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    Medium pKa and pKb Practice Questions

    March 29, 20267 min read14 views
    Medium pKa and pKb Practice Questions

    Concept Explanation

    The pKa and pKb values are the negative base-10 logarithms of the acid dissociation constant (Ka) and base dissociation constant (Kb), respectively, serving as quantitative measures of the strength of acids and bases in solution. Mastering pKa and pKb practice questions is essential for understanding how molecules behave in biological systems and chemical reactions. A lower pKa indicates a stronger acid, meaning the substance more readily donates a proton, while a lower pKb indicates a stronger base. These values are intrinsically linked through the self-ionization constant of water ( K w K_w ). At standard temperature (25Β°C), the relationship is defined by the equation p K a + p K b = 14 pK_a + pK_b = 14 . This relationship allows chemists to calculate the strength of a conjugate base if the strength of the parent acid is known, and vice versa. Understanding these logarithmic scales is vital for predicting the direction of equilibrium in strong acid vs weak acid comparisons and for performing complex pH calculation practice questions.

    Solved Examples

    Below are worked examples to demonstrate how to convert between K a / K b K_a/K_b and p K a / p K b pK_a/pK_b , and how to use the relationship between conjugate pairs.

    1. Example 1: Converting Ka to pKa
      An unknown weak acid has an acid dissociation constant ( K a K_a ) of 1.8 Γ— 1 0 βˆ’ 5 1.8 \times 10^{-5} . Calculate its p K a pK_a .

      1. Identify the formula: p K a = βˆ’ log ⁑ ( K a ) pK_a = -\log(K_a) .

      2. Substitute the value: p K a = βˆ’ log ⁑ ( 1.8 Γ— 1 0 βˆ’ 5 ) pK_a = -\log(1.8 \times 10^{-5}) .

      3. Calculate: p K a β‰ˆ 4.74 pK_a \approx 4.74 .

    2. Example 2: Finding pKb from pKa
      The p K a pK_a of hydrocyanic acid (HCN) is 9.21. Determine the p K b pK_b of its conjugate base, the cyanide ion ( C N βˆ’ CN^- ).

      1. Identify the relationship: p K a + p K b = 14.00 pK_a + pK_b = 14.00 (at 25Β°C).

      2. Rearrange for p K b pK_b : p K b = 14.00 βˆ’ p K a pK_b = 14.00 - pK_a .

      3. Calculate: p K b = 14.00 βˆ’ 9.21 = 4.79 pK_b = 14.00 - 9.21 = 4.79 .

    3. Example 3: Calculating Kb from pKb
      A base has a p K b pK_b of 3.30. What is the value of its base dissociation constant ( K b K_b )?

      1. Identify the formula: K b = 1 0 βˆ’ p K b K_b = 10^{-pKb} .

      2. Substitute the value: K b = 1 0 βˆ’ 3.30 K_b = 10^{-3.30} .

      3. Calculate: K b β‰ˆ 5.01 Γ— 1 0 βˆ’ 4 K_b \approx 5.01 \times 10^{-4} .

    Practice Questions

    Test your knowledge with these medium pKa and pKb practice questions. Ensure you have a scientific calculator ready for the logarithmic conversions.

    1. A specific organic acid has a K a K_a of 4.5 Γ— 1 0 βˆ’ 4 4.5 \times 10^{-4} . Determine its p K a pK_a .

    2. If the p K b pK_b of ammonia ( N H 3 NH_3 ) is 4.75, what is the p K a pK_a of its conjugate acid, the ammonium ion ( N H 4 + NH_4^+ )?

    3. Calculate the K b K_b for a base that has a p K b pK_b of 8.90.

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    4. Formic acid has a p K a pK_a of 3.75. Calculate the K a K_a value and determine the p K b pK_b of the formate ion.

    5. Arrange the following acids in order of increasing strength based on their p K a pK_a values: Acid A ( p K a = 2.1 pK_a = 2.1 ), Acid B ( p K a = 4.8 pK_a = 4.8 ), Acid C ( p K a = 1.2 pK_a = 1.2 ).

    6. A solution of methylamine has a K b K_b of 4.4 Γ— 1 0 βˆ’ 4 4.4 \times 10^{-4} . Calculate both the p K b pK_b and the p K a pK_a of its conjugate acid.

    7. Given that K a K_a for nitrous acid ( H N O 2 HNO_2 ) is 4.0 Γ— 1 0 βˆ’ 4 4.0 \times 10^{-4} , find the K b K_b for the nitrite ion ( N O 2 βˆ’ NO_2^- ).

    8. Which is a stronger base: Base X with p K b = 3.5 pK_b = 3.5 or Base Y with K b = 1.0 Γ— 1 0 βˆ’ 5 K_b = 1.0 \times 10^{-5} ?

    9. A hypothetical acid has a p K a pK_a of -2.0. Is this a strong or weak acid, and what does this imply about its K a K_a ?

    10. If the p K a pK_a of a conjugate acid increases, what happens to the strength of the conjugate base?

    Answers & Explanations

    1. Answer: 3.35.
      Explanation: Use the formula p K a = βˆ’ log ⁑ ( K a ) pK_a = -\log(K_a) . Thus, βˆ’ log ⁑ ( 4.5 Γ— 1 0 βˆ’ 4 ) = 3.346 -\log(4.5 \times 10^{-4}) = 3.346 , rounded to 3.35.

    2. Answer: 9.25.
      Explanation: Using p K a + p K b = 14 pK_a + pK_b = 14 , we get 14 βˆ’ 4.75 = 9.25 14 - 4.75 = 9.25 . This is a common calculation when working with buffer solution practice questions.

