Chapter 16
Acid-Base Equilibria and Solubility Equilibria
This is the third of the three-chapter sequence which discusses equilibrium concepts. Upon completion of this chapter, your students should be able to:
- Describe the common ion effect as a special case of Le Châtelier’s principle.
- Formulate the Henderson-Hasselbalch equation.
- Use the Henderson-Hasselbalch equation to determine the pH of a solution containing a weak acid (weak base) and its conjugate base (conjugate acid).
- Describe a buffer solution and its importance in chemical and biological systems.
- Calculate the pH of a buffer solution using the Henderson-Hasselbalch equation.
- Calculate the pH of a buffer solution after the addition of H+ or OH–.
- Describe how to prepare a buffer of a desired pH.
- Predict the pH profile of a strong acid-strong base titration and calculate the pH at any stage of the titration.
- Predict the pH profile of a strong acid-weak base (or strong base-weak acid) titration and calculate the pH at any stage.
- Distinguish between end point and equivalence point of a titration.
- Describe common acid-base indicators and name the correct method of selection for a specific titration.
- Use the concepts of equilibrium to relate ion product, Q, with Ksp to predict if a solution is unsaturated, saturated, or supersaturated.
- Explain the solubility of a substance in terms of molar solubility and solubility in terms of mass.
- Use the concept of fractional precipitation to predict concentration of insoluble ions in a solution.
- Calculate the solubility of an insoluble ion when a common ion is present.
- Describe how changing pH can affect solubility.
- Use the concepts of equilibrium, formation constant (Kf) and complex ion formation to predict solubility and ion concentration on solubility.
- Describe how solubility product principle is used in qualitative analysis.
Section 16.1 Homogeneous versus Heterogeneous Solution Equilibria
This chapter is split into two main topics: homogeneous equilibria and heterogeneous equilibria. Examples of heterogeneous equilibria include the solubility of slightly soluble salts such as AgCl and BaSO4.
Section 16.2 The Common Ion Effect
LeChâtelier’s principle states that a system at equilibrium will shift to relieve any stress that is applied to the system. For a weak acid
HA H+ +A–
the amount of H+ present in an aqueous solution is dependent upon the Kavalue for the specific acid. In many cases we can correctly assume that
[H+] < [HA]
or that the percent of ionization is very small. If the stress of adding the common ion A– is placed on the equilibrium system, then the equilibrium will shift in the direction of HA and even less of the weak acid will ionize. Therefore, if equal concentrations of a weak acid and its conjugate base are present in a solution, it is a safe assumption that the weak acid does not ionize and that
[HA] = [A–]
The Henderson-Hasselbalch equation
is easy to derive and is often referred to as the buffer equation. It should be noted that
pKa =–logKa
is consistent with what is stated in Chapter 15. Students may want to use the Henderson-Hasselbalch equation for weak bases too. However, to do so your students must be sure to use Kafor the conjugate acid of the weak base being studied. You can show that a similar equation to theHenderson-Hasselbalch equation can be derived using
to obtain
where the equilibrium
B− + H2O HB + OH−
is used.
Students may find it confusing to shift back to the Kafor the conjugate acid of the weak base being studied. They may question why we use Kaif we are working with a weak base. The pOH equation allows students to work with the pOH and Kbwhich may seem more logical to them since they are working with bases. The ultimate result should (hopefully) be the same.
Section 16.3 Buffer Solutions
A buffer solution is one that resists changes in pH when small amounts of acid or base are added to it. It is made by mixing a weak acid and its conjugate base or a weak base and its conjugate acid. When working with buffer solutions, the pH changes very little upon the addition of a small amount of acid as demonstrated in Example 16.3. It is worth mentioning that the addition of any amount of acid requires the pH of the solution to decrease. If, by chance, students do not recall the Henderson-Hasselbalch equation correctly and inadvertently place the concentration of the acid in the numerator and the concentration of the conjugate base in the denominator, the addition of H+ions will result in the pH increasing, which is obviously incorrect.
Example 16.4 demonstrates the method for preparing a “phosphate buffer” with a pH of 7.40. The solution requires 1.5 moles of Na2HPO4 and 1.0 mole NaH2PO4. As a matter of practice and as a review of some fundamental concepts, the amount of Na2HPO4 and NaH2PO4required could be reported in grams rather than moles.
Section 16.4 Acid-Base Titrations
Remind the students that the net ionic equation is
H+(aq) + OH–(aq) H2O()
for all reactions involving strong acids neutralizing strong bases. Therefore, it is reasonable that all strong acid-strong base titration curves should have the same shape. It is also explained that for the titration of 0.10 M HCl with 1.10 M NaOH, the pH approaches 13, which is the expected pH of 1.10 M NaOH.
