How do We Calculate Acid-Base Equilibria?
AP Chemistry Unit 15: Guided Notes
Unit 15: Acid-Base Equilibria
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Solutions of Acids or Bases Containing a Common Ion
Buffered Solutions
Buffering Capacity
Titrations and pH Curves
Acid-Base Indicators
Introduction
Most chemistry in the natural world takes place in an aqueous solution. One of the most
significant aqueous reactions is one with acids and bases. Most living systems are very
sensitive to pH and yet are subjected to acids and bases.
o The main idea of this unit is to understand the chemistry of a buffered system.
Buffered systems contain chemical components that enable a solution to be resistant
to change in pH.
15.1 Solutions of Acids or Bases Containing a Common Ion
Common Ions
Solutions that contain a weak acid and the salt of it conjugate base create a common ion
system.
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A common ion system can also be a weak base and the salt of its conjugate acid.
o Example:
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This dissociation increases the concentration of the conjugate ion and a shift in
equilibrium occurs according to Le Chatelier.
o Example:
NH 4Cl(s) ® NH +4(aq) + Cl(aq)
NH 3(aq) + H 2O(l ) « NH +4(aq) + OH (aq)
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Common Ion Effect
The shift left in the second equation is known as the common ion effect.
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Calculations for common ion equilibria is similar to weak acid calculations except that the
initial concentrations of the anion are not 0.
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Practice Problem #1: The equilibrium concentration of H+ in a 1.0 M HF solution is
2.7x10-2M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent
dissociation of HF in a solution containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M NaF.
15.2 Buffered Solutions
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The most important application of a common ion system is buffering.
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A buffered solution resists a change in its pH when either hydroxide ions or protons
are added.
o Blood is a practical example of a buffered solution; it is buffered with carbonic
acid and the bicarbonate ion (among others).
A solution can be buffered at any pH by choosing the appropriate components.
Practice Problem #2: A buffered solution contains 0.50 M acetic acid (Ka= 1.8 x 10-5) and
0.50 M sodium acetate. Calculate the pH of this solution.
How Does Buffering Work?
When OH- ions are added to a solution of a weak acid, the OH- ions react with the best source
of H+ (weak acid) to make water:
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When H+ ions are added to a solution of a weak acid, the H+ ions react with the strong
conjugate base.
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Buffered Solution Summary
Buffered solutions are simply solutions of weak acids or bases containing a common ion. The
pH calculations on buffered solutions require exactly the same procedures introduced in
Chapter 14.
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When a strong acid or base is added to a buffered solution, it is best to deal with the
stoichiometry of the resulting reaction first. After the stoichiometric calculations are
completed, then consider the equilbrium calculations.
Practice Problem #3: Calculate the change in pH that occurs when 0.010 mol solid NaOH
is added to the buffered solution in #2 (0.50 M acetic and 0.50 M sodium acetate).
Compare this pH change with the original (pH=4.74) and the pH that occurs with 0.010
mol of solid NaOH is added to 1.0 L of water.
Henderson-Hasselbalch Equation
Buffered solutions work well as long as the concentration of the salts and weak acids/bases
in solution are much larger than the amount of OH- and H+ being added.
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This equation,
, can be changed into another useful form by taking
the negative log of both sides:
This log form of the expression is called the Henderson-Hasselbalch Equation.
Henderson-Hasselbalch Equation: When using this equation it is often assumed that the the
equilibrium concentrations of A_ and HA are equal to their initial concentrations due to the
validity of most approximations.
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Practice Problem #4: Calculate the pH of a solution containing 0.75 M lactic acid
(Ka=1.4x10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common
constituent of biologic systems. For example, it is found in milk and is present in
human muscle tissue during exertion.
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Reminder
Buffered solutions can be formed from a weak base and the corresponding conjugate acid. In
these solutions, the weak base B reacts with any H+ added:
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The conjugate acid BH+ reacts with any added OH-:
Practice Problem #5: A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40
M NH4Cl. Calculate the pH of this solution.
Practice Problem #6: Calculate the pH of the solution that results when 0.10 mol
gaseous HCl is added to 1.0 L of the buffered solution from #5 (.25M NH3 and .40
NH4Cl).
