Dissociation Constant
The dissociation constant is a measure of the extent to which a compound dissociates into its ions in a solution. It is represented by the equilibrium constant for the dissociation reaction. A higher dissociation constant indicates a greater degree of dissociation, while a lower constant indicates less dissociation.
6 Key excerpts on "Dissociation Constant"
eBook - ePub- Juan Moreno, Rafael Peinado(Authors)
- 2012(Publication Date)
- Academic Press(Publisher)
...Both K a and K b are equilibrium constants. Therefore, the higher their value, the more the equilibrium is shifted to the right and the greater their tendency to produce H + or OH − ions. The strength scale is used only for weak acids and bases as the equilibrium of strong acids and bases is always displaced completely to the right. In such cases, the value of K a or K b is ∞, as the molecular species does not exist in solution. Example for a strong acid: Examples of strong acids are HClO 4, HCl, HNO 3 and H 2 SO 4, and strong bases are NaOH and KOH. TABLE 13.2. Dissociation Constants and pK a of a Range of Acids Acid K a (25°C) pK a HSO 4 − 1.2 × 10 −2 1.92 ClCH 2 -COOH 1.4 × 10 −3 2.85 HF 6.8 × 10 −4 3.17 HCOOH 1.8 × 10 −4 3.74 CH 3 -COOH 1.8 × 10 −5 4.74 CO 2 4.2 × 10 −7 6.38 HS − 10 −14 14 According to the values in the above table, the higher the equilibrium constant (K a), the lower the. pK a will be. Therefore, acids with a higher K a (and a lower pK a) will be stronger. The same occurs with bases (higher K b and lower pK b). 3.1 Degree of Dissociation The degree of dissociation (α) for weak acids and bases is defined as the fraction of 1 mole that dissociates at equilibrium. By definition, α is a unit factor but it is often also expressed as a percentage. If absolute concentrations are used instead of the degree of dissociation, the equilibrium constant is expressed as follows: On comparing the expressions at equilibrium, it is seen that: Two methods can be used to solve the K a equation: the first involves resolving the second-degree equation and the second involves performing successive approximations. In the second method, the following approximation is performed: c−x ≈ c, which gives the following equation: The value obtained for x is used to calculate a value for c−x, which, in turn, is used to calculate a new value for x...
eBook - ePubPhysicochemical 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 p K a 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...
- Kevin Reel, Derrick C. Wood, Scott A. Best(Authors)
- 2014(Publication Date)
- Research & Education Association(Publisher)
...For example, at 100°C, water exists as both a liquid and a gas in equilibrium with one another. TEST TIP If pressures are given for products and reactants in an equilibrium, be sure to write the expression for K p and not K c. EXAMPLE: The value of K p for the following reaction is 8.3 × 10 –3 at 700 K. What is the value for K c ? SOLUTION: Δn = 2 moles gaseous product – 4 moles gaseous reactants = –2 The Dissociation of Weak Acids and Bases • The water Dissociation Constant is shown in the following reaction: This relationship allows for computation of Dissociation Constants for conjugate acid–base pairs as well as concentrations of H 3 O + and OH – for use in pH and pOH calculations. • The equilibrium constants K a and K b can be used to calculate the pH of a weak acid or weak base in aqueous solution. EXAMPLE: What is the pH of a 0.10 molar solution of acetic acid, CH 3 COOH, which has a K a = 1.8 × 10 –5 ? SOLUTION: 1. Write the equation of the reaction that occurs when the weak acid or base is put in water, and its corresponding equilibrium expression. When the weak acid is placed in water, some number of moles will dissociate. For each mole of weak acid that dissociates, one mole of the proton and one mole of the weak base will be formed. 2. Construct a table to determine the amount of dissociation of the weak acid. For the AP Chemistry exam, the simplifying assumption can always be made for all types of equilibrium problems. 3. Write the equilibrium expression using the final concentrations from the preceding table. 4. Solve for x, which will be the [H 3 O + ] at equilibrium. 5. When calculating pH or % dissociation, always use the equilibrium [H 3 O + ]. TEST TIP For any problems that involve equilibrium, you must create a table as shown in step 2 of the previous problem in order to determine equilibrium concentrations...
