Chemistry

Weak Acid and Base Equilibria

Weak acid and base equilibria refer to the dynamic balance between the undissociated molecules and their respective ions in a solution. In the case of weak acids, the equilibrium lies more towards the undissociated form, while for weak bases, it lies more towards the dissociated form. The equilibrium constant (Ka for weak acids, Kb for weak bases) quantifies the extent of dissociation.

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10 Key excerpts on "Weak Acid and Base Equilibria"

  • Book cover image for: Chemistry: Atoms First
    • William R. Robinson, Edward J. Neth, Paul Flowers, Klaus Theopold, Richard Langley(Authors)
    • 2016(Publication Date)
    • Openstax
      (Publisher)
    748 Chapter 14 | Acid-Base Equilibria This OpenStax book is available for free at http://cnx.org/content/col12012/1.7 At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base). For example, a solution of the weak base trimethylamine, (CH 3 ) 3 N, in water reacts according to the equation: (CH 3 ) 3 N(aq) + H 2 O(l) ⇌ (CH 3 ) 3 NH + (aq) + OH − (aq), giving an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids. We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 14.10). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, K b , is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium: K b = [(CH 3 ) 3 NH + ][OH − ] [(CH 3 ) 3 N] Figure 14.10 pH paper indicates that a 0.1-M solution of NH 3 (left) is weakly basic. The solution has a pOH of 3 ([OH − ] = 0.001 M) because the weak base NH 3 only partially reacts with water. A 0.1-M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa) The ionization constants of several weak bases are given in Table 14.3 and in Appendix I.
  • Book cover image for: Chemistry
    eBook - PDF
    • Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
    • 2015(Publication Date)
    • Openstax
      (Publisher)
    782 Chapter 14 | Acid-Base Equilibria This OpenStax book is available for free at http://cnx.org/content/col11760/1.9 At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base). For example, a solution of the weak base trimethylamine, (CH 3 ) 3 N, in water reacts according to the equation: (CH 3 ) 3 N(aq) + H 2 O(l) ⇌ (CH 3 ) 3 NH + (aq) + OH − (aq), giving an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids. We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure 14.10). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, K b , is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium: K b = [(CH 3 ) 3 NH + ][OH − ] [(CH 3 ) 3 N] Figure 14.10 pH paper indicates that a 0.1-M solution of NH 3 (left) is weakly basic. The solution has a pOH of 3 ([OH − ] = 0.001 M) because the weak base NH 3 only partially reacts with water. A 0.1-M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base. (credit: modification of work by Sahar Atwa) The ionization constants of several weak bases are given in Table 14.3 and in Appendix I.
  • Book cover image for: Chemistry
    eBook - PDF

