Finding Ka

7 Key excerpts on "Finding Ka"

• 

  eBook - ePub

  Physicochemical and Biomimetic Properties in Drug Discovery

  Chromatographic Techniques for Lead Optimization

  • Klara Valko(Author)
  • 2013(Publication Date)
  • Wiley

    (Publisher)

  Chapter 8

  Molecular Physicochemical Properties that Influence Absorption and Distribution—Acid Dissociation Constant—pKa

  Definition of pKa

  The presence of charge on the molecules dramatically influences many of their physicochemical properties, such as lipophilicity, solubility, and permeability. The presence of charge depends on the acid dissociation constant of the ionizable groups and the pH of the solution/environment. The pH is defined as the negative logarithm of the proton or, more precisely, the hydronium ion concentration in aqueous solutions. The product of the concentrations of hydronium and hydroxide ions in water is constant ; thus, the pH normally ranges from 1 to 14. The acid dissociation constant, or , is defined as the pH where an ionizable group is 50% in ionized form. In other words, the acid dissociation constant, , is the equilibrium constant for the reaction in which a weak acid is in equilibrium with its conjugate base in aqueous solution. For example, for acetic acid, the following equilibrium takes place:

  8.1

  8.2

  When the acetate ion concentration is equal to the acetic acid concentration, equals the concentration. The negative logarithm of the concentration is the pH. The smaller the value of , the stronger is the acid. For basic compounds, Equation 8.3 and Equation 8.4 can be used.

  8.3

  8.4

  Again, the negative logarithm of equals the pH of the aqueous environment, where 50% of the basic group is in a protonated charged form, while 50% is in a neutral, unionized form. The percentage of the ionized molecules depends on the proton concentration (pH) and can be calculated at any pH using the Henderson–Hasselbalch equation [1]. Equation 8.5 describes the relationship between the percentage of ionized molecules and pH for a given

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  eBook - ePub

  Applied Theoretical Organic Chemistry

  • Dean J Tantillo(Author)
  • 2018(Publication Date)
  • WSPC (EUROPE)

    (Publisher)

  Chapter 16

  pK a Prediction

  Yijie Niu and Jeehiun K. Lee

  Department of Chemistry and Chemical Biology Rutgers, The State University of New Jersey, USA

  16.1Introduction

  Acidity is a key molecular feature that figures extensively in a wide range of organic reactivity and our understanding of it. In solution, acidity is represented most commonly by its pKa value. For an acid HA in water, the equilibrium for dissociation can be expressed as

  The equilibrium constant, Keq , is therefore

  Generally, since the concentration of water is ~55.5M, the Ka used in pKa calculations is defined as

  The measurement of pKa is often straightforward, but for many organic molecules of interest, with pKa values higher than that of water, the experimental determination can be more complex. Therefore, much work has gone into pKa prediction.

  1 –9

  Here, we will focus on practical methods by which experimental organic chemists might calculate pKa with some accuracy. First, we will review methodology and then we will highlight some organic chemical examples from the recent literature. Finally, we will proffer some practical advice on using some pKa methods.

  16.2Methodology

  16.2.1First-principles methods

  “First-principles” computations refer to those which utilize the relationship between the Gibbs free energy change of the acid dissociation reaction and Ka

  We will review two main strategies within the first-principles method, with and without thermodynamic cycles.

  16.2.1.1First-principles computations using thermodynamic cycles

  Most often, this type of calculation utilizes a thermodynamic cycle, a common example of which is shown in Fig. 16.1 .10

  Thermodynamic cycles are often invoked because accessible methodology (Hartree-Fock (HF) or Density Functional Theory (DFT); continuum solvation) does not yield accurate ΔGaq values.11 The gas-phase values of G(HA(g) ) and can be calculated using quantum mechanical methods. Most commonly, G(HA(aq) ) and are calculated using an “implicit solvation” method, or more specifically, continuum solvation models

  1 ,2 ,6 ,7

  Explicit solvation can also be used, but we will focus on the use of implicit solvation, which is practically simpler than explicit methods (see Chapter 4

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  eBook - ePub

  Survival Guide to General Chemistry

  • Patrick E. McMahon, Rosemary McMahon, Bohdan Khomtchouk(Authors)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  [

  OH −

  ] ;     1 ×

  10

  − 14

  = ( 0.015   M ) [

  OH −

  ] ;     [ O

  H −

  ] =

  (1 ×

  10

  − 14

  )

  (0 .015)

  = 6 . 7 × 1

  0

  − 13

    M

  Example: A certain solution has a [OH− (aq) ] = 4.0 × 10−5 M. Calculate [H+(aq) ].

