The acid dissociation constant k a
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The Acid Dissociation Constant, K a. The Acid Dissociation Constant. When a weak acid dissolves in water, it does not completely dissociate.

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The Acid Dissociation Constant, K a

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The acid dissociation constant k a

The Acid Dissociation Constant, Ka


The acid dissociation constant

The Acid Dissociation Constant

  • When a weak acid dissolves in water, it does not completely dissociate.

  • The concentration of the hydronium ions, and the concentration of the conjugate base of the acid that is formed in solution, depend on the initial concentration of the acid and the amount of acid that dissociates.


The acid dissociation constant1

The Acid Dissociation Constant

  • You can represent any weak monoprotic acid with the general formula HA.

  • Thus you can express the equilibrium of a weak monoprotic acid in aqueous solution as follows:

    HA(aq) + H2O(l) H3O+(aq) + A-(aq)

  • The equilibrium constant expression for this reaction is:

    Kc = [H3O+][A-]

    [HA][H2O]


The acid dissociation constant2

The Acid Dissociation Constant

  • In dilute solutions, the concentration of water is almost constant.

  • Multiplying both sides of the equilibrium expression by [H2O] gives the product of two constants on the left side.

  • This new constant is called the acid dissociation constant, Ka.

    Kc[H2O] = Ka = [H3O+][A-]

    [HA]


K a of a weak acids

Ka of a Weak Acids

  • Some Acid Dissociation Constants for Weak Acids at 25oC


Solving equilibrium problems that involve acids and bases

Solving Equilibrium Problems that Involve Acids and Bases

  • Problems that involve the concentrations of ions formed in aqueous solutions are considered to be equilibrium problems.

  • The steps for solving acid and base equilibrium problems are similar to the steps you learned in the previous unit for solving equilibrium problems.


Solving equilibrium problems that involve acids and bases1

Solving Equilibrium Problems that Involve Acids and Bases

  • Write the chemical equation.

  • Use the chemical equation to set up an ICE table for the reacting substances whenever the initial acid concentrations are involved.

  • Enter any values that are given in the problem.


Solving equilibrium problems that involve acids and bases2

Solving Equilibrium Problems that Involve Acids and Bases

  • For problems that give the initial concentration of the acid, [HA], compare the initial concentration of the acid with the acid dissociation constant, Ka.

  • If [HA]/Ka > 500, the change in the initial concentration is negligible and can be ignored.

  • If [HA]/Ka < 500, the change in the initial concentration may not be negligible. Solving the equilibrium constant expression for x will be more complex (possibly have to use quadratic equation).


Percent dissociation

Percent Dissociation

  • The percent dissociation of a weak acid is the fraction of acid molecules that dissociate compared with the initial concentration of the acid, expressed as a percent.

  • The percent dissociation depends on the value of Ka for the acid, as well as the initial concentration of the weak acid.


Sample problem

Sample Problem

  • Propanoic acid, CH3CH2COOH, is a weak monoprotic acid that is used to inhibit mould formation in bread. A student prepared a 0.10 mol/L solution of propanoic acid and found that the pH was 2.96. What is the acid dissociation constant for propanoic acid? What percent of its molecules were dissociated in the solution?

  • What is required?

    • Ka and percent dissociation for propanoic acid

  • What is given?

    • Initial [CH3CH2COOH] = 0.10 mol/L

    • pH = 2.96


Sample problem1

Sample Problem

  • Step 1

    Use the equation for the dissociation of propanoic acid in water to set up an ICE table.

    CH3CH2COOH(aq) + H2O(l) CH3CH2COO-(aq) + H3O+

    Initial 0.10 mol/L 0 0

    Change -x +x +x

    Equil 0.10 - x x x


Sample problem2

Sample Problem

  • Step 2

    Ka = [CH3CH2COO-][H3O+]

    [CH3CH2COOH]

    = (x)(x)

    (0.10 - x)

  • Step 3

    The value of x is equal to [H3O+] and [CH3CH2COO-].

    [H3O+] = 10-2.96

    = 1.1 x 10-3 mol/L


Sample problem3

Sample Problem

  • Step 4

    Ka = (1.1 x 10-3)2

    0.10 - (1.1 x 10-3)

    = 1.2 x 10-5

  • Step 5

    Percent dissociation = 1.1 x 10-3 mol/L x 100

    0.10 mol/L

    = 1.1 %


Practice problems

Practice Problems

  • Questions 5-10, pages 591-592


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