Easy Ka and Kb Calculations Practice Questions

Concept Explanation

Ka and Kb are equilibrium constants that measure the strength of an acid or a base in an aqueous solution. The acid dissociation constant ($K_a$) represents the extent to which an acid donates protons to water, while the base dissociation constant ($K_b$) represents the extent to which a base accepts protons from water. These values are essential for determining the concentration of ions in solution, which directly influences the pH calculation of a substance. For any conjugate acid-base pair, the relationship between these constants is governed by the ion-product constant of water ($K_w$), where $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ at 25°C. This mathematical link allows chemists to easily convert between the two values when studying chemical equilibria.

When a weak acid ($HA$) dissolves in water, it reaches a state of equilibrium: $HA + H_2O \rightleftharpoons A^- + H_3O^+$. The expression for $K_a$ is $[H_3O^+][A^-] / [HA]$. Similarly, for a weak base ($B$): $B + H_2O \rightleftharpoons BH^+ + OH^-$, the $K_b$ expression is $[BH^+][OH^-] / [B]$. Because these values are often very small, scientists frequently use logarithmic scales, leading to pKa and pKb values which are easier to manage in calculations. Understanding these constants is a fundamental step before moving on to more complex topics like buffer solution practice questions.

Solved Examples

Below are fully worked examples to demonstrate how to perform easy Ka and Kb calculations using standard formulas.

1. Calculating Ka from pH: A 0.10 M solution of a weak acid has a pH of 3.00. Find the $K_a$.

   1. Find $[H^+]$: $[H^+] = 10^{-pH} = 10^{-3} = 0.001 M$.

   2. Set up the $K_a$ expression: $K_a = [H^+][A^-] / [HA]$. Since $[H^+] = [A^-]$, $K_a = (0.001)^2 / (0.10 - 0.001)$.

   3. Assume $0.10 - 0.001 \approx 0.10$.

   4. $K_a = 1.0 \times 10^{-6} / 0.10 = 1.0 \times 10^{-5}$.

2. Calculating Kb from Ka: The $K_a$ of acetic acid is $1.8 \times 10^{-5}$. What is the $K_b$ of its conjugate base, the acetate ion?

   1. Use the relationship $K_a \times K_b = K_w$.

   2. $K_b = K_w / K_a = (1.0 \times 10^{-14}) / (1.8 \times 10^{-5})$.

   3. $K_b = 5.56 \times 10^{-10}$.

3. Finding OH- concentration from Kb: A 0.20 M solution of ammonia has a $K_b$ of $1.8 \times 10^{-5}$. Find $[OH^-]$.

   1. $K_b = [NH_4^+][OH^-] / [NH_3]$. Let $x = [OH^-]$.

   2. $1.8 \times 10^{-5} = x^2 / 0.20$.

   3. $x^2 = 3.6 \times 10^{-6}$.

   4. $x = \sqrt{3.6 \times 10^{-6}} = 1.9 \times 10^{-3} M$.

Practice Questions

1. A 0.050 M solution of a monoprotic weak acid has a $[H_3O^+]$ of $1.5 \times 10^{-4} M$. Calculate the $K_a$.

2. If the $K_b$ of a base is $4.0 \times 10^{-6}$, what is the $K_a$ of its conjugate acid?

3. Calculate the $[OH^-]$ of a 0.15 M weak base solution with a $K_b$ of $2.5 \times 10^{-7}$.

4. A weak acid has a $K_a = 6.2 \times 10^{-8}$. Calculate the $K_b$ of its conjugate base.

5. The pH of a 0.10 M solution of a weak base is 11.0. What is the $K_b$ of the base?

6. Find the $K_a$ for a 0.010 M solution of an acid that is 5% ionized.

7. What is the $K_b$ of the cyanide ion ($CN^-$) if the $K_a$ of $HCN$ is $4.9 \times 10^{-10}$?

8. A solution of 0.25 M hydrofluoric acid ($HF$) has a $K_a$ of $6.6 \times 10^{-4}$. Calculate the concentration of $H^+$.

9. If $pKa = 4.75$, calculate the $K_a$ value.

10. A base has a $pKb$ of 9.25. Calculate its $K_b$ value.

Answers & Explanations

1. Answer: $4.5 \times 10^{-7}$. $K_a = [H^+][A^-] / [HA] = (1.5 \times 10^{-4})^2 / 0.050 = 2.25 \times 10^{-8} / 0.050 = 4.5 \times 10^{-7}$.

2. Answer: $2.5 \times 10^{-9}$. $K_a = K_w / K_b = 1.0 \times 10^{-14} / 4.0 \times 10^{-6} = 2.5 \times 10^{-9}$.

3. Answer: $1.94 \times 10^{-4} M$. $K_b = x^2 / 0.15 \rightarrow x = \sqrt{2.5 \times 10^{-7} \times 0.15} = 1.94 \times 10^{-4} M$.

4. Answer: $1.61 \times 10^{-7}$. $K_b = 1.0 \times 10^{-14} / 6.2 \times 10^{-8} = 1.61 \times 10^{-7}$.

5. Answer: $1.0 \times 10^{-5}$. pH 11 means pOH = 3, so $[OH^-] = 10^{-3}$. $K_b = (10^{-3})^2 / 0.10 = 10^{-6} / 0.10 = 1.0 \times 10^{-5}$.

6. Answer: $2.63 \times 10^{-5}$. 5% of 0.010 M is $0.0005 M$ ($[H^+]$). $K_a = (0.0005)^2 / (0.010 - 0.0005) = 2.5 \times 10^{-7} / 0.0095 = 2.63 \times 10^{-5}$.

7. Answer: $2.04 \times 10^{-5}$. $K_b = 1.0 \times 10^{-14} / 4.9 \times 10^{-10} = 2.04 \times 10^{-5}$.

8. Answer: $1.28 \times 10^{-2} M$. $6.6 \times 10^{-4} = x^2 / 0.25 \rightarrow x = \sqrt{1.65 \times 10^{-4}} = 0.0128 M$.

9. Answer: $1.78 \times 10^{-5}$. $K_a = 10^{-pKa} = 10^{-4.75} = 1.78 \times 10^{-5}$.

10. Answer: $5.62 \times 10^{-10}$. $K_b = 10^{-pKb} = 10^{-9.25} = 5.62 \times 10^{-10}$.

Quick Quiz

Frequently Asked Questions

What is the difference between Ka and Kb?

Ka measures the strength of an acid by its ability to donate protons, whereas Kb measures the strength of a base by its ability to accept protons. Both are equilibrium constants that indicate the degree of ionization in water.

How do you calculate Kb if you only have Ka?

You can calculate Kb by dividing the ion-product constant of water ($K_w = 1.0 imes 10^{-14}$) by the given Ka value. This relationship only applies to conjugate acid-base pairs at a temperature of 25°C.

Why is Ka important in chemistry?

Ka is vital because it allows scientists to predict the pH of solutions and the direction of chemical reactions. It is a standard value used in Khan Academy chemistry tutorials to categorize weak versus strong electrolytes.

Does Ka change with concentration?

No, Ka is an equilibrium constant and remains the same regardless of the initial concentration of the acid, provided the temperature is constant. Only a change in temperature will alter the value of Ka or Kb.

Can Ka be greater than 1?

Yes, strong acids such as hydrochloric acid (HCl) have Ka values much greater than 1, indicating they ionize almost completely in water. You can learn more about these differences in our guide on strong acid vs weak acid practice questions.

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