Weak Acid-Base pH
We hinted at a process for weakacid base calculations involving an
assumption when we calculated the
Ka of weak acid.
When making weak acid-base
calculations the rule is to ALWAYS
think about the chemistry first, not
the problem.
Steps for weak acid-base problems
1. List all major species in acid-base
system
2. Write balanced chemical equations
for all reactions. (be certain to try
and identify any possible side
reactions)
3. Write the equilibrium expression
for the dominate species, including
Ka or Kb values when possible.
7. Use equilibrium concentrations to
calculate desired values; pH, change
in [ ], etc…
Review problem:
Calculate the Ka and % ionization of a
.005M unknown weak acid solution
having a pH of 3.45.
4. Develop an I.C.E. table with all given
concentrations and define the change in
concentration of all species.
5. Using the K<<1 assumption, substitute,
simplify and solve equilibrium
expression.
6. Apply the 5% rule to test the
assumption. (if greater than 5%, you
must us the quadratic formula to solve
the equilibrium expression)
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic
acid, HC2H3O2, at 25C.
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2(aq)
Ka for acetic acid at 25C is 1.8  105.
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1. Calculate the pH and % ionization
of a 1.00 M HF acid solution.
Weak Bases
Kb can be used to find [OH] and, through it, pH.
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2.What is the pH, pOH and percent
ionization of a 1.0 M solution of methyl
amine(CH3NH2)?
Two things to remember,
• Concentrations for acid-base
systems are reported in both M and
pH/pOH.
• Ka x Kb = Kw
Therefore, if we take the negative log
of the expression, we get
pKa + pKb = pKw = 14.00 = pH + pOH
(AP equation Sheet)
3.Lactic acid (HC3H5O3) is a waste
product of muscle use, leading to
pain and fatigue. If a 0.100 M lactic
acid solution is 3.7% dissociated,
calculate the Ka for this acid and pKb
for its conjugate base.
Polyprotic Acids
Polyprotic acids have more than one acidic proton.
If the difference between the Ka for the first
dissociation and subsequent Ka values is 103 or more,
the pH generally depends only on the first dissociation.
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Polyprotic Acids
Phosphoric acid is typical of most
polyprotic acids:
H3PO4 ⇄ H+ + H2PO4-
Ka1=7.5 x 10-3
H2PO4- ⇄ H+ + HPO42-
Ka2=6.2 x 10-8
HPO42- ⇄ H+ + PO43-
Ka3=4.2 x 10-13
Notice the extreme difference in
dissociation between Ka1 and Ka2
As long as successive Ka values
differ by a factor of 103 or more, it is
appropriate to calculate the
concentration, or pH, of polyprotic
acids considering only Ka1.
Ka and Kb of water, hydronium and hydroxide
This is the reason why we can negate the hydronium or
hydroxide concentration from the auto-ionization of water in
the presence of strong acids and most weak acids.
Ka H20 = Kb H20 = 1.8 e-16
Example:
Let’s calculate the total H+ and pH
from the complete deprotonation of
phosphoric acid.
Sulfuric acid is unique because it is
strong acid in its first ionization but
weak in its second. Does the same
relationship hold true?
Example:
Let’s calculate the pH of a 1.0 M
sulfuric acid solution.
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