16.5 Weak Acids And Acid Ionization Constants
Weak acids are only partially ionized in solution.
There is a mixture of ions and non-ionized acid in solution.
That is, weak acids are in equilibrium:
                    HA(aq) + H₂O(l) ---> H₃O⁺(aq) + A^-(aq)
                                                <---
           or       HA(aq) --->  H⁺(aq) + A^-(aq)
                                  <---

That is: K_a = [H₃O⁺][A^-]/[HA]    or  K_a = [H⁺][A^-]/[HA]
 

        K_a is the acid dissociation constant.

Note that [H₂O] is omitted from the K_a expression; H₂O is a
  pure liquid, and its concentration is essentially constant.
 The larger the K_a, the stronger the acid (i.e., the more
    ions are present at equilibrium relative to the unionized
    molecules).

If K_a >> 1, then  the acid is completely ionized and the
    acid is a strong acid.
In general, a weak acid has a K_a of less than 10^-3.
 

Percent ionization
Percent ionization is another method to assess acid
    strength.
All strong acids are 100% ionized!

   7 strong acids: HCl,HBr,HI,HClO₃,HClO₄,HNO₃,H₂SO₄
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Not so for weak acids.
       % ionization = ([H₃O⁺]_equil/[HA]₀) x 100

 Percent ionization relates the equilibrium [H₃O⁺] to the
    initial [HA] concentration.
 The higher percent ionization, the stronger the acid.
 Percent ionization of a weak acid decreases as the
    molarity of the solution increases.
 For acetic acid, 0.05 M solution is 2.0 % ionized, whereas
    a 0.15 M solution is only 1.0 % ionized.
 

Calculate the % ionization from K_a:
  The K_avalue for formic acid is 1.8 x 10^-4.  Calculate the
      % ionization for a 0.120 M formic acid solution.
      HCOOH is formic acid.

            HCOOH + H₂O ----> H₃O⁺ + HCOO^{-
                                    <----}

    K_a= 1.8 x 10^-4  =  ([H₃O⁺][HCOO^-])/HCOOH
                              = (x)(x)/(0.120-x)
            1.8 x 10^-4  =  x²/0.120
                           x = [1.8 x 10^-4 (0.120)]^1/2
                           x = 0.00465 M = [H₃O⁺]_equil = [HCOO^-]
    Therefore, % ionization = ([H₃O⁺]_equil/[HA]₀) x 100
                         = (0.00465M/0.120M) x 100 = 3.88 %
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Calculating pH for Solutions of Weak Acids:
  pH is the equilibrium concentration of H₃O⁺. By doing an
    equilibrium calculation using K_a , we can calculate
    [H₃O⁺] at equilibrium ([H₃O⁺]_equil).
Assume the initial [H₃O⁺]₀ and [A^-]₀ are zero; make
    x = change in moles of all species present, and
    [HA]₀= initial conc. of acid

For a monoprotic weak acid, e.g., acetic acid  HC₂H₃O₂:
                        K_a = [H₃O⁺]_equil[A^-]_equil/[HA]₀-x
                        K_a = (x)(x)/([HA]₀-x) = x²/([HA]₀-x)
            solve for x ===>  [H₃O⁺] ===> pH

HINT: If the value of x is small compared to the value of
       [HA]₀ , we can assume that ([HA]₀-x) = [HA]₀

 This simplifies the quadratic and saves a lot of math!

Always check the final result when assuming x is small
    by plugging all values back into the equation!!!

 Polyprotic acids have more than one ionizable proton.
 The protons are removed in steps, not all at once:
     H₂SO₃(aq) + H₂O(l) ----> H₃O⁺(aq) + HSO₃^-(aq)
                                       <----
                             K_a1 = 1.7 x 10^-2
     HSO₃^-(aq) + H₂O(l) ----> H₃O⁺(aq) + SO₃^2-(aq)
                                                         <----
                             K_a2 = 6.4 x 10^-8

It is always easier to remove the first proton in a
    polyprotic acid than the second.
That is, K_a1 > K_a2 > K_a3, etc.

Most H₃O⁺ at equilibrium usually comes from the first
    ionization (i.e., the K_a1 equilibrium).
Except for sulfuric acid, the second ionization is usually
    a weak acid problem since K_a2 values are less than 10^-3.
    In H₂SO₄, the 1st ionization is STRONG, and the 2nd
    ionization constant K_a2=1.2 x 10^-2
 

When calculating the concentrations for polyprotic acids,
    the problem is a little more complicated, since the initial
    concentrations for the second ionization are determined
    by K_a1, etc.

       [H₂SO₃] = 0.100M - 0.0336M = 0.066 M

For the 2nd step:
   Let y = [HSO₃^-] that ionizes = [H₃O⁺] = [SO₃^2-]
        formed in this reaction

   [H₃O⁺] = 0.0336 M + y;  [SO₃^-2] = y; 
            [HSO₃^-] = 0.0336-y

        K_a2 = 6.4 x 10^-8 = ([H₃O⁺][SO₃^-2])/[HSO₃^-]
        K_a2 = (0.0336 M + y)(y)/0.0336-y

    Since y << 0.0336,  then
                6.4 x 10^-8 = (0.0336 M)(y)/0.0336
    Therefore, y = 6.4 x 10^-8M = [H₃O⁺] = [SO₃^2-]
            formed in this reaction;
       [HSO₃^-] = 0.0336 - y  = 0.0336M.
 Therefore, ALL the [H₃O⁺] comes from the 1st reaction!
  The only concentration coming from the 2nd reaction
    is [SO₃^2-].
 

16.6 Weak bases and Base Ionization Constants
 Weak bases remove protons from substances.
 There is an equilibrium between the base and the
    resulting ions; note that a base can be neutral,
    e.g., NH₃, or anionic, e.g., O^2-

                B + H₂O(l) ----> HB⁺(aq) + OH^-(aq)
                                  <----

        or    B^- + H₂O(l) ----> HB(aq) + OH^-(aq)
                                   <----
 The base dissociation constant, K_b, is defined as
                K_b = [HB⁺][OH^-]/[B]
        or     K_b = [HB][OH^-]/[B^-]

Example:
          NH₃(aq) + H₂O(l) ---->  NH₄⁺(aq) + OH^-(aq)
                                        <----

 The base dissociation constant, K_b, is defined as
                      K_b = [NH₄⁺][OH^-]/[NH₃]

 The larger the value of K_b, the stronger the base.
 Bases generally have lone pairs or negative charges
    that attack protons, and weak bases often have a
    N atom (substances called amines), e.g., pyridine
    C₅H₅N, methylamine NH₂(CH₃), dimethylamine
    NH(CH₃)₂, triethylamine N(C₂H₅)₃

Anions of weak acids are also bases: e.g., OCl^-

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