    Chemistry

    The Molecular Nature of Matter

    • Neil D. Jespersen, Alison Hyslop(Authors)
    • 2021(Publication Date)
    • Wiley
      (Publisher)
    Bases, the other topic in this chapter, are often characterized as cleaning substances that will be discussed later in this chap- ter. In this chapter, we will discuss acids and bases, and expand the concepts of the chemical properties of acids and bases that we introduced in Chapter 15 and apply the principles of chemical equilibrium from Chapter 14 to acids and bases. Using the principles of molecular polarity and the concept of delo- calization of electrons, we were able to qualitatively compare the strengths of acids and bases. Our goal in this chapter is to bring these concepts together as we examine the quantitative aspects of acid–base chemistry. In particular we are interested in acids, bases, and their mixtures as they affect the concen- tration of hydronium ions in solution through the concept of pH. The principles we develop here have applications in laboratories that investigate environmental, forensic, and biochemical problems. High- tech laboratories interested in materials science and nanotechnology often use the principles of acid–base equilibria. The consumer-oriented industries that make products such as cosmetics, foods, beverages, and cleaning chemicals all employ chemists who know that control of pH is very important in safe and effective consumer products. CHAPTER OUTLINE 16.1 Water, pH, and “p” Notation 16.2 pH of Strong Acid and Base Solutions 16.3 Ionization Constants, K a and K b 16.4 Determining K a and K b Values 16.5 pH of Weak Acid and Weak Base Solutions 16.6 Acid–Base Properties of Salt Solutions 16.7 Buffer Solutions 16.8 Polyprotic Acids 16.9 Acid–Base Titrations Acid–Base Equilibria in Aqueous Solutions CHAPTER 16 PIXOLOGICSTUDIO/Getty Images 16.1 Water, pH, and “p” Notation 793 LEARNING OBJECTIVES After reading this chapter and working the problems, you should be able to: • define pH and explain the use of the “p” notation. • explain how to determine the pH of strong acids or bases in aqueous solution.
  • Book cover image for: Basics of Analytical Chemistry and Chemical Equilibria
    • Brian M. Tissue(Author)
    • 2023(Publication Date)
    • Wiley
      (Publisher)
    1 01 10 4 45 10 2 25 10 14 7 8 K 202 ACID–BASE EQUILIBRIA AND ACTIVITY ′ ′ > K K b a , so the second equilibrium predominates. In practice we use a sim- pler expression discussed in Section 6.4. You-Try-It 5.B The amphiprotic-salts worksheet in you-try-it-05.xlsx contains a ta- ble of salt solutions. For each salt determine if the solution will be neutral, acidic, or basic. For the weak bases, calculate K b values from the K a of the conjugate acid. Use Examples 5.9–5.11 as guides. 5.4 ACID STRENGTH The acidity or basicity (also called alkalinity) of a solution depends on both the concentration and on the intrinsic “strength” of the added weak acid or weak base. “Strength” is not an issue for strong acids and strong bases, which dissociate completely in water. p[H 3 O + ] and p[OH − ] of a solution of strong acid or strong base depend only on the concentration of strong acid or strong base in the solution after any neutralization reaction has occurred. Predicting p[H 3 O + ] and p[OH − ] for solutions of weak acids and weak bases is more involved than for those of the strong acids and strong bases because the weak electrolytes react only partially with water. However, keep in mind that solu- tions of weak acids and weak bases are not necessarily less acidic or less basic than solutions of strong acids and strong bases. For example, it is possible for a weak acid at a high concentration to produce a more acidic solution (a lower p[H 3 O + ]) than a strong acid at a low concentration. Our measure of the relative strength of different weak acids is the equilibrium constant, K a . By acid “strength,” we refer to the degree to which a weak acid dissociates in water to produce H 3 O + and its conjugate base. The larger the K a value, the more readily the acid gives up its proton to water. Table 5.6 provides K a and pK a (−log K a ) values for selected inorganic and organic weak acids.
  • Book cover image for: Chemistry
    eBook - PDF
    • John A. Olmsted, Gregory M. Williams, Robert C. Burk(Authors)
    • 2020(Publication Date)
    • Wiley
      (Publisher)
    752 CH A P TER 1 5 Aqueous Acid–Base Equilibria Salts of Weak Acids An aqueous solution of a soluble salt contains cations and anions, which may have acid–base properties. Anions that are conjugate bases of weak acids make a solution basic. For example, sodium fluoride dissolves in water to give Na + , F − , and H 2 O as major species. The fluoride anion is the conjugate base of the weak acid HF. This anion establishes a proton transfer equi- librium with water: F − (aq) + H 2 O(l ) ⟶ ⟵ HF(aq) + OH − (aq) K eq = K b,HF = [OH − ] eq [HF] eq _____________ [F − ] eq + + – – Collision and proton transfer The reaction generates hydroxide anions, so the solution is basic. Fluoride acts as a base, so the equilibrium constant is the base hydrolysis constant of HF, K b,HF . The fluoride ion equilibrium is linked to two other proton transfer equilibria. Combining the proton transfer equilibrium for HF with the proton transfer equilibrium for F reveals the relationship: HF(aq) + H 2 O(l ) ⟶ ⟵ F − (aq ) + H 3 O + (aq ) K a H 2 O(l ) + F − (aq ) ⟶ ⟵ OH − (aq ) + HF(aq ) K b ‾ 2 H 2 O(l ) ⟶ ⟵ OH − (aq ) + H 3 O + (aq ) K w Combining these two equilibria leads to cancellation of HF and F − , so the sum of the two equilibria is the water equilibrium. How are the equilibrium constants for these three equilib- ria related? Whenever two or more equilibria are added, the equilibrium constant for the net reaction is the product of the individual equilibrium constants of the summed reactions.
  • Book cover image for: Concepts & Calculations in Analytical Chemistry, Featuring the Use of Excel
    • Henry Freiser, Monika Freiser(Authors)
    • 1992(Publication Date)
    • CRC Press
      (Publisher)
    Equation 4-6a may be recognized as a result of subtracting the acid dissociation reaction ofNH4 + , from the dissociation reaction of water. Thus: H 2 0 ' H+ + OH-- (NH; ' H+ + NH 3 ) This subtraction of equations is equivalent to dividing the corresponding equilibrium expressions (Chap. 2). K w [H+] [OH-] K a [NH31 [H+]/[NH;] [NH; ] [OH-] [NH 3 1 Hence, we see that K b from 4-8 is identical to KJKa' (4-9) Although this relation was derived from a specific consideration of the NH4 + / NH3 conjugate pair, it will be recognized as a generally valid reaction, that is, (4-10) From this equation, it follows readily that if we list a series of acids in increasing strength, we have automatically listed the conjugate bases in the order of decreasing strength. The lewis Theory The Lewis Theory of acids and bases does not feature the special role for the proton that it has in the Bf0nsted-Lowry theory. Here an acid is any electron-pair deficient species. A base from this viewpoint is a species capable of furnishing electron pairs. Thus, acid-base reactions are considered as coordination reactions. This theory is of great value in understanding metal coordination complex formation. Chapter 4 Acid Base Equilibrium 57 In the following reactions: the Ag + ion* is seen to act in a manner that is similar to a proton, and the Ag + ion behaves as a dibasic acid. The subject of metal coordination com-plexes which involves reactions of this type will be considered at length in Chapter 5. The Lewis Theory has also found extensive use in explaining reactions such as the Friedel-Crafts reaction in nonaqueous media, involving nonprotonic acids such as BF 3' AICl3, etc. STRENGTHS OF ACIDS AND BASES The strengths of acids, that is, their proton donating tendencies, naturally depends upon the proton-accepting tendency of the base with which they react. An ordering of acids according to their strengths is obtained by the use of a reference base.
  • Book cover image for: Chemistry
    eBook - PDF

