Common Ion Effect.

How is the pH of a weak acid solution changed by the
presence of some extra conjugate base in the solution?

Consider the equilibrium:

     CH₃COOH(aq) + H₂O(l) ---> H₃O⁺(aq) + CH₃COO^-(aq)
                                             <---
If a strong electrolyte containing CH₃COO^- is added (e.g.,
Na⁺CH₃COO^- = sodium acetate; i.e., a salt containing the
same anion as the anion of the weak acid), the equilibrium
shifts to the left (Le Chatelier's principle).  In other words,
if the conjugate base CH₃COO^- is added, the H₃O⁺will
decrease, and pH will increase.

             This is called the common ion effect.

More precisely....
The dissociation of a weak electrolyte decreases when a
strong electrolyte is added that has an ion in common
with it.

To calculate the pH of such solutions:

  Ex.:  Calculate the pH of a buffer solution that is 0.050 M
           in HAc and 0.20 M in NaAc.  (K_a = 1.8 x 10^-5)

    Reaction 1:  NaAc + H₂O  --------> Na⁺ + Ac^-
           (No equilibrium!!!!)   100% to completion!!

    Reaction 2:  HAc + H₂O  -----> H₃O⁺ + Ac^-
                                            <-----

        K_a = 1.8 x 10^-5 = [H₃O⁺][Ac^-]/[HAc]

                    HAc          H₃O⁺               Ac^-
Init []        0.050          0                 0.20
D []               -x             +x               +x
Equil []       0.050-x      x                 0.20+x

        K_a = 1.8 x 10^-5 = (x)(0.20+x)/0.050-x)

We will make the assumption that x is SMALL as compared
    to 0.20M and to 0.050M.
    Therefore,     K_a = 1.8 x 10^-5 = (x)(0.20)/0.050)
                            x = 4.5 X 10^-6 M =  [H₃O⁺]
     ^{Check our assumption!!   Our assumption is valid!!}

Therefore, pH = -log (4.5 X 10^-6) = 5.35
 
 

17.2 Buffer Solutions
Composition and Action of Buffered Solutions:

A buffer consists of a mixture of a weak acid (HX) and
  its conjugate base (X^-), or a mixture of a weak base (B)
  and its conjugate acid (BH⁺).

 A buffer resists a change in pH when OH^- or H⁺ is added.

When OH^- is added to the buffer, the OH^- reacts with
    H₃O⁺ to produce more X^-.

But if the amounts of HX and X^- in the buffer are large
 compared to added hydroxide, the [HX]/[X^-] ratio
 remains more or less constant, so the pH is not
 significantly changed.
 

When H⁺ is added to the buffer, the H⁺ reacts with X^-
    to produce HX and water.

Once again, if the amounts of HX and X^- in the buffer
    are large compared to added H⁺, the [HX]/[X^-] ratio
    is more or less constant, so again the pH does not
    change significantly.

Contrast this with adding the same amounts of hydroxide
    or proton to water; the pH changes a lot!

Buffer capacity is the amount of acid or base a buffer
    can neutralize before significant pH changes occur. It
    depends on the amount of conjugate acid-base  present;
    the more HA & A^-, the more capacity.

pH of buffer depends on K_a or K_b; it is best to calculate the
    pH of a buffer using an equilibrium expression as shown
    in sample example 17.2.

Alternatively, if K_a or K_b are small (i.e., x is small compared
    to the original concentrations) we can use a simplified
    equation called:

  HENDERSON-HASSELBALCH equation.  This is derived as
            follows:

                               K_a = [H₃O⁺][A^-]/[HA]
                               [H₃O⁺] = K_a[HA]/[A^-]
  take negative logs:  -log[H₃O⁺] = -logK_a-log([HA]/[A^-])
                or                       pH = pK_a - log([acid]/[salt])
                or                       pH = pK_a + log([base]/[salt])

Calculate the pH of a solution that is 1.0 M HCN and 0.20 M
    KCN.  K_a = 4.0 x 10^-10

pH = pK_a+ log([base]/[acid])
pH = -log(4.0 x 10^-10) + log(0.20/1.0)
pH = 9.40 + (-0.70) = 8.70
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

There is another such equation for weak base buffer
      systems:        pOH = pK_b + log([salt]/[base])

- - - - - - - -  - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Weak Base & Salt of Weak Base Buffer Solution:

Calculate the pH of a solution that is 0.40 M aqueous NH₃
    and  0.15 M NH₄Cl.  K_b for NH₃ = 1.78 x 10^-5

     Reaction 1:  NH₄Cl + H₂O  -------> NH₄⁺ + Cl^-
               (no equilibrium!!!)    100% to completion!!

     Reaction 2:  NH₃ + H₂O  -----> NH₄⁺ + OH^-
                                            <-----

        K_b = 1.78 x 10^-5 = [NH₄⁺][OH^-]/[NH₃]

                    NH₃          OH^-                 NH₄⁺               
Init []        0.40            0                 0.15
D []               -x             +x               +x
Equil []      0.40-x         x                 0.15+x
 

        K_b = 1.78 x 10^-5 = (x)(0.15+x)/0.40-x)

We will make the assumption that x is SMALL as
    compared to 0.40M and to 0.15M.
    Therefore,     K_b = 1.78 x 10^-5 = (x)(0.15)/0.40)
                            x = 4.7 X 10^-5 M = [0H^-]
           ^{Our assumption is valid!!}

Therefore, pOH = -log (4.7 X 10^-5) = 4.33  and
                   pH = 14.00-4.33 = 9.67
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 

Calculate the pH of the buffer solution:  0.15 M aqueous
    NH₃ and 0.20 M (NH₄)₂SO₄.   K_b = 1.78 x 10^-5

Addition of Strong Acids or Bases to Buffers:

We break the calculation into two parts: stoichiometric
    and equilibrium.

The added strong acid or base added results in
    neutralization reactions:
                                X^- + H₃O⁺ -------> HX + H₂O
                                HX + OH^-  -------> X^- + H₂O.
By knowing how much H₃O⁺ or OH^- was added
    (stoichiometry), we know how much HX or X^- is
    formed.
With the concentrations of HX and X^- (note the change
    in volume of solution), we can calculate the pH either
    from the equilibrium expression or the Henderson-
    Hasselbalch equation.

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