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Ka and Kb Calculations: Mastering Acid-Base Chemistry
Dive into the world of acid-base chemistry with our comprehensive guide to Ka and Kb calculations. Learn to determine acid and base strengths, calculate pH levels, and understand buffer systems for success in chemistry.

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Now Playing:Ka and kb calculations – Example 0a
Intros
  1. Applying the Ka expression
  2. Applying the Ka expression
    Weak acids/bases at equilibrium.
  3. Applying the Ka expression
    Assumptions made in calculation.
Examples
  1. Find the pH of the weak acid solution and the percentage dissociation of the weak acid/base.
    1. What is the pH of a solution of 0.5 M ethanoic acid, CH3COOH? Find the percentage dissociation of this ethanoic acid solution.

    2. What is the pH of a solution of 0.1 M ammonia, NH3? Find the percentage dissociation of this ammonia solution.

Introduction to acid-base theory
Jump to:NotesConceptExampleFAQsPrerequisites
Notes

In this lesson, we will learn:

  • How to apply the Ka expression to find the pH of weak acid solutions.
  • The assumptions made in Ka calculations at equilibrium and how to justify them.
  • How to solve for Kb using Kw and find the pH of weak base solutions.

Notes:

  • We know that weak acids and bases are any acid/base species that does not completely dissociate in water. Dissolving a weak acid in water then has two chemical effects:
    • Some of the weak acid, HX, will interact with water and dissociate into H3O+ and X- ions.
    • The rest of the HX will stay un-dissociated.
    We can write an equilibrium for the dissociation of weak acid HX (with A concentration) in water, and show amounts in a table, in a ‘before and after’ format like below:

    HX + H2O ⇌\rightleftharpoons H3O+ + Cl-

    Start concentration (M)

    Equilibrium conc. (M)

    HX

    A

    A - B

    H3O+

    ≈\approx 0

    B

    X-

    0

    B


    The acid dissociation constant, Ka can be expressed in these terms:

    Ka = [X−][H3O+][HX]\frac{[X^{-}][H_3O^+]}{[HX]} = [X2][A]−B\frac{[X^2]}{[A] -B}


  • Taking an example with 0.1M methanoic acid HCOOH, we can write the following:

    Start concentration (M)

    Equilibrium conc. (M)

    HCOOH

    0.1

    0.1 - B

    H3O+

    0

    B

    X-

    0

    B



    We can apply the acid dissociation constant, Ka to this equilibrium. Methanoic acid1 has a Ka value of 1.8∗*10-4 so the Ka equation can be written fully:

    Ka = [HCOO−][H3O+][HCOOH]\frac{[HCOO^-][H_3O^+]}{[HCOOH]} = [H3O+]2[0.1]−[H3O+]\frac{[H_3O^+]^2}{[0.1] -[H_3O^+]} = 1.8 ∗* 10-4

    Some assumptions are made to complete this calculation:
    • The starting concentration of H3O+ ions, in neutral water is only 1*10-7 M (see Autoionization of water) and an equally tiny amount of hydroxide ions are also present. This is an incredibly small amount, so this H3O+ is not taken into the calculation; only H3O+ due to the weak acid is used in the calculation.
    • With weak acids, we assume that the acid is weak enough that the amount of dissociation doesn’t affect acid concentration. Using the table above, 0.1 M HCOOH added to neutral water will still have concentration of approximately 0.1M at equilibrium, and we can ignore the ‘– B’ in ‘A-B’. Where [HX]eq = equilibrium concentration, [HX]i = start concentration of HX:

      [HX]eq - B ≅\cong [HX]i

      YOU MUST STATE THIS ASSUMPTION IN CALCULATIONS.