An interesting point to determine the pH of a titration curve is when the titration is half way completed. Example 16.5 determines that the equivalence point is when 25 mL of 0.10 M NaOH is added. Therefore, when 12.5 mL of 0.1 M NaOH is added, the titration is half way completed. The reason this point is of interest is because of the following:
We started with 2.5 × 10–3mol acetic acid. Subtracting the 1.25 × 10–3mol of acetic acid that reacted with the NaOH that had been added, we ended up with 1.25 × 10–3mole of acetic acid left. We are at the point in the titration where
[CH3COO−] = [CH3COOH]
Using the Henderson-Hasselbalch equation,
we get pH = pKa at this point. Therefore, if we experimentally determine the value of pH at this point in titration, we have also determined the pKa for the acid.
Section 16.5 Acid-Base Indicators
An acid-base indicator is a weak acid or a weak base that changes color as the pH changes. If the concentration of this weak acid or weak base is very small (it needs to be concentrated enough so that our eyes can sense a color change), the addition of a couple of drops into a typical titration will not dramatically change the amount of base required to neutralize the acid. The point that our eyes sense a color change is the end point for that indicator. The goal is to select an indicator that has an end point that corresponds as closely as possible to the equivalence point–the point where the number of moles of base added equal the number of moles of acid in the original solution.
A simple and effective project to show the effect of pH on the staining ability of food coloring on eggs using solutions that are acidic, basic, and neutral. It becomes obvious why we use vinegar in the colored solutions to dye eggs.
Another useful demonstrationis to start with a few drops of phenolphthalein in a flask. Add about 10–15 mL of 0.1 M HCl to this flask. Since phenolphthalein is clear in an acid solution, it will remain clear. If we now add 0.1 M NaOH so that we go beyond the equivalence point, the solution turns pink. If we add acid, it will again turn clear. If we add base, it will turn pink; and so on if we add a few drops of bromothymol blue to 10–15 mL of deionized water, the color will be green. If we add acid, it turns yellow.
Now if the students are asked to predict what the color will be when base is added, they will likely answer either green or clear. However, when a base is added, the color changes to blue. This color change surprises most students. If one slowly adds acid to this blue solution, it is possible to catch the green color which is a combination of the yellow color of the acid from the indicator and the blue color, the basic form. This indicator works well to illustrate the difference between end point and equivalence point of a titration.
Finally, a non-toxic acid-base indicator can be made by simply using a blender to grind red cabbage in water. Once the cabbage has been blended into a slurry, strain the liquid into a container and you have a natural product indicator. Common household materials such as baking soda, lemon juice, vinegar, cola, etc. can also be tested. It is very important to warn the students not to arbitrarily mix household chemicals because certain combinations can be dangerous. Materials as drain cleaner and ammonia products should not be demonstrated because of their inherent danger to unsuspecting experimenters.
Section 16.6 Solubility Equilibria
Solubility equilibria are based on the assumption that the salts that dissolve are insoluble strong electrolytes. For some students using insoluble and strong electrolyte to describe the same material is contradictory. They have difficulty understanding how a salt can be insoluble and yet be a strong electrolyte. This is because all of the examples we give of strong electrolytes are soluble salts; therefore, it is assumed that all strong electrolytes must be soluble. It is important for students to understand that strong electrolyte solutions contain only the ions of the salt and no molecular species. Thus the concept of an insoluble salt dissolving slightly to give only ions can be explained. Once this is understood, then insoluble and strong electrolyte describing the same solution will be acceptable.
For the reaction
AgCl(s) Ag+(aq) + Cl–(aq)
The equilibrium expression can be written as
If we assume that silver has a density of approximately 10 grams per cubic centimeter, then it would contain about 100 moles of solid in a one liter block. We could then state that the denominator of the above expression is 100 moles per liter. If that solid chunk of silver was split so that it is one-half liter in volume, it would contain 50 moles. The ratio of moles to liters remains at 100. If that half liter of solid is split again, the ratio of moles to liters still remains at 100. This same logic can be used for the smallest piece of solid – the ratio of moles to volume will remain constant. Therefore, it can be incorporated into the equilibrium constant and the more useful expression of
Ksp = [Ag+][Cl–]
becomes effective.
It is important students understand that Kspvalues by themselves do not necessarily correspond directly to solubility. For example, Mg(OH)2has aKspvalue of 1.2 × 10–11while AgCl has a Kspvalue of 1.6 ×10–10. A complicating factor is that solubility is expressed in grams of solute that will dissolve per liter of solution versus molar solubility which is expressed in moles per liter. If two compounds happen to have the same molar solubility, then the compound with the largest molar mass will have the larger quantity of compound with the larger solubility measured in grams that dissolve per liter of solution.