    Chemistry

    An Atoms First Approach

    • Steven Zumdahl, Susan Zumdahl, Donald J. DeCoste, , Steven Zumdahl, Steven Zumdahl, Susan Zumdahl, Donald J. DeCoste(Authors)
    • 2020(Publication Date)
    as possible 1. What is meant by the presence of a common ion? How does the presence of a common ion affect an equilib- rium such as HNO 2 saqd m H 1 saqd 1 NO 2 2 saqd What is an acid–base solution called that contains a common ion? 2. Define a buffer solution. What makes up a buffer solu- tion? How do buffers absorb added H 1 or OH 2 with little pH change? Is it necessary that the concentrations of the weak acid and the weak base in a buffered solution be equal? Explain. What is the pH of a buffer when the weak acid and conjugate base concentrations are equal? A buffer generally contains a weak acid and its weak conjugate base, or a weak base and its weak conjugate acid, in water. You can solve for the pH by setting up the equilibrium problem using the K a reaction of the weak acid or the K b reaction of the conjugate base. Both reactions give the same answer for the pH of the solu- tion. Explain. Review Questions Answers to the Review Questions can be found on the Student Web site (accessible from www.cengagebrain.com). 615 For Review Copyright 2021 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 7. Sketch the titration curve for a weak acid titrated by a strong base. When performing calculations concerning weak acid–strong base titrations, the general two-step procedure is to solve a stoichiometry problem first, then to solve an equilibrium problem to determine the pH.
  • Book cover image for: Analytical Chemistry
    • Gary D. Christian, Purnendu K. Dasgupta, Kevin A. Schug(Authors)
    • 2013(Publication Date)
    • Wiley
      (Publisher)
    That is, we have a mixture of a weak acid and its salt or a weak base and its salt. Consider an acetic acid–acetate buffer. The equilibrium that governs this system is HOAc  H + + OAc − But now, since we have added a supply of acetate ions to the system (e.g., from sodium acetate), the hydrogen ion concentration is no longer equal to the acetate ion concentration. The hydrogen ion concentration is [H + ] = K a [HOAc] [OAc − ] (7.40) Taking the negative logarithm of each side of this equation, we have −log[H + ] = − log K a − log [HOAc] [OAc − ] (7.41) pH = pK a − log [HOAc] [OAc − ] (7.42) Upon inverting the last log term, it becomes positive: pH = pK a + log [OAc − ] [HOAc] (7.43) This form of the ionization constant equation is called the Henderson – Hasselbalch The pH of a buffer is determined by the ratio of the conjugate acid–base pair concentrations. 3 This is an unedited student video and contains some errors of statement, e,g., it talks about K a of NH 3 whereas it should really refer to it as K a of NH 4 + ; it mistakenly states that bases like to take up electrons whereas bases of course like to take up hydrogen ions, etc. But despite these errors of statement, it is a nicely set up example of a correctly solved problem that shows the use of Excel Goal Seek! 7.8 BUFFERS——KEEPING THE PH CONSTANT (OR NEARLY SO) 239 equation. It is useful for calculating the pH of a weak acid solution containing its salt. A general form can be written for a weak acid HA that ionizes to its salt, A − , and H + : HA  H + + A − (7.44) pH = pK a + log [A − ] [HA] pH = pK a + log [conjugate base] [acid] pH = pK a + log [proton acceptor] [proton donor] (7.45) (7.46) (7.47) Example 7.11 Calculate the pH of a buffer prepared by adding 10 mL of 0.10 M acetic acid to 20 mL of 0.10 M sodium acetate. Solution We need to calculate the concentration of the acid and salt in the solution.
  • Book cover image for: Physical Chemistry and Its Biological Applications
