What is the percent ionization of a 0.300 M solution of formic acid (HCOOH), given that Ka = 1.78 × 10^{-4}?

• A. 0.24%
• B. 8.7%
• C. 24%
• D. 2.4%

Answer: (D)

Verified step by step guidance

1. Write the ionization equilibrium expression for formic acid (HCOOH): it partially ionizes in water as HCOOH \( \rightleftharpoons \) H\(^+\) + HCOO\(^-\).
2. Set up an ICE table (Initial, Change, Equilibrium) for the concentrations: Initial concentration of HCOOH is 0.300 M, and initial concentrations of H\(^+\) and HCOO\(^-\) are 0.
3. Define the change in concentration of HCOOH that ionizes as \( x \). At equilibrium, the concentration of HCOOH is \( 0.300 - x \), and the concentrations of H\(^+\) and HCOO\(^-\) are both \( x \).
4. Write the expression for the acid dissociation constant \( K_a \): \( K_a = \frac{{[\text{H}^+][\text{HCOO}^-]}}{{[\text{HCOOH}]}} = \frac{{x^2}}{{0.300 - x}} \)
5. Assuming \( x \) is small compared to 0.300, approximate the denominator as 0.300 and solve for \( x \). Then calculate the percent ionization using the formula: \( \text{Percent ionization} = \frac{x}{0.300} \times 100\% \)