pKa and pKb Practice Questions with Answers

Concept Explanation

pKa and pKb are logarithmic measures that quantify the strength of an acid or a base, respectively, providing a convenient way to express very small equilibrium constants. The pKa (acid dissociation constant) is defined as the negative base-10 logarithm of the acid dissociation constant (Ka), or pKa = -log(Ka). Similarly, the pKb (base dissociation constant) is defined as the negative base-10 logarithm of the base dissociation constant (Kb), or pKb = -log(Kb).

A lower pKa value indicates a stronger acid, meaning it dissociates more extensively in solution, while a higher pKa indicates a weaker acid. Conversely, a lower pKb value indicates a stronger base, meaning it accepts protons more readily, and a higher pKb indicates a weaker base. The relationship between pKa and pKb for a conjugate acid-base pair in aqueous solution at 25°C is given by pKa + pKb = 14, which derives from the ion product of water (Kw = 1.0 x 10-14 at 25°C) where Kw = Ka * Kb.

Understanding these values is crucial for predicting reaction direction, buffer preparation, and analyzing acid-base behavior in various chemical and biological systems. For more on understanding fundamental chemical concepts, you might find our guide on how to study effectively helpful.

Solved Examples

Example 1: Calculating pKa from Ka

Calculate the pKa of acetic acid (CH₃COOH) if its Ka is 1.8 x 10-5.

1. Identify the given value: Ka = 1.8 x 10-5.

2. Recall the formula for pKa: pKa = -log(Ka).

3. Substitute the Ka value into the formula: pKa = -log(1.8 x 10-5).

4. Calculate the logarithm: log(1.8 x 10-5) ≈ -4.74.

5. Multiply by -1: pKa = -(-4.74) = 4.74.

6. State the answer: The pKa of acetic acid is 4.74.

Example 2: Calculating pKb from Kb

Given that the base dissociation constant (Kb) for ammonia (NH₃) is 1.8 x 10-5, determine its pKb.

1. Identify the given value: Kb = 1.8 x 10-5.

2. Recall the formula for pKb: pKb = -log(Kb).

3. Substitute the Kb value into the formula: pKb = -log(1.8 x 10-5).

4. Calculate the logarithm: log(1.8 x 10-5) ≈ -4.74.

5. Multiply by -1: pKb = -(-4.74) = 4.74.

6. State the answer: The pKb of ammonia is 4.74.

Example 3: Calculating Ka from pKa

The pKa of hydrocyanic acid (HCN) is 9.21. What is its Ka?

1. Identify the given value: pKa = 9.21.

2. Recall the inverse relationship: Ka = 10-pKa.

3. Substitute the pKa value: Ka = 10-9.21.

4. Calculate the value: Ka ≈ 6.17 x 10-10.

5. State the answer: The Ka of hydrocyanic acid is 6.17 x 10-10.

Example 4: Calculating pKb from pKa of a Conjugate Acid-Base Pair

If the pKa of the ammonium ion (NH₄⁺) is 9.25, what is the pKb of its conjugate base, ammonia (NH₃), at 25°C?

1. Identify the given value: pKa (NH₄⁺) = 9.25.

2. Recall the relationship for conjugate acid-base pairs: pKa + pKb = 14 (at 25°C).

3. Rearrange the formula to solve for pKb: pKb = 14 - pKa.

4. Substitute the pKa value: pKb = 14 - 9.25.

5. Calculate the result: pKb = 4.75.

6. State the answer: The pKb of ammonia (NH₃) is 4.75.

Practice Questions

1. Calculate the pKa of hypochlorous acid (HClO) if its Ka is 3.0 x 10-8.

2. A weak base, methylamine (CH₃NH₂), has a Kb of 4.4 x 10-4. What is its pKb?

3. The pKa of benzoic acid (C₆H₅COOH) is 4.20. Determine its acid dissociation constant (Ka).

4. Given that the pKb of the fluoride ion (F⁻) is 10.83, calculate the pKa of its conjugate acid, hydrofluoric acid (HF), at 25°C.

5. Which is a stronger acid: an acid with pKa = 3.5 or an acid with pKa = 5.0? Explain your reasoning.

6. If the Ka of phenol (C₆H₅OH) is 1.0 x 10-10, what is the pKb of its conjugate base, phenoxide (C₆H₅O⁻), at 25°C?

7. A solution has a pH of 9.5. Is it more likely to contain a weak acid with pKa = 7.0 or a weak base with pKb = 4.0? Justify your choice.

8. Calculate the Kb for a base whose pKb is 2.80.

9. A biochemist is studying two amino acids. Amino acid A has an alpha-carboxyl group with pKa = 2.34, and amino acid B has an alpha-carboxyl group with pKa = 2.10. Which amino acid's alpha-carboxyl group is a stronger acid?

10. The pKa of a certain indicator is 7.2. What will be the ratio of its conjugate base form to its acid form ([In⁻]/[HIn]) when the pH of the solution is 8.2? (Hint: Use the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]))

Answers & Explanations

1. pKa of hypochlorous acid (HClO):

• Given Ka = 3.0 x 10-8.