    M. Jones, Data for Biochemical Research, 2nd ed., Oxford University Press, New York, 1969. John T. Edsall and Jeffries Wyman, Biophysical Chemistry, Vol. 1, Academic Press, New York, 1958. Chapters 8 and 9 include extensive accounts of biochemical aspects of acid-base equilibria. E. J. King, Acid-Base Equilibria, Pergamon Press, Elmsford, N.Y., 1965. A comprehensive review at an intermediate level. R. Bruce Martin, Introduction to Biophysical Chemistry, McGraw-Hill, New York, 1964. Chapter 4 includes a good account of multiple acid-base equilibria and methods for analyzing them, and Chapter 5 describes protein titrations. 212 SIX ACID-BASE EQUILIBRIA Charles Tanford, Physical Chemistry of Macromolecules, Wiley, New York, 1961. Chapter 8 presents a comprehensive, fairly advanced account of multiple equilibria, such as those between macromolecules and small ions. Charles Tanford, The Interpretation of Hydrogen Ion Titration Curves of Proteins, in Advances in Protein Chemistry, Vol. 17, Academic Press, New York, 1962. C. A. VanderWerf, Acids, Bases and the Chemistry of the Covalent Bond, Reinhold, New York, 1961. A very good elementary introduction. A. White, P. Handler, and E. L. Smith, Principles of Biochemistry, 5th ed., McGraw-Hill, New York, 1973. Acid-base equilibrium and its biochemical applications are well covered. Journal Articles C. R. Allen and P. G. Wright, Entropy and Equilibrium, J. Chem. Educ. 41, 251 (1964). Interpretation of ionization data for organic acids. R. G. Bates, R. A. Robinson, and A. K. Covington, pK Values for D 2 0 and H 2 0 , J. Chem. Educ. 44, 635 (1967). J. D. Burke, On Calculating [H+], J. Chem. Educ. 53, 79 (1976). G. E. Clement and T. P. Hartz, Determination of the Microscopic Ionization Constants of Cysteine, J. Chem. Educ. 48, 395 (1971). H. L. Clever, The Ion Product Constant of Water, /. Chem. Educ. 45, 231 (1968). H.-L. Fung and L. Cheng, Linear Plots in the Determination of Microscopic Dissociation Constants, J.
  • Book cover image for: The Chemistry Companion
    • Anthony C. Fischer-Cripps(Author)
    • 2011(Publication Date)
    • CRC Press
      (Publisher)
    9.7 Acid  Base Reactions A few acids and bases are stron g dissociation of acetic acid HC 2 H 3 O 2 , a 116 2 3 2     2 3 2 2 3 2 H C O H O H O H HC The concentration of H 3 O + is found f r                   2 3 2 3 2 5 3 2 2 3 2 2 3 2 3 O H C O H HC 10 74 . 1 O H O H O H HC O H C O H K If a solution is made with 0.1 mol e [HC 2 H 3 O 2 ] is 0.1M. What then is the small since the value of K A implies a l us add 0.1 moles of solid NaC 2 H 3 O 2 almost completely, the concentration 0.1M. Thus   75 . 4 pH 1 . 0 1 . 0 10 74 . 1 O H 5 3      If to this solution we now add a sm a moles (a drop) of 2M NaOH, then molecules: 2 3 2 NaOH O H HC   The quantity [HC H O ] is reduce The quantity [HC 2 H 3 O 2 ] is reduce [C 2 H 3 O 2  ] increases to 0.11M (since this new solution is 74 . 1 log pH      The HC 2 H 3 O 2 / C 2 H 3 O 2  solution is c ability to resist changes in pH. If a solution, this reacts with the acetate i the other way, and the pH remains al m If the concentration of the introduc e becomes too large then the buffering becomes too large , then the buffering the solution is exhausted and the pH c appreciably. g , but most are weak. Consider the a weak acid with K A = 1.74  10  5 . The Chemistry Companion A Strong acids: HCl, HBr, HI, HNO 3 , H 2 SO 4 , HClO 4  2 3 O H r om:    2 2 O e s of HC 2 H 3 O 2 , then the concentration concentration of [C 2 H 3 O 2  ]? It must be l ow incidence of dissociation. N ow, le t . Since the salt, NaC 2 H 3 O 2 dissociates of the acetate ion is now approximately 1 1 a ll amount of a strong base, say 0.001 the NaOH reacts with the acetic acid 2 3 2 2 O H NaC O H   ed from 0 1 M to 0 099 M The quantity ed from 0 . 1 M to 0 . 099 M . The quantity the NaC 2 H 3 O 2 dissociates). The pH of 8 . 4 11 . 0 099 . 0 10 5      c alled a buffer solution , because of its small amount of acid is added to the i on and the concentrations shift slightly m os t unchanged.
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