Chapter 16 – Acid – Base Equilibria I. The Common Ion Effect a. Here, we will discuss the acid-­‐base properties of a solution with two solutes containing the same (i.e. a common) ion. This ion can be a cation or an anion. b. Consider two solutes, NaF and HF (for HF, Ka = 7.2 x 10-­‐4). NaF is a salt and it dissociates completely into ions when mixed with water. This is expressed by the reaction: NaF(s)  Na+(aq) + F-­‐(aq) c. In the same solution, we have HF (aq), a weak acid. The ionization of the weak acid is expressed as: HF(aq)  H+(aq) + F-­‐(aq) Ka = 7.2 x 10-­‐4 d. Note: The fact that Ka > Kw indicates that in this solution, H+ is produced predominantly by the dissociation of HF. e. In this solution, F-­‐ is the common ion produced from NaF and HF. The presence of the common ion, F-­‐, may have a significant effect on the equilibrium dissociation of the weak HF acid. f. Let us focus on the ionization reaction of the weak acid, HF. HF(aq)  H+(aq) + F-­‐(aq) g. What is the consequence of adding NaF to an equilibrium solution of HF? i. Adding NaF to the HF solution increases [F-­‐] and disrupts the equilibrium. ii. Recall Le Chatelier’s Principle: “When a disturbance is imposed on a reaction system at equilibrium, the equilibrium shifts in the direction that reduces the effects of the disturbance.” iii. Hence, increasing [F-­‐] by addition of NaF results in a shift of the equilibrium to the left. In turn, this shift reduces [H+], and the pH. iv. This is called the common ion effect h. In summary, the common ion effect is the shift of an equilibrium caused by addition of a solute having an ion in common with the equilibrium system. i. The common ion effect has two important roles: i. It allows one to change the pH of a solution. ii. It allows one to change the solubility of salts that are slightly soluble in water. j. Consider a general solution of a weak acid, HA, and a salt, NaA. The salt dissociates completely into ions as expressed by the reaction: NaA(s)  Na+(aq) + A-­‐(aq) k. Note: The complete dissociation of the salt into its ions means that if we are given 0.1 M NaA, then [Na+] = 0.1 M, and [A-­] = 0.1 M. l.
The ionization reaction of the weak acid, HA, is expressed as: HA(aq)  H+(aq) + A-­‐(aq) m. Recall that A-­‐ is the weak conjugate base of the weak acid HA. The equilibrium constant for the ionization reaction is expressed as: Ka =
[H+ ][A− ]
[HA]
[H+ ] =
−log [H+ ] = − log€K a − log
€
Thus, pH = pK a − log
€
€
[K a ][HA]
[A− ]
[HA]
[A− ]
[HA]
[A− ]
Hence, pH = pK a + log
[A− ]
[HA]
n. A-­‐ is the weak conjugate base of the weak acid HA. o. This € equation is called the Henderson-­Hasselbalch equation [conjugate base]
[acid]
p. One can use this equation to calculate the pH for solutions of compounds having a common ion. q. In problems that involve the common ion effect, the initial €