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Summary
1. Buffered solutions contain relatively large concentrations of a weak acid and the
corresponding weak base. They can involve a weak acid HA and the conjugate base Aor a weak base B and the conjugate acid BH+.
2. When H+ is added to a buffered solution, it reacts essentially to completion with the
weak base:
3. When OH- is added to a buffered solution, it reacts essentially to completion with the
weak acid present:
4. 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. This will be the case as long as the concentration of the
buffering materials (HA and A- or B and BH+) are large compared with the amounts of
H+ or OH- added.
15.3 Buffering Capacity
The buffering capacity of a buffered solution represents the amount of protons or hydroxide
ions the buffer can absorb without a significant change in pH.
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The pH of a buffered solution is determined by the ratio [A-] / [HA]. The capacity of a
buffered solution is determined by the magnitudes of [HA] and [A-].
Practice Problem #7: Calculate the change in pH that occurs when 0.010 mol of gaseous
HCl is added to 1.0 L of each of the following solutions:
Soln A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
Soln B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
Ka=1.8x10-5.
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Optimal Buffering
The optimal buffering occurs when [HA] is equal to [A-].
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o For example, if a buffered solution is needed with a pH of 4.0. The most effect
buffering will occur when [HA] is equal to [A-] and the pKa of the acid is close to
4.0 or Ka=1.0x10-4.
Practice Problem #8: A chemist needs a solution buffered at pH=4.30 and can choose
from the following acids and their sodium salts. Calculate the ratio [HA]/[A-] required
for each system to yield a pH of 4.30. Which system will work best?
a.
b.
c.
d.
chloroacetic acid (Ka=1.35x10-3)
propanoic acid (Ka=1.3x10-5)
benzoic acid (Ka=6.4x10-5)
hypochlorous acid (Ka=3.5x10-8)
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15.4 Titrations and pH Curves
Titration Curve
An acid-base titration is often graphed by plotting the pH of the solution (y-axis) vs. the
amount oftitrant added (x-axis).
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Strong Acid-Strong Base Titrations
The net ionic reaction for a strong acid- strong base titration is:
To compute [H+] at any point in a titration, the moles of H+ that remains at a given point must
be divided by the total volume of the solution.
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Titrations often include small amounts. The mole is usually very large in comparison.
Molarity with mmols
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Strong Acid-Base Titration Example Calculation
50.0 ml of 0.200 M HNO3 is titrated with 0.100 M NaOH. Calculate the pH of the solution at
the following points during the titration.
a. NaOH has not been added.
b. 10.0 ml of 0.100 M NaOH has been added.
c. 20.0 mL of 0.100 M NaOH has been added.
d. 50.0 mL of 0.100 M NaOH has been added.
e. 100.0 mL of 0.100 M NaOH has been added.
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f. 150.0 mL of 0.100 M NaOH has been added.
g. 200.0 mL of 0.100 M NaOH has been added.
The results of these calculations are summarized by this graph:
The pH changes very gradually until the titration is close to the equivalence point, then
dramatic change occurs. Near the E.P., small changes produces large changes in the OH-/H+
ratio.
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Strong Acid-Base Summary
1. Before the equivalence point, [H+] can be calculated by dividing the number of
millimoles of H+ remaining by the total volume of the solution in mL.
2. At the equivalence point, pH=7.0
3. After the equivalence point, [OH-] can be calculated by dividing the number of
millimoles of excess OH- by the total volume of the solution. [H+] is calculated from Kw.
Titrations of Weak Acids with Strong Bases
When acids don’t completely dissociate, calculations used previously need adjusted.
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To calculate [H+] after a certain amount of strong base has been added, the weak acid
dissociation equilibrium must be used.
Remember:
Calculating the pH Curve for a Weak Acid-Strong Base Titration
Calculating the pH Curve for a Weak Acid-Strong Base Titration is broken down into two
steps:
1. A stoichiometric problem. The reaction of hydroxide ion with the weak acid is
assumed to run to completion, and the concentrations of the acid remaining and the
conjugate base formed are determined.
2. An equilibrium problem: The position of the weak acid equilibrium is determined, and
the pH is calculated.
Weak Acid-Strong Base Titration Example Calculation
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
a. NaOH has not been added.
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Weak Acid-Strong Base Titration Example Calculation contd..