  K w

  = [

  H +

  ] [

  OH −

  ] ;     1 ×

  10

  − 14

  = [

  H +

  ] ( 4.0 ×

  10

  − 5

    M )     [

  H +

  ] =

  (1 ×

  10

  − 14

  )

  (4 .0 ×

  10

  − 5

  )

  = 2 .5×1

  0 −10

    M

  II	RELATIONSHIP BETWEEN ACID (pKa ) AND CONJUGATE BASE STRENGTH

  The acid strength of an acid is inversely proportional to the base strength of its conjugate base. Very strong acids have very weak conjugate bases; very weak acids have strong conjugate bases. This conclusion is based on the spontaneity of the forward and reverse reactions. The degree of spontaneity (size of the negative value for ∆G) for the forward acid reaction correlates to the strength of the acid. The stronger the acid, the less spontaneous (size of the positive value for ∆G) the reverse conjugate base reaction will be, correlating to the weaker base. Conversely, the weaker the acid (unfavorable ∆G), the stronger conjugate base (favorable ∆G). In anthropomorphic terms, a strong acid that desperately wants to get rid of its proton has no desire to get it back through its conjugate base. A weak acid that desperately wants to keep its proton has a great desire to get it back through its conjugate base if it loses that proton.

  A property of equilibrium constants states that if two component equations add to produce a specific required equation, then the equilibrium constant for the required equation is equal to the multiplication of the equilibrium constants for the two component equations.

  The relationship between Ka of an acid and Kb of its conjugate base can be expressed through the related reference reactions with water as the base or the acid. HA represents the acid and (A)− represents the conjugate base.

  To derive the reference reaction for Kb in terms of the reference reaction for Ka , first reverse the reference reaction for Ka . Recall that K(reverse) = 1/K(forward)

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  eBook - ePub

  Rapid Review of Chemistry for the Life Sciences and Engineering

  With Select Applications

  • Armen S. Casparian, Gergely Sirokman, Ann Omollo(Authors)
  • 2021(Publication Date)
  • CRC Press

    (Publisher)

  b , respectively.

  For example, acetic acid (found in vinegar) is a weak acid. Its equilibrium is represented as follows:

  CH 3

  COOH

  ( aq )

  +

  H 2

  O

  ( l )

  ⇋

  H 3

  O +

  ( aq )

  +

  CH 3

  COO −

  ( aq )

  (4.6)

  Here, H3 O+ is the active acid species, and CH3 COO− is called the conjugate base of acetic acid. Recall that the double arrows indicate an equilibrium process, meaning that this reaction does not go to completion and that all four chemical species in the reaction are present at any given time in varying concentrations.

  The equilibrium acid constant expression, Ka , is as follows:

  K a

  =

  [

  H 3

  O +

  ]

  [

  CH 3

  COO −

  ]

  [

  CH 3

  COOH

  ]

  (4.7)

  The Ka value for acetic acid at room temperature is 1.8 × 10−5 .

  The average percent ionization, depending on initial concentration of the parent acid and temperature, is about 3%–5%. Generally speaking, the smaller the value of the Ka , the weaker the acid. Table 4.1 (A) gives Ka values for a number of weak acids at 25°C. It is evident then, from comparing Ka values of acetic acid with nitrous acid, that acetic acid is weaker than nitrous acid. Phenol, in turn, is much weaker than either nitrous acid or acetic acid. Polyprotic acids have more than one hydrogen ion to dissociate and ionize and so have more than one dissociation constant, i.e., Ka 1 , Ka 2 , and Ka 3 , which get successively smaller, indicating progressive weakness in acidity. For example, oxalic acid has two Ka values, while citric acid has three.

  TABLE 4.1

  Equilibrium or Dissociation Constants for (A) Weak Acids K

  a

  (Monoprotic and Polyprotic) and (B) Weak Bases K

  b

  Monoprotic Acids	Ka
  HC2 O2 CI3	Trichloroacetic acid	(Cl3 CCO2 H)	2.2 × 10−1
  HIO3	Iodic acid	.	1.69 × 10−1
  HC2 HO2 Cl2	Dichloracetic acid	(Cl2 CHCO2 H)	5.0 × 10−2
  HC2 H2 O2 CI	Chloroacetic acid	(ClH2 CCO2 H)	1.36 × 10−3
  HNO2	Nitrous acid	7.1 × 10−4

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  No longer available |Learn more

  AP Chemistry Premium, 2024: 6 Practice Tests + Comprehensive Review + Online Practice

  • Neil D. Jespersen, Pamela Kerrigan(Authors)
  • 2023(Publication Date)
  • Barrons Educational Services

    (Publisher)

  Hydrofluoric acid is a weak acid that dissociates in water according to the equation

  Equation 13.11 may also be written in a shorter, more convenient form by eliminating water:

  Either Equation 13.11 or Equation 13.12 can be used to write the equilibrium expression:

  The two forms are identical, and either may be used in the calculations that follow.