eBook - ePubChemistry
Concepts and Problems, A Self-Teaching Guide
- Richard Post, Chad Snyder, Clifford C. Houk(Authors)
- 2020(Publication Date)
- Jossey-Bass(Publisher)
...Most of the equilibria discussed involved reactants and products that were all in the gaseous state or equilibria between gases and pure solids and liquids. Chemists have learned that many reactions that occur between ions in aqueous solutions also do not go to completion and reach a state of equilibrium. Ionic equilibria are also affected by stresses according to Le Chatelier's Principle. We now discuss several special cases of ionic equilibria and how they are used to determine the degree of dissociation of weak electrolytes and the concentration of ions in aqueous solutions. We also discuss how slightly soluble compounds behave and how their solubility is determined in aqueous solutions and in the presence of other ions. IONIC EQUILIBRIA You have used Le Chatelier's Principle and the equilibrium constant for reversible reactions. Equilibria also exist for such things as salts and their ions in solution, and acids and bases and their dissociated ions. Other substances, even though they may dissociate only very slightly into ions, are at equilibrium with those ions. An ionic equilibrium constant expression is written just like those of the reversible reactions previously encountered. Write the ionic equilibrium constant for the following reaction. K eq = ____________ Answer: Instead of K eq, the equilibrium constant may be K a if it represents the dissociation of an acid, K b if it represents the dissociation of a base, K i for a general ionization constant, or K sp for a solubility product constant. We will discuss K sp now and these other constants later. The equilibrium constant expression is the same for any of these as for the reactions dealt with previously. Write the equilibrium constant expression for the solubility product of the very slightly soluble salt AgBr. K sp = ____________ Answer: [Ag + ][Br − ] (The solid is not included since its concentration is constant at constant temperature.) The abbreviation K sp stands for the solubility product constant...
eBook - ePub- Kevin R. Reel(Author)
- 2013(Publication Date)
- Research & Education Association(Publisher)
...Determination of the equilibrium constant for a chemical reaction Example problem: A student is given a solution of acetic acid and is asked to find the equilibrium constant for the following equilibrium. (a) Describe how the student can use this information and a pH meter to find the K a of acetic acid. The student can make a standardized solution of acetic acid, then measure its pH. The pH can tell the student how much the acetic acid has dissociated. From the amount dissociated, the student can determine the concentrations of all species after equilibrium, and then calculate the equilibrium constant using the relationship: (b) Calculate the initial concentration (before equilibrium is established) of the sample of acetic acid if 2.0 mL of an initial 12.0 M stock solution is diluted to make 400 mL of solution. (c) After equilibrium is established, the student measures the pH to be 2.98. What is the Dissociation Constant (K a) for this reaction? Plugging into the equilibrium expression: (d) What is a reasonable approximation for the standard free energy, ΔG°, for this reaction? 9. Determination of the rate of a reaction and its order Example problem: A student runs an experiment that measures the rate of reaction between iodine and acetone at varied concentrations, according to the following reaction: The student measures the time it takes for the iodine to disappear when varying the concentrations in the following four trials. (a) Determine the order of the reaction with respect to each chemical species tested. The reaction is first order with respect to acetone and H +, and zero order with respect to iodine. (b) What is the rate equation for this reaction? Rate = k [C 3 H 6 O] [H + ] (c) What are the value and units for the rate constant, k ? 2.6 × 10 −5 L / M sec. 10...
eBook - ePub- James F. Pankow(Author)
- 2019(Publication Date)
- CRC Press(Publisher)
...Early chemists looking at the solubility of ionic solids adopted the instability point of view: “how strongly does a solid dissociate to yield ions in solution?”, and hence the K s0 a.k.a. K sp (“solubility product”) format. The reader, having by now considered the chemistry of acids with a range of Dissociation Constants, will understand that acids with Dissociation Constants of ≳10 –2, ~10 –5, and ≲10 –8 may be characterized as strong, weak, and very weak, respectively. We can make this understanding useful in considering what types of metal–ligand formation (i.e., stability) constant values characterize very weak, weak, and strong complexes. Consider: Ca + 2 + OH − = CaOH + K H1 = 20. (10.16) For the value expressed in the same way as an acidity constant, that is, as an instability constant: CaOH + = Ca 2 + + OH − 1 K H1 = 1 20 = 5 × 10 − 2. (10.17) An acid with a Dissociation Constant of 5 × 10 –2 = 10 –1.30 is fairly strong, that is, unstable. Thus, a metal–ligand formation constant of 20 corresponds to a weak complex. Overall, for K values of ≲10 2, ~10 5, and ≳10 8, formation constants may be characterized as being weak, strong, and very strong respectively (Table 10.1). TABLE 10.1 General Values of Metal–Ligand Equilibrium Constants as Both Formation and Dissociation Constants Type of Constant Strength of Complex Weak Strong Very Strong Metal–ligand formation constant (accepted format) <10 2 ~10 5 >10 8 Metal–ligand Dissociation Constant (format used for acids) >10 –2 (strong acid) ~10 –5 (weak acid) <10 –8 (very weak acid) Metal–ligand formation constants span a wide range of values. Complexes between metal ions from the first column of the periodic table (e.g., Na +, K +, etc.) with ligands binding using one pair of electrons (“monodentate ligands”) are generally very weak. For example, the complexes NaOH° and NaCl° are extremely weak and can essentially always be neglected (Na + and Cl − have been labeled spectator ions)...
