APPLICATIONS OF AQUEOUS EQUILIBRIA
ACID – BASE EQUILIBRIA
SOLUTIONS OF ACIDS OR BASES CONTAINING A COMMON ION
1.
We are looking at a solution that contains the weak acid and its salt.
2.
Look at the weak acid HF (Ka = 7.2 x 10-4) and its salt, sodium fluoride. When a salt dissolves in water, it
completely ionizes because it is a strong electrolyte.
Hydrofluoric acid dissolves as follows:
3.
The major species in the solution are:
4.
The ______________________________ in this solution is _______.
5.
When looking at the effect the F- ion will have on the dissociation of HF, one must use Le Chatelier’s
Principle.
6.
If there are F- ions present from the NaF, then the dissociation of HF will shift ________. In other words,
the extent of the dissociation will be less in the presence of NaF.
7.
The shift in equilibrium position that occurs because of the addition of an ion already involved in the
equilibrium reaction is called the ___________________________________.
8.
This makes a solution of NaF and HF less acidic than a solution of HF alone.
9.
For a base:
Equilibrium shifts left and less OH- is produced.
***** Calculate the [H+] and percent dissociation of HF in a solution containing 1.0 M HF (Ka = 7.2 x10-4) and
1.0 M NaF.
OVER
BUFFERED SOLUTIONS
1.
A ______________________________ is one that resists a change in its pH when either hydroxide ions or
protons are added.
2.
A buffered solution may contain a weak acid and its salt or a weak base and its salt.
3.
By choosing the appropriate components, a solution can be buffered at virtually any pH.
***** Calculate the pH of a solution that is 0.60 M HF and 1.00 M KF. K a = 7.2 x 10-4.
***** Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of a buffered solution
containing 0.50 M acetic acid (HC2H3O2, Ka = 1.8 x 10-5) and 0.50 M sodium acetate (NaC2H3O2).
4.
Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH
calculations are the same as in Chapter 14.
5.
When a strong acid or base is added to a buffered solution, it is best to dea with the stoichiometry of the
resulting reaction first.
6.
After the stoichiometric calculations are completed, then consider the equilibrium calculations.
BUFFERING: HOW DOES IT WORK?
1.
Look at a buffer that contains relatively large quantities of a weak acid _______ and its conjugate base
_______.
OH- ions are not allowed to accumulate. They are replaced by A- ions.
2.
Look at the equilibrium expression:
Rearranging:
3.
The equilibrium concentration of H+, and thus the pH, is determined by the ratio
4.
When OH- ions are added, HA is converted to A- and the ratio decreases.
5.
If the amounts of HA and A- originally present are very large compared to the amount of OH- added, the
change in the ratio will be small.
6.
Buffering, then, is that [HA] and [A-] are large compared to the amount of OH- added. Thus, when the OHis added, the concentrations of HA and A- change, but only by small amounts.
7.
The ratio and [H+] remain virtually constant.
8.
Similar reasoning applies when protons are added to a buffered solution of a weak acid and a salt of its
conjugate base.
9.
We can calculate [H+] of a buffered solution using:
OVER
Taking the –log of both sides yields:
10. This is called the ________________________________________.
11. For a particular buffering system (acid-conjugate base pair), all solutions tht have the same ratio [A-]/[HA]
will have the same pH.
***** Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25 M sodium lactate.
12. Buffered solutions can be formed from a weak base and the corresponding conjugate acid.
BUFFER CAPACITY
1. The ______________________________ of a buffered solution represents the amount of protons or hydroxide
ions the buffer can absorb without a significant change in pH.
2. The pH of a buffered solution is determined by the ratio [A-]/[HA]. The capacity of a buffered solution is
determined by the magnitude of [HA] and [A-].
***** Calculate the change in pH that occurs when 0.010 mol gaseous HCl is added to 1.0 L of each of the
following solutions:
Solution A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
Solution B: 0.050 M HC2H3O2 and 0.050 M NaC2H3O2
For acetic acid, Ka = 1.8 x 10-5
Add the HCl:
3. Optimal buffering occurs when [HA] = [A-]. The ratio [A-]/[HA] is most resistant to change.
4. When choosing the buffering components for a specific application, the pK a of the weak acid to be used in the
buffer should be as close as possible to the desired pH.
