-arrange                 [H⁺(aq)]            =     [HX(aq)]  K_a

                                                              [X¯(aq)]

The solutions have been mixed; the volume is now 1 dm³

therefore          [HX]   =   0.05 mol dm^-3          and    [X¯]        =   0.10 mol dm^-3

Substituting                  [H⁺(aq)]    =    0.05 x 4 x 10^-5          =    2 x 10^-5  mol dm^-3

                                                          0.1

pH        =     - log₁₀ [H⁺(aq)]       =    4.699