concentrations of the salt and the weak acid are usually given. r. As long as the initial concentrations of the salt and weak acid are greater than 0.1 M, we can neglect the ionization of the acid and the hydrolysis of the salt (reaction of the salt with water). s. Under such conditions, we can use the initial concentrations for the acid and the conjugate base in the Henderson-­‐Hasselbalch equation. II. Equilibrium Calculations for Bases with the Common Ion Effect a. The common ion effect can also be applied to a solution containing a weak base and a salt of the base. b. Consider a solution of a weak base, NH3, and a salt, NH4Cl. The salt, NH4Cl, dissociates completely in H2O according to the reaction: NH4Cl (s)  NH4+ (aq) + Cl -­‐ (aq) c. We know that NH4+ is an acid, and that it can donate a proton to water. Thus, NH4+ (aq)  NH3 (aq) + H+ (aq) d. The equilibrium constant expression for this reaction is: Therefore, pH = pK a + log
Ka =
€
[NH3 ][H+ ]
+
[NH4 ]
e. Rearrange this expression to solve for [H+]. +
+
K [NH4 ]
[H ] = a
[NH3 ]
[NH4 ]
−log[H ] = −logK a − log
[NH3 ]
+
+
+
€
€
[NH4 ]
pH = pK a − log
[NH€
3]
pH = pK a + log
[NH3 ]
+
[NH4 ]
[base]
pH = pK a + log
[conjugate acid]
€
III. Buffer Solutions €
a. Buffer solutions have the ability to resist changes in pH upon addition of small amounts of either acid or base. b. Buffers are very important in chemical and physiological processes. c. The pH in the human body varies from one fluid to another. i. For example, the pH of blood is 7.4, whereas the pH of gastric juices in the stomach is about 1.5. ii. Blood can absorb the acids and bases produced in biological reactions without changing its pH. iii. Thus, blood is a buffer solution. d. A buffer solution consists of a weak acid and its salt or a weak base and its salt. e. In other words, a buffer solution follows the concepts of common ion solutions. f. An example of an acidic buffer solution is a mixed solution of the weak acid, HF, and its salt, NaF. g. Similarly, an example of a basic buffer solution is a mixed solution of the weak base, NH3, and its salt, NH4Cl. h. By choosing appropriate solutes and concentrations, a buffer solution can be created for any desired pH. i. Consider a buffer solution of a weak acid, HA, and its salt, NaA. The salt dissociates as: i. NaA(s)  Na+(aq) + A-­‐(aq) and the weak acid as: ii. HA(aq)  H+(aq) + A-­‐(aq) j. When the buffer solution is at equilibrium, we can calculate the pH using the Henderson-­‐Hasselbalch equation. [A− ]
k. Now, let’s disturb this buffer solution by pH = pK a + log
adding some NaOH. NaOH dissociates [HA]
completely in solution: i. NaOH(s)  Na+(aq) + OH-­‐(aq) €
l.
OH-­‐ is a strong base and accepts a proton from the acid. In solution, we have the acid, HA. Thus, OH-­‐ reacts with HA as: i. OH-­‐(aq) + HA (aq)  H2O (l) + A-­‐(aq) m. Hence, adding a strong base to an acid buffer leads to the formation of more A-­‐. Now, look at the equilibrium dissociation of the weak acid. i. HA(aq)  H+(aq) + A-­‐(aq) n. The equilibrium constant for this reaction is expressed as: Ka =
[H+ ][A− ]
[HA]
or
[H+ ] =
K a [HA]
[A− ]
o. HA(aq)  H+(aq) + A-­‐(aq) p. According to Le Chatelier’s principle, increasing [A-­‐] leads to a shift of the equilibrium to the left in the direction that consumes A-­‐. Thus, the €
shift of the equilibrium results in an increase in [HA]. Thus, the ratio
[HA]
changes very little.
