Question

The pH of a 0.01 M solution of a weak acid HA is 4. Calculate its dissociation constant (Ka).

• A. $1.0 \times 10^{-6}$
• B. $1.0 \times 10^{-8}$
• C. $1.0 \times 10^{-4}$
• D. $1.0 \times 10^{-5}$

Hint:

For weak acids, use the approximation $[\text{H}^+] = \sqrt{Ka \cdot C}$ or $Ka = [\text{H}^+]^2 / C$.

Answer 1

The Correct Option is A

To determine the dissociation constant (K_a) of the weak acid HA, given the pH of the solution, follow these steps:

Step 1: Calculate the hydrogen ion concentration [H⁺].

pH is given as 4, so:

pH = -log[H⁺]

[H⁺] = 10^-pH = 10^-4 M

Step 2: Write the equilibrium expression for the dissociation of HA.

HA ⇌ H⁺ + A^-

Step 3: Express the K_a in terms of concentration.

K_a = \(\frac{[H^+][A^-]}{[HA]}\)

Assuming x M of HA dissociates, an equal amount of H⁺ and A^- is formed at equilibrium:

[H⁺] = x = 10^-4 M

[HA] = Initial concentration of HA - x

Given [HA] = 0.01 M, so remaining [HA] ≈ 0.01 M (since x is small).

Step 4: Substitute into the K_a expression.

K_a ≈ \(\frac{(10^{-4})(10^{-4})}{0.01}\)

K_a = \(\frac{10^{-8}}{10^{-2}}\) = 10^-6

Therefore, K_a is \(1.0 \times 10^{-6}\).

The correct choice is \(1.0 \times 10^{-6}\).

Answer 2

Given: \[ \text{pH} = 4 \Rightarrow [\text{H}^+] = 10^{-4} \ \text{mol/L}, \quad C = 0.01 \ \text{mol/L} \] For a weak acid HA: \[ \text{Ka} = \frac{[\text{H}^+]^2}{C} = \frac{(10^{-4})^2}{0.01} = \frac{10^{-8}}{10^{-2}} = 10^{-6} \] Answer: $\boxed{1.0 \times 10^{-6}}$