15.5 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)
Since we use H+ and H3O+ interchangeably:
= [H3O+][A-] /
Ka = [H+][A-] /
Ka is the acid dissociation constant.
[H2O] is omitted from the Ka
expression; H2O is a
liquid, and its concentration is essentially constant.
The larger the Ka, the stronger the acid (i.e., the more ions are present at equilibrium relative to the un-ionized molecules).
>> 1, then the acid is completely ionized and the acid
a weak acid has a Ka of
less than ~10-3.
Percent ionization is another method to assess acid strength.
All strong acids are 100% ionized!
acids: HCl, HBr, HI, HClO3, HClO4, HNO3, H2SO4
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Not so for weak acids.
% ionization = ([H3O+]equil / [HA]0) x 100
the equilibrium [H3O+]
to the INITIAL
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
The Ka value for formic acid is 1.8 x 10-4. Calculate the % ionization for a 0.120 M formic acid solution. HCOOH is formic acid.
H2O ----> H3O+
1.8 x 10-4 = ([H3O+][HCOO-])/[HCOOH]
1.8 x 10-4 = x2/0.120 (Assuming that x is small compared to 0.120)
[1.8 x 10-4 (0.120)]1/2 = x
x = 0.00465 M = [H3O+]equil = [HCOO-]
Therefore, % ionization = ([H3O+]equil / [HA]0) x 100
% ionization = (0.00465M / 0.120M) x 100 = 3.88 %
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Calculating pH for
of Weak Acids:
pH is the equilibrium concentration of H3O+. By doing an equilibrium calculation using Ka , we can calculate [H3O+] at equilibrium: ([H3O+]equil).
Assume the initial [H3O+]0 and [A-]0 are zero; let x = change in moles of all species present, and [HA]0 = initial conc. of acid
For a monoprotic weak acid,
acetic acid, HC2H3O2:
Ka = [H3O+]equil[A-]equil / [HA]0 - x
Ka = (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!
FOR HELP SOLVING THESE KINDS
check out chapter
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., O2-
H2O(l) ----> HB+(aq)
B- + H2O(l)
----> HB(aq) + OH-(aq)
The base dissociation constant, Kb, is defined as
(1) Kb = [HB+][OH-]/[B]
or (2) Kb = [HB][OH-]/[B-]
NH3(aq) + H2O(l) ----> NH4+(aq) + OH-(aq)
Kb = [NH4+][OH-] / [NH3]
Kb, the stronger
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
bases: e.g., OCl-