Question

At 25\(^\circ\)C, the percentage of ionization of 'x' M acetic acid is 4.242. What is the value of x? \((K_a = 1.8 \times 10^{-5})\)

• A. 0.05
• B. 0.04
• C. 0.02
• D. 0.01

Hint:

For weak acids, use the percent ionization formula involving \(K_a\) and concentration to find unknown values like molarity or ionization percentage.

Answer 1

The Correct Option is D

Step 1: Use the formula for percent ionization.
\[ % Ionization} = \frac{\sqrt{K_a \cdot C}}{C} \times 100 = \sqrt{\frac{K_a}{C}} \times 100 \] Given: \[ % Ionization} = 4.242, K_a = 1.8 \times 10^{-5} \] Step 2: Rearranging and solving for \(C = x\): \[ 4.242 = \sqrt{\frac{1.8 \times 10^{-5}}{x}} \times 100 \] \[ \frac{4.242}{100} = \sqrt{\frac{1.8 \times 10^{-5}}{x}} \Rightarrow 0.04242^2 = \frac{1.8 \times 10^{-5}}{x} \] \[ x = \frac{1.8 \times 10^{-5}}{(0.04242)^2} \approx 0.01 \]