  The constant Ka is called the acid dissociation constant; values for selected weak acids are tabulated in Appendixes 4 and 5 . For the example above, since we know the initial concentration of HF and the value of K

  a

  , the hydrogen ion concentration can be calculated. The technique used is similar to the equilibrium calculations in Chapter 9 .

  EXAMPLE 13.1

  If 0.100 mol of HF is diluted with distilled water to a volume of 500 mL, what is the pH of the solution?

  Solution

  To solve this problem, we set up an equilibrium table with the reaction on the first line, and we enter the initial concentration on the Initial Conc. line. The initial concentration of HF is calculated as

  In the Change row we enter x’s to represent the stoichiometric relationships between the reactants and the products. In this table we assign a positive value to the x’s under the products because they start at zero and cannot possibly decrease. As a consequence any x’s under the reactant(s) must be negative.

  The Equilibrium row is then the sum of the Initial Conc. and Change rows of the table:

  At this point we enter the terms in the Equilibrium row into the equilibrium expression, along with the value of Ka :

  If this equation is solved for x by ordinary means, a quadratic equation results. As before, we can try a simplifying assumption to solve the problem more quickly. The only term that can be simplified is 0.200 - x. We use the assumption that x ≪ 0.200 and conclude that 0.200 - x = 0.200. (Remember that when we finally calculate x

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  eBook - ePub

  General Chemistry for Engineers

  • Jeffrey Gaffney, Nancy Marley(Authors)
  • 2017(Publication Date)
  • Elsevier

    (Publisher)

  Chapter 7 . The acid ionization constant is equal to the ratio of the molar concentrations of the ionized products (conjugate acid and conjugate base) to the molar concentration of the unionized acid at equilibrium;

  K a

  =

  H 3

  O +

  A −

  HA w

    (6)

  Although water is a reactant in the ionization of a weak acid, the concentration of water is not used in the expression of the acid ionization constant. This is because, as the solvent, the concentration of water is in great excess compared to the concentrations of the acid and the ionization products. So, the concentration of water remains constant during the reaction. Only the concentrations of the species changed by the reaction are included in the expression for the acid equilibrium constant.

  The stronger the acid and the more it is ionized, the higher the concentrations of the conjugate acid and conjugate base (numerator of K a ) and the smaller the concentration of the unionized acid (denominator of K a ), resulting in a larger K a . The weaker the acid and the less it is ionized, the lower the concentrations of the conjugate acid and conjugate base and the higher the concentration of the unionized acid, resulting in a smaller K a . This means that the larger the K a the stronger the acid and the smaller the K a

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  eBook - ePub

  Aquatic Chemistry Concepts, Second Edition

  • James F. Pankow(Author)
  • 2019(Publication Date)
  • CRC Press

    (Publisher)

  because K a is finite. The reaction of A − with water is A − + H 2 O = HA + OH −. (5.48) We obtain the equilibrium constant for Eq.(5.48) as follows. The reverse of the equilibrium for K a involves the conversion of A − to HA: H + + A − = HA (K a) − 1. (5.49) The dissociation reaction for water is H 2 O = H + + OH − K w. (5.50) Adding these two reactions gives A − + H 2 O = HA + OH − K = K w K a ≡ K b = [ HA ] [ OH − ] [ A − ] (5.51) in which A − is acting as. a base and the equilibrium constant is given the symbol K b. The relation between any K a and the corresponding K b is therefore K a × K b = K w (5.52) log K a + log K b = log K w (5.53) p K a + p K b = p K w. (5.54) At 25 °C and 1 atm, p K a + p K b = 14.0. (5.55) As an example use of “corresponding”, when HA represents acetic acid, then A − is the acetate ion. Since we know the value of K w, there is no information contained in the value of K b that is not in K a. K a tells us how much A − “likes” a proton from the. perspective of the willingness of HA to give up an H + (think of a dollar) and become A −. So, K a is analogous to a measure of generosity. Alternatively, K b tells us how much A − “likes” a proton from the point of view of the tendency of A − to take a proton (dollar) from water (the reference person with a dollar), and become HA, while forming OH − in the process. K b is then analogous to a measure of miserliness. Carrying this anthropomorphic analogy one step further, if you happen to know that your Aunt Fatima is a quite generous person, you automatically know that after she has readily given up her one dollar to the community, she will not be prone to take the dollar back. Fatima with a dollar is like HF (rather large K a, about 10 −3); Fatima without a dollar is like F − (rather small K b, about 10 −11). The given-up dollar is like the solvated proton in solution. There is much symmetry between the problem of a solution of HA and that for a solution of NaA

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