[A− ]
q. This implies that [H+] changes very little. Hence, the pH of an acid buffer solution does not change much when a strong base is added. €
IV. Preparing a Buffer Solution with a Specific pH a. In order to calculate the pH of a buffer solution, one uses the Henderson-­‐Hasselbalch equation. b. HA(aq)  H+(aq) + A-­‐(aq) c. Here, A-­‐ is the base and HA is the acid. Thus, pH = pK a + log
[A− ]
[HA]
d. Recall, it is the ratio [A-­‐]/[HA] that is the most resistant to change when either H+ or OH-­‐ is added to the buffer solution. The concentrations of A-­‐ and HA in the buffer solution are usually chosen €
in such a way that the ratio [A-­‐]/[HA] is as close as possible to 1. i. Thus, pH = pKa + log (1) ii. log (1) = 0 iii. Hence, pH = pKa e. Thus, when preparing a buffer solution of a desired pH, one should choose a weak acid of pKa values as close as possible to the desired pH. V. Acid-­‐Base Titrations and pH Curves a. Titration is used to determine the amount of acid or base in a solution. b. The solution being analyzed, called the analyte, is the acid or the base solution in an Erlenmyer flask or a beaker. c. The titrant is the acid or the base solution in the buret. d. The progress of an acid-­‐base titration is often shown on a graph of the pH of the solution being analyzed vs. the amount of titrant added. e. Such a graph is called a pH curve, or a titration curve. f. In this section, we will consider three types of reactions: i. Strong acid-­‐strong base titrations ii. Weak acid-­‐strong base titrations iii. Strong acid-­‐weak base titrations VI. pH Curve for Strong Acid -­‐ Strong Base Titrations a. The plot of “pH vs. Volume of Titrant Added” is called the Titration Curve, or the pH Curve. b. In the previous example, the titrant was NaOH. c. Hence, the pH curve for a strong acid-­‐strong base titration is: d. In this curve, the equivalence point is at pH = 7.00. i. Before the equivalence point, [H+] and pH can be calculated by dividing the moles of [H+] remaining by the total volume of the solution (acid + base) in Liters. ii. After the equivalence point, [OH-­‐] and the pOH can be calculated by dividing the moles of OH-­‐ remaining by the total volume of the solution (acid + base) in Liters. e. Note: In order to calculate pH, use the relation pH + pOH = 14.00. f. If the titrant is HCl, then the pH curve for a strong acid-­‐strong base titration is: g. In this curve, the equivalence point is at pH = 7.00. h. Before the equivalence point, [OH-­‐] and the pOH can be calculated by dividing the moles of OH-­‐ remaining by the total volume of the solution (acid + base) in Liters. i. After the equivalence point, [H+] and the pH can be calculated by dividing the moles of H+ remaining by the total volume of the solution (base + acid) in Liters. j. Note: In order to calculate pH, use the relation pH + pOH = 14.00. VII. Weak Acid -­‐ Strong Base Titrations a. Calculating the pH during the titration of a weak acid by a strong base is a two-­‐step process. i. The stoichiometric step: The reaction of OH-­‐ with a weak acid is assumed to be complete. The concentrations of acid remaining in solution and conjugate base formed are determined. ii. The equilibrium step: The equilibrium constant for dissociation of the weak acid, the acid and the conjugate base concentrations are used to calculate [H+], then the pH. b. The plot of pH vs. Volume of Titrant added is called the Titration Curve, or pH Curve. The pH curve for a weak acid-­‐strong base titration is shown to the right. c. The equivalence point in an acid-­‐base titration is defined by the stoichiometry. d. Why? e. Because the equivalence point occurs when sufficient titrant has been added to react exactly with all the acid or base present. f. In the titration curve of a weak acid by a strong base, the equivalence point is located at pH > 7.00. g. At the equivalence point, the solution contains only a soluble salt and water. h. The soluble salt consists of the cation from the strong base and the anion of the weak acid. i. The former does not affect the pH, while the latter makes the solution basic (see Section in the Chapter on Acids and Bases). j. The pH value at the equivalence point is affected by the strength of the acid. k. The weaker the acid, the stronger its conjugate base and the higher the pH value at the equivalence point. VIII. Strong Acid -­‐ Weak Base Titrations a. Consider the titration of HCl, a strong acid, with