    With the assumptions, we have a final expression:

    Ka = [HCOO−][H3O+][HCOOH]\frac{[HCOO^-][H_3O^+]}{[HCOOH]} = [H3O+]2[0.1]\frac{[H_3O^+]^2}{[0.1] } = 1.8 ∗* 10-4

    0.1 ∗* 1.8 ∗* 10-4 = [H3O+]2

    1.8∗10−5\small\sqrt{1.8 * 10^{-5}} = [H3O+] = 4.24 ∗* 10-3

    pH = -log [H3O+] = 2.37


    Note that the assumption we made was that [HX] – B is approximately equal to [HX]. We now know that B = 4.24*10-3, so our assumption in this example was to say that 0.1 – 4.24*10-3 = 0.0958.
    The assumption can be justified if percentage dissociation is less than 5% which we can work out:

    % dissociation = [H3O+]eq[HX]i\frac{[H_3O^+]_{eq}}{[HX]_i} ∗* 100

    Therefore:

    % dissociation = [4.24∗10−3][0.1]\frac{[4.24 *10^{-3}]}{[0.1]} ∗* 100 = 4.24%

    As the calculation shows, the assumption was justified as only 4.24% dissociation occurs.

  • Kb calculations are similar to Ka calculations with some changes:
    • Because acidity strength tables give only Ka, Kb of a weak base will need to be found by the calculation in the autoionization of water expression. You will need to find the Ka of the conjugate acid in the acidity strength table to do this.
    • The equilibrium concentrations you obtain using Kb will give you [OH-], so pH will need to be found by solving: pH = 14 – pOH.


    Taking an example with 0.5M of the weak base ammonia, NH3, we can write the following:

    NH3 + H2O ⇌\rightleftharpoons NH4+ + OH-

    Start concentration (M)

    Equilibrium conc. (M)

    NH3

    0.5

    0.5 - B

    NH3+

    0

    B

    OH-

    0

    B



    The conjugate acid of ammonia is the ammonium ion, NH4+ which has a Ka value1 of 5.6 ∗* 10-10. Solving the autoionization expression for Kb(NH3) gives:

    Kb = KwKa\frac{K_w}{K_a} = 10−145.6∗10−10\frac{10^{-14}}{5.6 * 10^{-10}} = 1.79 ∗* 10-5


    Using our value for Kb (now rounding to 1.8 ∗* 10-5 or 2 significant figures) we can solve for the equilibrium concentration of hydroxide ions. Again, we make the assumption that the concentration of NH3 isn’t significantly affected by the dissociation into NH4+:

    [NH3]eq - B ≅\cong [NH3]i

    Now we can find the hydroxide ion concentration:

    Kb = [NH4+][OH−][NH3]\frac{[NH_{4}^{+}][OH^-]{}} {[NH_3]} = [OH−]2[0.5]\frac{[OH^-]^2}{[0.5]} = 1.8 ∗* 10-5


    [OH-] = ((1.8∗10−5)∗0.5)\sqrt{((1.8 * 10^{-5}) * 0.5)} = 3 ∗* 10-3


    pOH = -log[OH-] = -log (3 ∗* 10-3) = 2.52


    pH = 14 - pOH = 14 - 2.52 = 11.48

    Testing the assumption can now be done:

    % dissociation = [OH−]eq[NH3]i\frac{[OH^-]_{eq}}{[NH_3]_i} ∗* 100 = 3∗10−30.5\frac{3 * 10^{-3}}{0.5} ∗* 100 = 0.6 %

    With only 0.6% dissociation, the assumption is justified and the pH has been found.

  • Another calculation that shows the difference between strong and weak acids and bases is the effect of dilution on pH.
    The effect on strong acids is straightforward, because we assume 100% dissociation:
    • If a solution of strong acid, e.g. HCl is 1M, then [H3O+ (aq)] = 1M. Taking the negative log of this:
      pH = -log[1] = 0
      Diluting this by a factor of 10 will give a concentration of 0.1M
      pH = -log[0.1] = 1
      A further 10, or 100 fold from the original:
      pH = -log[0.01] = 2
      In short, diluting a strong acid or base has a direct logarithmic effect on pH.
    The effect of dilution on weak acids and bases is different:
    • If a solution of weak acid, e.g. CH3COOH is 1M, then pH and [H3O+ (aq)] is worked out using the Ka expression:
      Ka (CH3COOH) = 1.4*10-5

      Ka = [H3O+][CH3COO−][CH3COOH]\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]}

      The calculation to find pH using this expression has been explained above, so moving forward to an answer (using the assumptions needed)

      1.4 * 10-5 = [H3O+][CH3COO−][1]\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[1]} where [H3O+] = [CH3COOH-]