TITRATION AND pH CURVES
1.
A titration is commonly used to determine the amount of acid or base in a solution.
2.
The process involves a solution of known concentration (the titrant) delivered from a buret into the
unknown solution until the substance being analyzed is just consumed.
3.
The stoichiometric (_________________________) point is often signaled by the color change of an
indicator.
OVER
4.
The progress of an acid – base titration is often monitored by plotting the pH of the solution being analyzed
as a function of the amount of titrant added.
5.
Such a plot is called a _________________________ or ______________________________.
STRONG ACID – STRONG BASE TITRATIONS
1.
The net ionic equation for a strong acid – strong base titration is:
2.
To compute [H+] at a given point in the titration, we must determine the amount of H + that remains at that
point and divide by the total volume of the solution.
3.
A unit used for titrations is the _______________________________
4.
Case study of a strong acid – strong base titration:
50.0 mL of 0.200 M HNO3 with 0.100 M NaOH
Reaction before any NaOH is added:
Other species present:
10.0 mL NaOH is added, but before it reacts:
Major species:
Reaction:
After the reaction, the major species are:
20.0 mL (total) NaOH has been added:
50.0 mL (total) NaOH has been added:
100.0 mL (total) NaOH has been added:
OVER
150.0 mL (total) NaOH has been added:
5.
The pH curve can summarize all of these calculations. pH changes gradually at first because there is a
relatively large amount of H+ in solution compared to the OH-. This produces small changes in pH.
6.
As the titration nears the equivalence point, [H+] is relatively small and adding OH- produces large pH
changes.
7.
Before the equivalence point, [H+] (and thus pH), can be calculated by dividing the number of millimoles
of H+ remaining by the total volume of the solution in milliliters.
8.
At the equivalence point, the pH is 7.00.
9.
After the equivalence point, [OH-] can be calculated by dividing the number of millimoles of excess OH - by
the total volume of solution. Then [H+] is obtained from Kw.
10. Titration of a strong base with a strong acid is similar except that OH- is in excess before the equivalence
point and H+ is in excess after the equivalence point.
TITRATIONS OF WEAK ACIDS WITH STRONG BASES
1.
To calculate [H+] after a certain amount of strong base has been added, we must deal with the weak acid
dissociation equilibrium.
2.
Calculation of the pH curve for a titration of a weak acid with a strong base really amounts to a series of
buffer problems.
3.
Even though the acid is weak, it reacts _____________________________________ with hydroxide ion, a
very strong base.
4.
Calculating the pH curve for a weak acid – strong base titration involves two steps:
5.
It is essential to do these steps separately.
WEAK ACID – STRONG BASE TITTRATION
1.
Consider the titration of 50.0 mL of 0.100 M acetic acid with 0.10 M NaOH.
2.
No NaOH has been added.
3.
10.0 mL NaOH has been added.
Major species before the reaction:
After the reaction has taken place, the major species are:
The equilibrium problem becomes:
4.
Each addition of the base is calculated the same way; always go back to the original reaction.
5.
At the point that is halfway to the equivalence point,
OVER
6.
50.0 mL (total) of 0.10 M NaOH has been added. At this point, 5.0 mmol OH - has been added which will
react with 5.0 mmol HC2H3O2 originally present.
The major species are:
7.
The pH at the equivalence point of a titration of a weak acid with a strong base is always greater than 7.00.
8.
After the equivalence point, the excess OH- controls the pH.
9.
When comparing pH curves one can see that the shapes of the plots are different before the equivalence
point and similar afterwards.
10. Near the beginning of the titration of a weak acid, the pH increases more rapidly than it does in the case of
a strong acid.
11. The pH levels off near the equivalence point because of buffering effects.
12. After the equivalence point, the curves are similar in shape because excess OH - controls the pH in this
region.
13. For the titration of a strong acid, the equivalence point occurs at a pH of 7.00. For the titration of a weak
acid, the pH at the equivalence point is greater than 7.00 because of the basicity of the conjugate base of the
weak acid.
14. The equivalence point in an acid – base titration is defined by the _________________________, not by
the pH.
***** Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (K a =
6.2 x 10-10) 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.
