HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) If HA is a weak acid,
identify which side dominates at equilibrium. Identify and the
bases in solution. Which base is stronger? Which base is weaker?
HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) If HA is a strong acid,
identify which side dominates at equilibrium. Which base is
stronger? Which base is weaker? 2 Review!
Slide 3
How good is Cl (aq) as a base? Is A (aq) a good base? The bases
from weakest to strongest are: Cl , H 2 O, A 3 Review!
Slide 4
Buffers Solutions that resist changes in pH Made of weak acids
or weak bases containing a common ion. Common ion is the conjugate
base/conjugate acid
Slide 5
Common Ion Effect Shift in equilibrium position that occurs
because of the addition of an ion already involved in the
equilibrium reaction. An application of Le Chteliers
principle.
Slide 6
Common Ion Effect HCN(aq) + H 2 O(l) H 3 O + (aq) + CN - (aq)
Addition of NaCN will shift the equilibrium to the left because of
the addition of CN -, which is already involved in the equilibrium
reaction. A solution of HCN and NaCN is less acidic than a solution
of HCN alone. Ensures that concentrations of both HA and A - are
high Allows for absorption of both H + and OH - Resists change in
pH
Slide 7
What happens when you add acid/base to a buffered solution?
Strong base added to buffered solution Strong acid added to
buffered solution
Slide 8
Solving Problems with Buffered Solutions
Slide 9
Example 15.1 Calculate [H + ], the pH, and the percent
dissociation of HF in a solution containing 1.0 M HF (K a = 7.2 x
10 -4 ).
Slide 10
Example 15.1 b Calculate [H + ], the pH, and the percent
dissociation of HF in a solution containing 1.0 M HF (K a = 7.2 x
10 -4 ) and 1.0 M NaF.
Slide 11
Example 15.2 A buffered solution contains 0.50M acetic acid (HC
2 H 3 O 2, K a = 1.8 x 10 -5 ) and 0.50M sodium acetate (NaC 2 H 3
O 2 ). Calculate the pH.
Slide 12
What are major species? 12 To play movie you must be in Slide
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Slide 13
Example 15.3 Calculate the change in pH that occurs when 0.010
mol solid NaOH is added to 1.0L of the buffered solution described
in the previous example. Compare this pH change with that which
occurs when 0.10 mol solid NaOH is added to 1.0L water.
Slide 14
Example 15.3 After reaction with NaOH, main controller of pH is
the dissociation of acetic acid.
Slide 15
Characteristics of Buffers Contain large amounts of weak acid
and corresponding conjugate base. Added H + reacts to completion
with the conjugate base. Added OH - reacts to completion with the
weak acid.
Slide 16
Characteristics of Buffers The pH in the buffered solution is
determined by the ratio of the concentrations of the weak acid and
weak base. As long as this ratio remains virtually constant, the pH
will remain virtually constant. [A ] / [HA] [B] / [BH + ] This will
be the case as long as the concentrations of the buffering
materials (HA and A or B and BH + ) are large compared with amounts
of H + or OH added. A consequence of this is
Slide 17
Henderson-Hasselbalch For a particular buffering system
(conjugate acidbase pair), all solutions that have the same ratio
[A ] / [HA] will have the same pH. Can find pH directly from A -
and HA concentrations Notice what happens when these are equal! pK
a of the weak acid to be used in the buffer should be as close as
possible to the desired pH.
Slide 18
Practice practice practice!
Slide 19
AP Exam, 1986 In water, hydrazoic acid, HN 3, is a weak acid
that has an equilibrium constant, K a, equal to 2.8 10 5 at 25C. A
0.300 litre sample of a 0.050 molar solution of the acid is
prepared. (a)Write the expression for the equilibrium constant, K
a, for hydrazoic acid. (b)Calculate the pH of this solution at 25C.
(c)To 0.150 litre of this solution, 0.80 gram of sodium azide, NaN
3, is added. The salt dissolved completely. Calculate the pH of the
resulting solution at 25C if the volume of the solution remains
unchanged. (d)To the remaining 0.150 litre of the original
solution, 0.075 litre of 0.100 molar NaOH solution is added.
Calculate the [OH ] for the resulting solution at 25C.
Slide 20
AP Exam, 1992 The equations and constants for the dissociation
of three different acids are given below. HCO 3 H + + CO 3 2 K a =
4.210 7 H 2 PO 4 H + + HPO 4 2 K a = 6.210 8 HSO 4 H + + SO 4 2 K a
= 1.310 2 (a)From the systems above, identify the conjugate pair
that is best for preparing a buffer with a pH of 7.2. Explain your
choice. (b)Explain briefly how you would prepare the buffer
solution described in (a) with the conjugate pair you have chosen.
(c)If the concentrations of both the acid and the conjugate base
you have chosen were doubled, how would the pH be affected? Explain
how the capacity of the buffer is affected by this change in
concentrations of acid and base. (d)Explain briefly how you could
prepare the buffer solution in (a) if you had available the solid
salt of the only one member of the conjugate pair and solution of a
strong acid and a strong base.
Slide 21
ACID-BASE TITRATIONS
Slide 22
Titrations Determine amount of an acid or base in solution.
Graphically understand chemistry Graphically determine K a and
compare acid strength (Equivalence point)
Slide 23
Titration curves Plotting the pH of the solution being analyzed
as a function of the amount of titrant added. Equivalence
(Stoichiometric) Point point in the titration when enough titrant
has been added to react exactly with the substance in solution
being titrated. (can determine how much acid or base was in your
original solution).
Slide 24
Titration of: 50.0 mL of 0.200 M HNO 3 with 0.100 M NaOH Enough
OH - added to react with all H+ pH = 7 @ e.p. for strong acidstrong
base titrations Relatively little pH change in beginning: early on
there is a large amount of H+
Slide 25
Titration of: 100.0 mL of 0.50M NaOH with 1.0 M HCl
Slide 26
Weak acid-Strong base titrations pH @ E.P is NOT 7. Always
greater than 7 Why? Amount of acid, not strength, determines
equivalence point. Strength of acid, not amount, determines pH at
the equivalence point.
Slide 27
Weak acid- Strong base titrations pH increases rapidly in
beginning (b/c weak acid). pH levels off because of buffering
Occurs where [HA] = [A-] Titration of: 50.0 mL of 0.100 M HC 2 H 3
O 2 with 0.100 M NaOH
Slide 28
Titration of: 50.0 mL of 0.100 M HC 2 H 3 O 2 with 0.100 M NaOH
For weak acid or weak base titrations: At half the equivalence
point [H+] = K a pH = pK a
Slide 29
K a trends in weak acid titrations
Slide 30
Acid-Base Indicators Marks the end point of a titration by
changing color (endpoint). The equivalence point is not necessarily
the same as the end point (but they are ideally as close as
possible). Acid and Base forms of phenolphthalein
Slide 31
Choosing the best indicators
Slide 32
Different indicators change color and a specific pH 0.10 M HCl
with 0.10 M NaOH0.1 M H 3 COOH with 0.1 M NaOH
Slide 33
Determining the effective range of an indicator The color
transition (endpoint) occurs over a range of pH determined by: When
titrating an acid: pH = (pK a(indicator) -1) When titrating a base:
pH = (pK a(indicator) +1) It is best to choose an indicator whose
endpoint is as close as possible to the equivalence point.