• pKa = -log(Ka) = -log(3.0 x 10-8).

• pKa ≈ 7.52.

2. pKb of methylamine (CH₃NH₂):

• Given Kb = 4.4 x 10-4.

• pKb = -log(Kb) = -log(4.4 x 10-4).

• pKb ≈ 3.36.

3. Ka of benzoic acid (C₆H₅COOH):

• Given pKa = 4.20.

• Ka = 10-pKa = 10-4.20.

• Ka ≈ 6.31 x 10-5.

4. pKa of hydrofluoric acid (HF):

• Given pKb (F⁻) = 10.83.

• For a conjugate acid-base pair, pKa + pKb = 14 (at 25°C).

• pKa (HF) = 14 - pKb (F⁻) = 14 - 10.83.

• pKa (HF) = 3.17.

5. Stronger acid comparison:

• An acid with pKa = 3.5 is a stronger acid than an acid with pKa = 5.0.

• A lower pKa value corresponds to a higher Ka value, which indicates a stronger acid because it dissociates more completely in solution.

6. pKb of phenoxide (C₆H₅O⁻):

• Given Ka (phenol) = 1.0 x 10-10.

• First, calculate pKa (phenol): pKa = -log(1.0 x 10-10) = 10.00.

• For a conjugate acid-base pair, pKa + pKb = 14.

• pKb (phenoxide) = 14 - pKa (phenol) = 14 - 10.00.

• pKb (phenoxide) = 4.00.

7. Solution pH analysis:

• A solution with pH 9.5 is basic.

• A weak acid with pKa = 7.0 would have its conjugate base present in significant amounts at pH 9.5, but the solution would only be basic if the conjugate base was strong enough.

• A weak base with pKb = 4.0 corresponds to a relatively strong weak base (Kb = 10-4 = 1.0 x 10-4). This base would produce OH⁻ ions, making the solution basic.

• Therefore, it is more likely to contain a weak base with pKb = 4.0.

8. Kb for a base with pKb = 2.80:

• Given pKb = 2.80.

• Kb = 10-pKb = 10-2.80.

• Kb ≈ 1.58 x 10-3.

9. Stronger acid comparison (amino acids):

• Amino acid A: pKa = 2.34.

• Amino acid B: pKa = 2.10.

• A lower pKa indicates a stronger acid.

• Therefore, amino acid B's alpha-carboxyl group (pKa = 2.10) is a stronger acid than amino acid A's (pKa = 2.34).

10. Ratio of conjugate base to acid form:

• Given pKa = 7.2 and pH = 8.2.

• Using the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]).

• 8.2 = 7.2 + log([In⁻]/[HIn]).

• 1.0 = log([In⁻]/[HIn]).

• [In⁻]/[HIn] = 101.0 = 10.

• The ratio of conjugate base to acid form is 10:1.

Quick Quiz

Frequently Asked Questions

What is the difference between Ka and pKa?

Ka is the acid dissociation constant, representing the equilibrium constant for the dissociation of a weak acid into its conjugate base and a proton. pKa is the negative logarithm of Ka (pKa = -log Ka), providing a more convenient scale for expressing acid strength, especially for very small Ka values.

Why is pKa used instead of Ka?

pKa is used because Ka values for weak acids are often very small numbers (e.g., 10-5, 10-10), making them cumbersome to compare. The logarithmic pKa scale converts these small values into more manageable positive numbers, simplifying comparisons of acid strengths. For example, a pKa of 5 is easier to grasp than a Ka of 10-5.

How does temperature affect pKa and pKb?

pKa and pKb values are temperature-dependent because the underlying equilibrium constants (Ka and Kb) are temperature-dependent. The ion product of water (Kw) also changes with temperature, which in turn affects the relationship pKa + pKb = pKw. This relationship is exactly 14 only at 25°C.

What is the significance of the pKa + pKb = 14 relationship?

This relationship, valid at 25°C, highlights the inverse strength of a conjugate acid-base pair. If an acid is strong (low pKa), its conjugate base will be weak (high pKb), and vice-versa. It is derived from the autoionization of water, where Kw = Ka * Kb for a conjugate pair.

How do pKa and pKb relate to pH?

pKa and pKb relate to pH through the Henderson-Hasselbalch equation (pH = pKa + log([A⁻]/[HA])) for acids and an analogous equation for bases (pOH = pKb + log([BH⁺]/[B])). These equations show that when the concentrations of the acid/base and its conjugate are equal, pH = pKa or pOH = pKb, respectively, which is critical for understanding buffer solutions.

Can pKa or pKb be negative?

Yes, pKa or pKb can be negative. A negative pKa indicates a very strong acid (Ka > 1), meaning it dissociates almost completely. For example, the pKa of hydrochloric acid (HCl) is approximately -7. Similarly, a negative pKb would indicate a very strong base.

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