NH3, a weak base. HCl ionizes as: i. HCl(aq)  H+(aq) + Cl-­‐(aq) b. NH3 is a base; thus, it reacts with the acid by accepting a proton. The reaction is: i. NH3 (aq) + H+(aq)  NH4+(aq) c. At the stoichiometric point, the solution contains H2O, NH4+, and Cl-­‐. d. Since NH4+ is a weak acid and Cl-­‐ is an extremely weak base, the solution is acidic at the stoichiometric point. e. This means that NH4+ donates a proton to H2O, a weak base. f. Therefore, i. NH4+(aq) + H2O (l)  NH3 (aq) + H3O+(aq) g. Often, H2O is ignored and the reaction is written as: i. NH4+(aq)  NH3 (aq) + H+(aq) h. The important point to remember is that at the stoichiometric point, the solution is acidic and is characterized by Ka. i. However, because we start off with a weak base (NH3) and the Kb value, we need to calculate Ka from Kb. j. The plot of “pH vs. Volume of Titrant Added” is called the Titration Curve or pH Curve. The pH curve for a strong acid-­‐weak base titration is shown to the right. k. In this curve, the equivalence point is located at pH < 7.00. This is because of the acidic nature of the conjugate acid of a weak base. l. The equivalence point in an acid-­‐base titration is defined by the stoichiometry. i. In this case, the pH value at the equivalence point is determined by the strength of the base. ii. The weaker the base, the stronger its conjugate acid and the lower the pH value at the equivalence point. IX. Acid – Base Indicators a. In acid-­‐base titrations, the end point is observed by the change in color of an indicator. i. An indicator is an organic dye whose color depends on the pH of the solution. b. For example: The indicator methyl red is red at pH below 4. i. This means that if the pH of a solution is 4 or below and a drop or two or methyl red indicator is added, then the solution turns red. c. The indicator methyl red is yellow at pH above 7. i. This means if the pH of a solution is 7 or higher and a drop or two of methyl red indicator is added, then the solution turns yellow. d. As the pH changes from 4 to 7, the color of the indicator changes from red, to reddish-­‐orange, to orange and to yellow. i. Thus, through changes in color, an indicator displays the acidity, or the basicity of a solution. e. Another very common acid-­‐base indicator is phenolphthalein. i. Phenolphthalein is colorless in solutions with a PH below 8. ii. Phenolphthalein turns bright pink when the pH of a solution is above 10. f. Acid-­‐base indicators are generally weak organic acids. i. Hence, an indicator is often represented as HIn. HIn is a weak acid and its dissociation is represented as: ii. HIn (aq)  H+(aq) + In-­‐(aq) g. Since HIn is a weak acid, it is characterized by its Ka value. Assume that the Ka value for the indicator HIn is 1.0 x 10-­‐8. Thus, the equilibrium constant is expressed as: [H + ][In − ]
Ka =
[HIn]
h. Remember: HIn is a weak acid; thus, its conjugate base, In-­, is a weak base. i. Rearrange this equation as €
j.
Ka
[In− ]
=
[H+ ] [HIn]
Assume that we add one or two drops of indicator to a solution of pH = 1.0. If pH = 1.0, then [H+] = 0.10 M. €
Ka
1.0 x 10-8
Thus,
=
= 1.0 x 10-7
+
0.10
[H ]
1
[In− ]
=
10,000,000 [HIn]
€
[HIn] = 10,000,000 × [In− ]
k. This means that the concentration of HIn is much greater than that of €
In-­‐, and that the predominant species in solution is HIn. l.€ Hence, if HIn was methyl red, we know that the indicator would turn the solution red since the pH is below 4. m. Thus, a solution of pH = 1 with a few drops of methyl red appears red. n. As a base is added, the concentration of H+ decreases. o. This causes the equilibrium to shift to the right: i. HIn (aq)  H+(aq) + In-­‐(aq) p. At some point during the titration, [In-­‐] would increase to an extent that a change in color will be noticed. q. The question is what should be the concentration of In-­‐ present in solution for the human eye to detect a color change. r. In practice, the color of In-­‐ is observed if it is present in much higher concentration (ten fold) than HIn: [In− ] 10
=
[HIn] 1
s. Conversely, the color of HIn is observed if the concentration of HIn is ten times larger than that of In-­‐. €
[In− ]
1
=
[HIn] 10
t. Note: When choosing an indicator for an acid-­base titration, make sure that the indicator end point (i.e. the point at which the color changes) is as close as possible to the equivalence point of the titration (i.e. point €where the number of moles of acid and base are the same).