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 calculate the dissociation constant (K_a) for the weak acid HA, given the solution's pH, perform the following steps:

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

With a pH of 4:

pH = -log[H⁺]

[H⁺] = 10^-pH = 10^-4 M

Step 2: State the equilibrium expression for HA dissociation.

HA ⇌ H⁺ + A^-

Step 3: Formulate the K_a expression using concentrations.

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

If x M of HA dissociates, then equilibrium concentrations are:

[H⁺] = x = 10^-4 M

[A^-] = x = 10^-4 M

[HA] = Initial [HA] - x

Given an initial [HA] of 0.01 M, the remaining [HA] is approximately 0.01 M because x is small.

Step 4: Substitute values into the K_a expression.

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

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

The dissociation constant, K_a, is therefore \(1.0 \times 10^{-6}\).

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