      1.4∗10−5\sqrt{1.4 * 10^{-5}} = [H3O+] = 3.74 * 10-3

      pH = -log[ 3.74 * 10-3] = 2.42

      A dilution of this weak acid solution to make it 0.1M would have the following effect on the calculation:

      1.4 * 10-5 = [H3O+][CH3COO−][0.1]\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[0.1]} where [H3O+] = [CH3COOH-]

      1.4∗10−6\sqrt{1.4 * 10^{-6}} = [H3O+] = 1.18 * 10-3

      pH = -log[ 1.18 * 10-3] = 2.92

      Another dilution by a factor of ten:

      1.4 * 10-5 = [H3O+][CH3COO−][0.01]\frac{[H_{3}O^{+}][CH_{3}COO^{-}]}{[0.01]} where [H3O+] = [CH3COOH-]

      1.4∗10−7\sqrt{1.4 * 10^{-7}} = [H3O+] = 3.74 * 10-4

      pH = -log[ 3.74 * 10-4] = 3.42

      In short, diluting a weak acid has a lesser effect on pH than in strong acids.
Concept

Introduction to Ka and Kb Calculations

Ka and Kb calculations are fundamental concepts in chemistry, dealing with acid dissociation constants for weak acids and bases. The introduction video provides a comprehensive overview of these calculations, serving as a crucial starting point for understanding the topic. Ka represents the acid dissociation constant, while Kb is the base dissociation constant. These values are essential in determining the strength of weak acids and bases, respectively. By mastering Ka and Kb calculations, students can predict the behavior of acids and bases in solution, calculate pH levels, and understand buffer systems. These calculations play a vital role in various chemical processes, from industrial applications to biological systems. The ability to perform and interpret Ka and Kb calculations is indispensable for chemists, allowing them to analyze and manipulate acid-base equilibria in diverse scientific and practical contexts.

Example

Applying the Ka expression Weak acids/bases at equilibrium.

Step 1: Introduction to Ka and Kb Calculations

Welcome to the section on Ka and Kb calculations. In this guide, we will walk through the process of finding the pH of a solution given the acid dissociation constant (Ka) and the concentration of a weak acid. We will also touch upon the assumptions made during these calculations and how to solve for Kb using the Kw expression.

Step 2: Understanding Weak Acids and Bases at Equilibrium

Weak acids and bases do not fully dissociate in water. This partial dissociation means that the equilibrium expression for weak acids must account for both the undissociated acid and the ions produced. For weak acids, the general dissociation can be represented as:

HX + H2O H3O+ + X-

Here, HX is the weak acid, H3O+ is the hydronium ion, and X- is the conjugate base. The Ka expression for this equilibrium is:

Ka = (H3O+)(X-) / (HX)

Step 3: Setting Up the Equilibrium Expression

To find the pH of a weak acid solution, we start by setting up the equilibrium expression. Consider a weak acid HX with an initial concentration of 'a'. At equilibrium, the concentration of HX will be 'a - b', where 'b' is the amount that dissociates. The concentrations of H3O+ and X- at equilibrium will both be 'b'. Thus, the Ka expression becomes:

Ka = (H3O+)(X-) / (HX) = b2 / (a - b)

Step 4: Making Assumptions for Simplification

Given that weak acids do not dissociate completely, 'b' is typically much smaller than 'a'. This allows us to make the assumption that 'a - b' is approximately equal to 'a'. This simplifies the Ka expression to:

Ka b2 / a

This assumption holds true for most weak acids, where the degree of dissociation is less than 50%.

Step 5: Solving for the Hydronium Ion Concentration

To find the concentration of H3O+ (b), we rearrange the simplified Ka expression:

b = (Ka * a)

By substituting the known values of Ka and 'a', we can calculate the concentration of H3O+.

Step 6: Calculating the pH

Once we have the concentration of H3O+, we can find the pH of the solution using the formula:

pH = -log(H3O+)

Substitute the value of (H3O+) obtained from the previous step to calculate the pH.

