Weak acids

Weak Acids and Bases
Today’s Class
• Weak Acids
• Weak bases
• Percent dissociation
Acids (from last class)
• When an acid dissolves in water, a proton
(hydrogen ion) is transferred to a water
molecule to produce a hydroxonium ion
and a negative ion depending on what
acid you are starting from.
• In the general case . . .
HA + H2O
H3O+ + A• The strength of an acid is defined by the
equilibrium position of the dissociation
reaction shown above
Weak Acids
• A weak acid is an acid whose equilibrium
lies far to the left.
• Most of the acid originally present in the
solution is still present as HA at
• In other words, a weak acid only
dissociates to a small extent in aqueous
• In contrast to a strong acid, the conjugate
base of a weak acid is a much stronger
base than H2O
Weak acids
• Acetic acid is a typical weak acid. It reacts with water
to produce hydronium ions and ethanoate ions, but
the back reaction is more successful than the forward
one. The ions react very easily to reform the acid and
the water.
CH3COO- + H3O+
• At any one time, only about 1% of the acetic acid
molecules have converted into ions. The rest remain
as simple acetic acid molecules
• Most organic acids are weak. Hydrofluoric acid is a
weak inorganic acid that you may come across
a) Strong acid
b) Weak acid
Weak Acids
• Because of the equilibria involved, the
treatment of weak acid solutions with
regard to [H+] and pH is not as
straightforward strong acids
• Keep in mind that the solution is always in
equilibrium, so we use another form of K,
(equilibrium constant) called an acid
dissociation constant, denoted by Ka, is
an equilibrium constant for the
dissociation of a weak acid
Weak Acids
• Recall the expression for
dissociation of an acid is:
HA + H2O
H3O+ + ASo the Ka expression would be
Ka= [H3O+][A-]
Calculating pH of a weak
• Let’s do an example to demonstrate
Calculate the pH of 1.0M solution of HF the
Ka can be found on the table on page 639
and equals 7.1x10-4
The first step is always write the major
species in the solution. Since the Ka for
HF is so small we know it is a weak acid
and will be dissociated only to a small
So the major species in solution are:
HF and H2O
Example continued
• The second step is to decide which of the
major species can furnish H+ ions. In this
case both species can do this:
HF(aq) H+(aq) + F-(aq) Ka=7.1x10-4
H+(aq) + OH-(aq) Kw=1.0x10-14
• In aqueous solutions, one source of H+
ions can usually be singled out as
• By comparing Kw and Ka we can see that
HF, although weak, is still a much
stronger acid than H2O.
• We ignore the contriution of water
because it is so tiny
Example Continued
• From previous slide we found that
the dissociation of HF will determine
the equilibrium concentration of H+
and the pH:
• HF(aq)
H+(aq) + F-(aq)
• The equilibrium expression is:
• Ka=7.1x10-4=[H+][F-]
Example Continued
• To solve the equilibrium problem we follow the same
procedure we used in the previous chapter on
equilibrium. (ICE box way)
• First list the initial concentrations:
[HF]0=1.0M, [F-]0=0, and [H+]0≈0
• Second step is to determine the change required to
reach equilibrium
• Since some of HF will dissociate to come to
equilibrium (amount presently unknown) we’ll let x be
the change in concentration of HF needed to reach
eqilibrium. So x mol/L of HF will dissociate to produce
x mol/L of H+ and x mol/L of F-. Defined in terms of x
the equilibrium concentrations are:
[HF]= [HF]0-x= 1.00-x
[F-]=[F-]o+x= 0+x= x
[H+]=[H+]o+x= 0+x= x
Example Continued
• Now we can substitute equilibrium concentrations
into equilibrium expression, which gives:
Ka=7.1x10-4=[H+][F-]= (x)(x)
[HF] 1.00-x
This produces a quadratric equation that can be solved,
but since the Ka for HF is so small, HF will only
dissociate slightly, thus x is very small compared to
1.00 so the term in the denominator can be
approximate as:
So the equilibrium expression becomes:
Which gives
x2=7.1x10-4(1.00), taking the square root of the right
side we obtain:
Example Continued
• How valid is the approimation
• We use the following test:
Compare the sizes of x and [HF]0
If the expression: x X100% is less than
Or equal to 5%, the value of x is small
enough for the approximation.
Example Continued
• So testing it for our example we have
x X100%=2.7x10-2 X100%=2.7%
Therefore the approximation made was
valid so the value of x can be used.
x=[H+]=2.7x10-2M and
Summary: Solving weak
acid equilibrium problems
• List major species in the solution
• Choose species that can produce H+, and write
balanced equations for them
• Compare the equilibrium constant for the
reactions you have written, decide which
reaction will dominate in production of H+
• Write equilibrium expression for dominant
• Define the change needed to reach equilibrium
(define x)
• Write equilibrium concentration in terms of x
• Substitute the equilibrium concentrations into
• Solve for x the easy way(assume [HA]0-x≈[HA]0)
• Verify the approximation(5% rule)
• Calculate [H+] and pH
Weak Bases
• Much the same as weak acids:
• A weak base is a base whose equilibrium lies
far to the left.
• Most of the base originally present in the
solution is still present as original base at
• In other words, a weak base only dissociates
to a small extent in aqueous solution.
• In contrast to a strong base, the conjugate
acid of a weak base is a much stronger acid
than H2O
Weak Bases
• Weak base calculations are done in
the exact same way as weak acids
• The only difference is instead of
using Ka as the equilibrium constant
we use Kb
Percent Dissociation
• It is often useful to specify the amount of
weak acid that has dissociated in
reaching equilibrium
• It is defined as
% dissocation= amount dissociated(mol/L) X100%
initial concentration(mol/L)
• For a given weak acid, the % dissociation
Increases as the acid becomes more
• The % dissociation Increases as [HA]0
• Taking the values obtained from the
previous example for weak acid
% dissociation=2.7x10-2M x 100%
= 2.7%

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