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The Ka and Kb of it

The Ka and Kb of it. http://www.chem1.com/acad/webtext/abcon/abcon-2.html. Autoionization of Water. Because water is amphiprotic, one water molecule can react with another to form an OH – ion and an H 3 O + ion in an autoionization process:

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The Ka and Kb of it

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  1. The Ka and Kb of it

  2. http://www.chem1.com/acad/webtext/abcon/abcon-2.html

  3. Autoionization of Water • Because water is amphiprotic, one water molecule can react with another to form an OH– ion and an H3O+ ion in an autoionization process: 2H2O(l)⇋H3O+ (aq) + OH– (aq) • Equilibrium constant K for this reaction can be written as [H3O+] [OH–] [H2O]2 • 1 L of water contains 55.5 moles of water. In dilute aqueous solutions: • The water concentration is many orders of magnitude greater than the ion concentrations. The concentration is essentially that of pure water. Recall that the activity of pure water is 1. • When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25ºC, the concentrations of hydronium ion and hydroxide ion are equal: [H3O+]=[OH–] = 1.0 x 10–7 M [H3O+][OH–] = 1.0 x 10–14 M = Kw pH = pOH = 7 pH + pOH = pKw = 14 Kc = Kc [H2O]2 = Kw = [H3O+][OH–] = 1.0 x 10–14

  4. Leveling Effect It’s all because of Gibbs Free Energy No acid stronger than H3O+ and no base stronger than OH– can exist in aqueous solution, leading to the phenomenon known as the leveling effect. • Any species that is a stronger acid than the conjugate acid of water (H3O+) is leveled to the strength of H3O+ in aqueous solution because H3O+ is the strongest acid that can exist in equilibrium with water. • In aqueous solution, any base stronger than OH– is leveled to the strength of OH– because OH– is the strongest base that can exist in equilibrium with water • Any substance whose anion is the conjugate base of a compound that is a weaker acid than water is a strong base that reacts quantitatively with water to form hydroxide ion http://www.chem1.com/acad/webtext/abcon/abcon-2.html

  5. Acid-Base Equilibrium Constants: Ka, Kb, pKa, and pKb • The magnitude of the equilibrium constant for an ionization reaction can determine the relative strengths of acids and bases • The general equation for the ionization of a weak acid in water, where HA is the parent acid and A– is its conjugate base, is HA(aq)+ H2O(l)⇋ H3O+(aq)+ A–(aq) • The equilibrium constant for this dissociation is K = [H3O+] [A–] [H2O] [HA] • The concentration of water is constant for all reactions in aqueous solution, so [H2O] can be incorporated into a new quantity, the acid ionization constant (Ka): Ka = K[H2O] = [H3O+] [A–] [HA]

  6. Ionization Constants for Weak Monoprotic Acids and Bases • Strong acids and bases ionize essentially completely in water; the percent ionization is always approximately 100%, regardless of the concentration • The percent ionization in solutions of weak acids and bases is small and depends on the analytical concentration of the weak acid or base; percent ionization of a weak acid or a weak base actually increases as its analytical concentration decreases and percent ionization increases as the magnitude of the ionization constants Ka and Kb increases

  7. Ionization Constants for Weak Acids and Bases When is a 1M solution not a 1 M solution? • A 1 M solution is prepared by dissolving 1 mol of acid or base in water and adding enough water to give a final volume of exactly 1 L. • If the actual concentrations of all species present in the solution were listed, it would be determined that none of the values is exactly 1 M because a weak acid or a weak base always reacts with water to some extent. • Only the total concentration of both the ionized and unionized species is equal to 1 M. • The analytical concentration (C) is defined as the total concentration of all forms of an acid or base that are present in solution, regardless of their state of protonation. • Thus; a 1 M solution has an analytical concentration of 1 M, which is the sum of the actual concentrations of unionized acid or base and the ionized form.

  8. Let’s look at the dissolution of acetic acid, a weak acid, in water as an example. The equation for the ionization of acetic acid is: The equilibrium constant for this ionization is expressed as: Ionization Constants for Weak Monoprotic Acids and Bases

  9. Ionization Constants for Weak Monoprotic Acids and Bases • The water concentration in dilute aqueous solutions is very high. • 1 L of water contains 55.5 moles of water. • Thus in dilute aqueous solutions: • The water concentration is many orders of magnitude greater than the ion concentrations. • Thus the water concentration is essentially that of pure water. • Recall that the activity of pure water is 1.