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
b. 10.0 mL of 0.10 M NaOH has been added.
c. 25.0 mL of 0.10 M NaOH has been added.
d. 40.0 mL of 0.10 M of NaOH has been added.
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Weak Acid-Strong Base Titration Example Calculation contd..
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
e. 50.0 mL of NaOH is added.
f. 60.0 mL of 0.10 M NaOH is added.
g. 75.0 mL of 0.10 M NaOH is added.
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Weak Acid-Strong Base Titration
Some of the differences in the the titration curves:
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The pH increases more rapidly in the beginning of the weak acid titration.
Some of the differences in the the titration curves:
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The titration curve levels off near the halfway point due to buffering effects.
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The value of the pH at the equivalence point is not 7. The value of the e.p. is greater
than 7.
Practice Problem #9: Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is
highly toxic. It is a very weak acid (Ka=6.2x10-10) when dissolved in water. If a 50.0 mL
sample of 0.100 M HCN is titrated with 0.100 M NaOH, calculate the pH of the solution.
a. after 8.00 mL of 0.100 M NaOH has been added.
b. at the halfway point of the titration
c. at the equivalence point of the titration
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Important Conclusions
When comparing the example calculation and the practice problem with weak acids, two
major conclusions can be made:
1. The same amount of 0.10 M NaOH is required to reach the equivalence point even
though HCN is a much weaker acid.
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2. The pH value at the equivalence point is affected by the acid stregth. The pH at the e.p.
for acetic acid is 8.72. The pH at the e.p. for hydrocyanic acid is 10.96.
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Weak Acid Titration Curves
The strength of a weak acid has a significant effect on the shape of its pH curve. The e.p.
occurs at the same point, but the shapes of the curve is dramatically different.
Practice Problem #10: A chemist has synthesized a monoprotic weak acid and wants
to determine its Ka value. To do so, the chemist dissolves 2.00 mmol of the solid acid
in 100.0 mL water and titrates the resulting solution with 0.0500 M NaOH. After 20.0
mL NaOH has been added, the pH is 6.00. What is the Ka value for the acid?
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Titration of Weak Bases with Strong Acids Example Calculation
Find the pH of a solution of 100.0 mL of 0.050 M NH3 is titrated with 0.10 M HCl (Kb=1.8x105).
a. HCl has not been added.
b. 10.0 mL of HCl is added.
c. 25.0 mL of HCl is added.
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Titration of Weak Bases with Strong Acids Example Calculation cond..
Find the pH of a solution of 100.0 mL of 0.050 M NH3 is titrated with 0.10 M HCl (Kb=1.8x105).
d. 50.0 mL of HCl is added.
e. 60.0 mL of HCl is added.
The Titration Curve for a Weak Base with a Strong Acid
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15.5 Acid-Base Indicators
Acid-Base Indicators
There are two common methods for determining the equivalence point of a titration:
1. Use a pH meter to monitor the pH and then plot the titration curve. The center of the
vertical region of the pH curve indicates the equivalence point.
2.
The equivalence point of a titration is defined by the stoichiometry, but it is not necessarily
the same as the end point where the indicator changes color).
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Indicators
The most common acid-base indicators are complex molecules that are weak acids (HIn).
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o Example:
Indicator Example
(RED)
(BLUE)
The color will depend on the ratio of [In-] to [HIn]. For most indicators, approximately 1/10th
of the initial form must be converted to the final form before a color is perceived by the
human eye.
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Practice Problem #11: Bromthymol blue, an indicator with Ka=1.0x10-7, is yellow in its
HIn form and blue in its In- form. Suppose we put a few drops of this indicator in a
strongly acidic solution. If the solution is then titrated with NaOH, at what pH will the
indicator color change first be visible?
Henderson-Hasselbalch and Indicators
The H-H equation is very useful in determining the pH at which an indicator changes color.
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Example:Bromthymol blue (Ka=1x10-7, or pKa=7), the pH at the color change is
pH=7-1=6
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Indicator Titrations
When a basic solution is titrated, the indicator will initially exist as In- in solution, but as acid
is added more HIn is formed. Color change will occur at:
Substituting this reciprocal into the H-H equation gives us:
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Bromthymol blue=pH 7+1=8;
Indicator Ranges
THE END
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