OVER
15. If one compares the titration of acetic acid to hydrocyanic acid one can see that it is the _______________
of acid, not its strength,that determines the equivalence point.
16. The pH value at the equivalence point _______ affected by the acid strength.
17. The strength of a weak acid has a significant effect on the shape of its pH curve. The weaker the acid, the
greater is the pH value at the equivalence point.
CALCULATION OF Ka
***** A chemist has synthesized a monoprotic acid and wants to determine its K a 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 of the acid?
TITRATION OF WEAK BASES WITH STRONG ACIDS
1.
Use the same procedures already discussed.
2.
Stoichiometry first, then equilibrium.
ACID – BASE INDICATORS
1.
There are two common methods for determining the equivalence point of an acid – base titration.
2.
One method makes use of a pH meter to monitor pH and then plot the titration curve. The center of the
vertical region of the pH curve indicates the equivalence point.
3.
Another is to use an _________________________ which marks the end-point of a titration by changing
color.
4.
Although the equivalence point of a titration, defined by the stoichiometry, is not necessarily the same as
the end point (where the indicator changes color), careful selection of the indicator will ensure the error is
negligible.
5.
The most common acid-base indicators are complex molecules that are themselves weak acids, ________.
6.
They exhibit one color when the proton is attached to the molecule and a different color when the proton is
absent.
7.
For most indicators, about one-tenth of the original form must be converted to the other form before a new
color is apparent.
8.
For the titration of an acid with a base, the color change will occur at a pH where:
***** Bromthymol blue, an indicator with a Ka value of 1.0 x 10-7, is yellow in its HIn form and blue in its Inform. A few drops are added to a strongly acidic solution which is titrated with NaOH. At what pH will the
indicator color change be visible?
9. Color change for an acid is predicted to occur at:
10. Color change for a base is predicted to occur at:
11. For a strong acid – strong base titration, indicator color changes will be sharp. They will occur with the
addition of a single drop of titrant.
12. The titration of a weak acid is different. The weaker the acid being titrated, the smaller is the vertical area
around the equivalence point. We must choose an indicator whose useful pH range has a midpoint as close
as possible to the pH at the equivalence point.
OVER
SOLUBILITY EQUILIBRIA AND THE SOLUBILITY PRODUCT
1. Two competing processes occur: the dissolution reaction and its reverse:
2. When dynamic equilibrium is reached, the solution is said to be ____________________.
3. The equilibrium expression is:
4. __________ is the ______________________________ or _________________________.
5. The amount of excess solid present does not affect the position of the solubility equilibrium.
6. More surface area of the solid means the rate of dissolution will increase, but the rate of re-formation of the
solid will increase by the same amount.
7. Neither the amount of excess solid nor the size of the particles present will shift the ____________ of the
solubility equilibrium.
8. The solubility product is an _________________________ and has only __________ value for a given solid
at a given temperature.
9. Solubility is an _________________________.
***** Copper (I) bromide has a measured solubility of 2.0 x 10 -4 mol/L at 25.00C. Calculate its Ksp value.
***** Calculate the Ksp for bismuth sulfide (Bi2S3) which has a solubility of 1.0 x 10-15 mol/L at 25.00C.
***** The Ksp value for copper (II) iodate, Cu(IO3)2, is 1.4 x 10-7 at 25.00C. Calculate its solubility at this
temperature.
RELATIVE SOLUBILITIES
1. A salt’s Ksp value gives us information about its solubility.
2. But, be careful when using Ksp values to predict the relative solubilities of a group of salts.
3. Relative solubilities can be predicted by comparing Ksp values only for salts that produce the same total
number of ions.
***** For each of the following pairs of solids, determine which solid has the smallest molar solubility.
a. FeC2O4 (Ksp = 2.1 x 10-7) or Cu(IO4)2 (Ksp = 1.4 x 10-7)
b. Ag2CO3 (Ksp = 8.1 x 10-12) or Mn(OH)2 (Ksp = 2.0 x 10-13).
COMMON ION EFFECT
1. We need to consider what happens when an ionic solid dissolves in water that already contains an ion in
common with the dissolving salt.
2. Let’s look at dissolving Ag2CrO4 in a 0.100M solution of AgNO3.
Before any Ag2CrO4 dissolves the major species are:
And
OVER
At equilibrium:
Solubility of Ag2CrO4:
The solubility of Ag2CrO4 in pure water is 1.3 x 10-4 mol/L.