Students sometimes worry about what they should assign the variable x to when they solve these types of problems. They fail to see that it doesn’t matter. For example, if one is to determine the molar solubility of Mg(OH)2, it doesn’t matter if we let x equal the amount of Mg(OH)2that dissolves or the amount of Mg2+that forms or the amount of OH–that forms. It is true that the algebra may be a bit easier, but the results must be the same if done correctly. It may be worthwhile to demonstrate this point to your students.
Section 16.7 Separation of Ions by Fractional Precipitation
To make the importance of solubility calculations show relevance, it may be useful to demonstrate how such calculations can be important in industry. For example,
Ksp= [Ag+][Br –] = 7.7 × 10–13
therefore,
[Ag+] = 8.8 × 10–7 mol/L
The question that is posed is what is the Ag+ concentration in parts per million. To do this calculation it is recalled that parts per million (ppm) is 1 × 106grams of solution.
Section 16.8 The Common Ion Effect and Solubility
To show the common ion effect and its affect on solubility, we can take the 1.095 ppm Ag+determined above and show we can reduce it to reach a desired level. For instance, let us assume that a local environmental regulation will allow only 1.1010 ppmAg+to beexpelled into a local river from our plant. By adding NaBr to our waste stream we can lower that Ag+below this restriction. This example is shown below.
The problem is to determine the moles per liter of NaBr that must be added to lower the Ag+concentration to below 0.010 ppm.
[Ag+][Br–] = 7.7 × 10–13
(9.3 × 10–8 mol Ag+ /L)[Br–] = 7.7 × 10–13
[Br–] = 8.3 × 10–6 mol/L
Therefore, we need to add NaBr on the order of 1×10–5 mol/L to maintain the Ag+concentration below 0.010 ppm.
Section 16.9 pH and Solubility
Just as we demonstrated the common ion effect by adding NaBr to the aqueous solution of AgBr, we could add OH–to a solution containing Mg2+. The stress placed on the system will drive the equilibrium in the direction to produce more solid Mg(OH)2. A classical exercise is to ask your students to calculate the pH that would result from preparing two different buffer solutions and then determine the maximum of Mg2+ concentration in those two solutions. This exercise brings together several concepts including pH, Henderson-Hasselbalch equation, and solubility.
Section 16.10 Complex Ion Equilibria and Solubility
Complex ion formation equilibria are, in many respects, the opposite of solubility equilibria. The ability of ions to form complexes allows the solubility of insolubility of insoluble materials to increase dramatically. The classical example is the beautiful solution that is formed when ammonia complexes with copper to form . The blue solution is spellbinding.
Your students should note that complex ion formation favors the formation of products, thus the value of the equilibrium constant will have a positive exponent. Students could easily fall into the trap of assuming that complex ion formation equilibria will have negative exponents because up to this time,they have been using equilibrium constants with negative exponents.
Chemistry in Action on page 762 discusses “How an Eggshell is Formed.”
Section 16.11 Application of the Solubility Product Principle to Qualitative Analysis
Over the past few years, the use of qualitative analysis using wet chemical methods has lost favor. It can be argued that the multi-step qualitative analysis scheme can be used to teach students not only several laboratory techniques, but also an appreciation for solubility and the beauty of the precipitates that form.
a)Group 1 cations are the insoluble chlorides.
b)Group 2 cations are the insoluble sulfides in acidic solution.
c)Group 3 cations are the insoluble sulfides in basic solution.
d)Group 4 cations are the insoluble carbonates.
e)Group 5 cations are the soluble cations that are left.
It should be noted that the cations in Group 1 must be removed before Group 2 can be determined and Group 1 and 2 must be removed before Group 3, etc. It should be pointed from Table 16.2 that silver sulfide has a Kspvalue of 6.0 ×10–51 suggesting that it would be very insoluble in the presence of sulfide ions. Therefore, if any silver ions remain in the solution to be tested for Group 2 cations, they will precipitate giving rise to confusing results.
It should be emphasized that these Group numbers have nothing to do with the Group number designations of the periodic table.
It is certainly true that the use of chemical instrumentation to determine both qualitative and quantitative analysis has driven many chemical educators to downplay this qualitative analysis scheme. On the other hand,allowing students the opportunity to use the wet-chemical method to solve an unknown, and experience the beauty of the many colored precipitates that form when performing the wet-chemical method,is valuable because those experiences leave a lasting impression on most students.