Step 7: Example Calculation

Let's consider an example with 0.1 M methanoic acid (HCOOH) with a Ka of 1.8 x 10-4. Using the simplified Ka expression:

b = (1.8 x 10-4 * 0.1) = (1.8 x 10-5) 4.24 x 10-3

Then, the pH is calculated as:

pH = -log(4.24 x 10-3) 2.37

Step 8: Conclusion

By following these steps, you can apply the Ka expression to find the pH of a weak acid solution. Remember to consider the assumptions made during the calculations and verify if they hold true for your specific case. This method can also be adapted for weak bases using the Kb expression and the relationship between Ka, Kb, and Kw.

FAQs

Here are some frequently asked questions about Ka and Kb calculations:

1. How do you calculate pH from Ka?

To calculate pH from Ka, follow these steps:

  1. Write the acid dissociation equation
  2. Set up the Ka expression
  3. Use an ICE table to solve for (H+)
  4. Calculate pH using the formula: pH = -log(H+)

For example, if Ka = 1.8 × 10-5 for a 0.1 M weak acid solution:

Ka = (H+)(A-) / (HA) = 1.8 × 10-5 = x2 / (0.1 - x)

Solve for x (assuming x << 0.1), then calculate pH = -log(x)

2. What is the relationship between Ka and Kb?

The relationship between Ka and Kb is expressed by the equation:

Ka × Kb = Kw

Where Kw is the ion product constant of water (1.0 × 10-14 at 25°C). This relationship allows you to calculate Kb if you know Ka, and vice versa.

3. How do you convert from Ka to Kb?

To convert from Ka to Kb, use the following formula:

Kb = Kw / Ka

For example, if Ka = 1.8 × 10-5, then:

Kb = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10

4. How is pKa related to Ka?

pKa is the negative logarithm of Ka:

pKa = -log(Ka)

This relationship is useful for comparing acid strengths. A lower pKa indicates a stronger acid.

5. How do you calculate the percentage dissociation of a weak acid?

To calculate the percentage dissociation:

  1. Find the equilibrium concentration of H+ ions
  2. Divide by the initial concentration of the acid
  3. Multiply by 100

Percentage dissociation = ((H+) / (HA)initial) × 100

For example, if (H+) = 1.34 × 10-3 M for a 0.1 M acid solution:

Percentage dissociation = (1.34 × 10-3 / 0.1) × 100 = 1.34%

Prerequisites

Understanding Ka and Kb calculations is crucial in chemistry, particularly when dealing with acids and bases. However, to truly grasp these concepts, it's essential to have a solid foundation in certain prerequisite topics. Two key areas that are fundamental to mastering Ka and Kb calculations are the acid dissociation constant and strong and weak acids and bases.

The acid dissociation constant, also known as Ka, is a fundamental concept that quantifies the strength of an acid in solution. This constant is directly related to the equilibrium constant for base dissociation, which is essential for understanding Kb calculations. By mastering the acid dissociation constant, students can more easily comprehend how acids behave in solution and how this relates to their strength and dissociation properties.

Similarly, a thorough understanding of strong and weak acids and bases is crucial for Ka and Kb calculations. This topic provides the foundation for distinguishing between different types of acids and bases based on their dissociation behavior. It's particularly important when dealing with weak bases, as pOH calculations for weak bases are directly related to Kb values.

The relationship between these prerequisite topics and Ka and Kb calculations is intricate and multifaceted. For instance, the acid dissociation constant (Ka) is used to determine the strength of an acid, which in turn affects its behavior in solution. This knowledge is crucial when performing calculations involving Ka values. Similarly, understanding the characteristics of strong and weak acids and bases helps in predicting their behavior and applying the appropriate formulas in Ka and Kb calculations.

Moreover, these prerequisite topics provide the conceptual framework necessary for interpreting Ka and Kb values. For example, knowing how to classify acids and bases as strong or weak based on their dissociation constants allows students to make informed predictions about their behavior in various chemical reactions and equilibrium situations.

In conclusion, a solid grasp of the acid dissociation constant and strong and weak acids and bases is essential for success in Ka and Kb calculations. These prerequisite topics provide the necessary foundation for understanding the underlying principles, applying the correct formulas, and interpreting results accurately. By mastering these concepts, students will be well-equipped to tackle more complex problems involving acid-base equilibria and related calculations in their chemistry studies.