3. The solubility of a solid is lowered if the solution contains ions common to the solid.
pH AND SOLUBILITY
1. The pH of a solution can greatly affect a salt’s solubility.
2. If the anion _____ is an effective base – that is, if HX is a weak acid – the salt _____ will show an increased
solubility in the acidic solution.
3. Common anions that are effective bases are:
4. Salts containing these anions are much more soluble in an acidic solution than in pure water.
PRECIPITATION AND QUALITATIVE ANALYSIS
1. To predict whether a precipitate will form when two solutions are mixed, use the ____________________.
2. The ion product is defined the same way as Ksp except the ______________________________ are used
instead of equilibrium concentrations.
3. For solid CaF2, the ion product expression is:
4. If we add a solution of Ca2+ ions to a solution containing F- ions, a precipitate may, or may not, form. It will
depend on the concentrations of these ions in the resulting mixed solution.
5. To predict whether a precipitate will form, consider the relationship between Q and K sp.
6. If _________ the precipitation will occur and will continue until the concentrations are reduced to the point
that satisfy Ksp.
7. If __________ then no precipitation occurs.
***** Will a precipitate form when 100.0 mL of 0.020 M Pb(NO 3)2 is added to 100.0 mL of 0.020 M NaCl?
***** A solution is prepared by mixing 50.0 mL of 0.100 M Pb(NO 3)2 with 50.0 mL of 1.00 M KCl. Calculate
the concentrations of Pb2+ and Cl- at equilibrium.
Ksp = 1.6 x 10-5
First determine whether a precipitate forms:
If the reaction goes to completion, then:
But, at equilibrium some of the PbCl2 re-dissolves:
At equilibrium:
OVER
SELECTIVE PRECIPITATION
1. Mixtures of metal ions in aqueous solutions are often separated by a process known as
______________________________.
2. Use a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture.
***** A solution contains 0.259 M Ni(NO3)2 and 0.250 M Cu(NO3)2. Can the metal ions be separated by
slowly adding Na2CO3? Assume that for successful separation 99% of the metal ion must be precipitated
before the other metal ion begins to precipitate. Assume no volume change with the addition of Na 2CO3.
Ksp for NiCO3 = 1.4 x 10-4
Ksp for CuCO3 = 2.5 x 10-10
For NiCO3, the Ksp expression is:
This is the concentration at which Ni2+ begins to precipitate.
For CuCO3, the Ksp expression is:
This is the concentration at which Cu2+ begins to precipitate.
To find the [Cu2+] when the NiCO3 begins to form:
For successful separation, 1% Cu2+ or less of the initial amount of Cu2+ may be present before the NiCO3 begins
to precipitate. The percent Cu2+ present when NiCO3 begins to precipitate is:
3. Since metal sulfides differ dramatically in their solubilities, the sulfide ion is often used to separate metal
ions by selective precipitation.
4. One real advantage of the sulfide ion as a precipitating reagent is that because it it basic, its concentration
can be controlled by regulating the pH of the solution.
5. The most insoluble sulfide salts, such as CuS and HgS, can be precipitated from an acidic solution leaving
the more soluble ones behind. These can then be precipitated by making the solution slightly basic.
QUALITATIVE ANALYSIS
1. The classic scheme for separating a mixture containing all the common cations involves first separating them
into five major groups based on their solubilities.
2. Each group is then treated further to separate and identify the individual ions.
3. See Figure 15.12.
OVER
COMPLEX ION EQUILIBRIA
EQUILIBRIA INVOLVING COMPLEX IONS
1. A ____________________ is a charged species consisting of a metal ion surrounded by ________________.
2. A ligand is simply a Lewis base. It is a molecule or ion having a lone electron pair that can be donated to an
empty orbital on the metal ion to form a covalent bond.
3. some common ligands are:
4. The number of ligands attached to a metal ion is called the ______________________________. The most
common coordination numbers are ________________ and ______, but others exist.
Examples:
5. Metal ions add ligands one at a time in steps characterized by equilibrium constants called
______________________________ or ______________________________.
6. In solution, all species will exist at equilibrium.