  10. Ionization Constants for Weak Monoprotic Acids and Bases • We can define a new equilibrium constant for weak acid equilibria that uses the previous definition. • This equilibrium constant is called the acid ionization constant. • The symbol for the ionization constant is Ka.

  11. Ionization Constants for Weak Monoprotic Acids and Bases • The ionization constant values for several acids are given below. • Which acid is the strongest? • Are all of these acids weak acids? • What is the relationship between Ka and strength? • What is the relationship between pKa and strength? • What is the relationship between pH and strength?

  12. Determining Ka and Kb • The ionization constants Ka and Kb are equilibrium constants that are calculated from experimentally measured concentrations. • What does the concentration of an aqueous solution of a weak acid or base exactly mean? – A 1 M solution is prepared by dissolving 1 mol of acid or base in water and adding enough water to give a final volume of exactly 1 L. – If the actual concentrations of all species present in the solution were listed, it would be determined that none of the values is exactly 1 M – Only the total concentration of both the ionized and unionized species is equal to 1 M. – The analytical concentration (C) is defined as the total concentration of all forms of an acid or base that are present in solution, regardless of their state of protonation. – 1 M solution has an analytical concentration of 1 M, the sum of the actual concentrations of unionized acid or base and the ionized form. Two common ways to obtain the concentrations • By measuring the electrical conductivity of the solution, which is related to the total concentration of ions present • By measuring the pH of the solution, which gives [H+] or [OH–]

  13. Determining Ka and Kb • RICEProcedure for determining Ka for a weak acid and Kb for a weak base 1. The analytical concentration of the acid or base ionization Reactionis the Initialconcentration 2. The stoichiometry of the reaction with water determines the Change in concentrations 3. The final (Equilibrium) concentrations of all species are calculated from the initial concentrations and the changes in the concentrations 4. Inserting the final concentration into the equilibrium constant expression enables the value of Ka or Kb to be calculated

  14. Ionization Constants for Weak Monoprotic Acids: the math In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid. • Use the concentrations that were just determined in the ionization constant expression to get the value of Ka.

  15. Ionization Constants for Weak Monoprotic Acids the MATH • The pH of a 0.10 M solution of a weak monoprotic acid, HA, is found to be 2.97. What is the value for its ionization constant? • Use the [H3O+] and the stoichiometry of the ionization reaction to determine concentrations of all species. Simplifying Assumption: Is the change significant? Later we will find that in general, if the Ka/[] is < 1x10-3 you can apply the simplifying assumption. • Calculate the ionization constant from this information.

  16. Ionization Constants for Weak Monoprotic Acidsthe MATH Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution. • It is always a good idea to write down the ionization reaction and the ionization constant expression. Next we combine the basic chemical concepts with some algebra to solve the problem

  17. Ionization Constants for Weak Monoprotic Acids and Bases • Substitute these algebraic quantities into the ionization expression. • Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations. • Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations.

  18. Ionization Constants for Weak Monoprotic Acids and Bases • Complete the algebra and solve for the concentrations of the species. • Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

  19. Ionization Constants for Weak Monoprotic Acids: the MATH • Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Ka= 4.0 x 10-10 for HCN • Substitute these algebraic quantities into the ionization expression. • Solve the algebraic equation, using the simplifying assumption that is appropriate for all weak acid and base ionizations.

  20. Let’s look at the percent ionization of two weak acids as a function of their ionization constants. Note that the [H+] in 0.15 M acetic acid is 200 times greater than for 0.15 M HCN. Ionization Constants for Weak Monoprotic Acids and Bases [ionized HY] % ionization = x 100% [unionized HY]

  21. Ionization Constants for Weak Monoprotic Acids and Bases • All of the calculations and understanding we have at present can be applied to weak acids and weak bases. Calculate the concentrations of the various species in 0.15 M aqueous ammonia.

  22. Ionization Constants for Weak Monoprotic Bases: the MATH • All of the calculations and understanding we have at present can be applied to weak acids and weak bases. Calculate the concentrations of the various species in 0.15 M aqueous ammonia. Kb = 1.8E-5

  23. Ionization Constants for Weak Monoprotic Bases: the MATH • The pH of an aqueous ammonia solution is 11.37. Calculate the molarity (original concentration) of the aqueous ammonia solution • Examination of the last equation suggests that our simplifying assumption can be applied. In other words (x-2.3x10-3)  x. • Making this assumption simplifies the calculation.

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