    3. Answer: 1.26 Γ— 1 0 βˆ’ 9 1.26 \times 10^{-9} .
      Explanation: K b = 1 0 βˆ’ p K b = 1 0 βˆ’ 8.90 β‰ˆ 1.26 Γ— 1 0 βˆ’ 9 K_b = 10^{-pKb} = 10^{-8.90} \approx 1.26 \times 10^{-9} .

    4. Answer: K a = 1.78 Γ— 1 0 βˆ’ 4 K_a = 1.78 \times 10^{-4} ; p K b = 10.25 pK_b = 10.25 .
      Explanation: K a = 1 0 βˆ’ 3.75 = 1.78 Γ— 1 0 βˆ’ 4 K_a = 10^{-3.75} = 1.78 \times 10^{-4} . For the conjugate base, p K b = 14 βˆ’ 3.75 = 10.25 pK_b = 14 - 3.75 = 10.25 .

    5. Answer: Acid B < Acid A < Acid C.
      Explanation: Strength increases as p K a pK_a decreases. Acid C (1.2) is the strongest, and Acid B (4.8) is the weakest.

    6. Answer: p K b = 3.36 pK_b = 3.36 ; p K a = 10.64 pK_a = 10.64 .
      Explanation: p K b = βˆ’ log ⁑ ( 4.4 Γ— 1 0 βˆ’ 4 ) = 3.356 pK_b = -\log(4.4 \times 10^{-4}) = 3.356 . p K a = 14 βˆ’ 3.356 = 10.644 pK_a = 14 - 3.356 = 10.644 .

    7. Answer: 2.5 Γ— 1 0 βˆ’ 11 2.5 \times 10^{-11} .
      Explanation: K a Γ— K b = 1 0 βˆ’ 14 K_a \times K_b = 10^{-14} . So, K b = 1 0 βˆ’ 14 / ( 4.0 Γ— 1 0 βˆ’ 4 ) = 2.5 Γ— 1 0 βˆ’ 11 K_b = 10^{-14} / (4.0 \times 10^{-4}) = 2.5 \times 10^{-11} .

    8. Answer: Base X.
      Explanation: Convert Base Y's K b K_b to p K b pK_b : βˆ’ log ⁑ ( 1.0 Γ— 1 0 βˆ’ 5 ) = 5.0 -\log(1.0 \times 10^{-5}) = 5.0 . Since Base X has a lower p K b pK_b (3.5), it is the stronger base.

    9. Answer: Strong acid; K a = 100 K_a = 100 .
      Explanation: A negative p K a pK_a indicates a very strong acid that dissociates completely or nearly completely in water. K a = 1 0 βˆ’ ( βˆ’ 2 ) = 1 0 2 = 100 K_a = 10^{-(-2)} = 10^2 = 100 .

    10. Answer: The conjugate base becomes stronger.
      Explanation: As p K a pK_a increases, the acid becomes weaker. Because p K a + p K b = 14 pK_a + pK_b = 14 , a higher p K a pK_a means a lower p K b pK_b , which signifies a stronger conjugate base.

    Quick Quiz

    Interactive Quiz 5 questions

    1. Which of the following pKa values represents the strongest acid?

    • A 10.5
    • B 4.2
    • C 1.5
    • D 7.0
    Check answer

    Answer: C. 1.5

    2. If an acid has a pKa of 5.0, what is the pKb of its conjugate base?

    • A 5.0
    • B 14.0
    • C 9.0
    • D -5.0
    Check answer

    Answer: C. 9.0

    3. How is Ka related to pKa?

    • A pKa = Ka / 14
    • B pKa = -log(Ka)
    • C pKa = 10^Ka
    • D pKa = ln(Ka)
    Check answer

    Answer: B. pKa = -log(Ka)

    4. A base with a Kb of 1.0 x 10^-4 has what pKb?

    • A 4.0
    • B 10.0
    • C -4.0
    • D 1.0
    Check answer

    Answer: A. 4.0

    5. In a conjugate acid-base pair, if the acid is very strong (pKa < 0), the conjugate base is:

    • A Also very strong
    • B Extremely weak
    • C Neutral
    • D Unstable
    Check answer

    Answer: B. Extremely weak

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    Frequently Asked Questions

    What is the difference between Ka and pKa?

    Ka is the acid dissociation constant that measures the equilibrium concentration of ions, while pKa is the negative logarithm of that constant. The pKa scale is more convenient for comparing acid strengths because it uses smaller, more manageable numbers rather than scientific notation.

    Can a pKa value be negative?

    Yes, extremely strong acids like hydrochloric acid or sulfuric acid have negative pKa values. A negative pKa indicates that the acid dissociation constant (Ka) is greater than 1, meaning the acid dissociates almost completely in aqueous solution.

    How does temperature affect pKa and pKb?

    Since pKa and pKb are derived from equilibrium constants, they are temperature-dependent. As temperature increases, the self-ionization of water ( K w K_w ) typically increases, which shifts the p K a + p K b = 14 pK_a + pK_b = 14 relationship to a lower sum.

    Why is the sum of pKa and pKb always 14?

    This relationship is derived from the water auto-ionization constant, K w = [ H + ] [ O H βˆ’ ] = 1.0 Γ— 1 0 βˆ’ 14 K_w = [H^+][OH^-] = 1.0 \times 10^{-14} at 25Β°C. Taking the negative log of both sides of the equation K a Γ— K b = K w K_a \times K_b = K_w results in p K a + p K b = 14 pK_a + pK_b = 14 .

    How do pKa values help in predicting reaction direction?

    Acid-base reactions favor the side with the weaker acid and weaker base. By comparing the pKa values of the acids on both sides of the equation, you can predict that the equilibrium will shift toward the species with the higher pKa value.

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