Question

The percentage of ionization of \( 1 \, \text{L} \) of \( x \, \text{M} \) acetic acid is \( 4.242\% \) and is called solution "A". The percentage of ionization of \( 1 \, \text{L} \) of \( y \, \text{M} \) acetic acid is \( 3\% \) and is called solution "B". Solution "A" is mixed with solution "B". What is the concentration of acetic acid in the resultant solution? (\( K_a \) of acetic acid = \( 1.8 \times 10^{-5} \))

• A. $ 0.05 \, \text{M} $
• B. $ 0.015 \, \text{M} $
• C. $ 0.02 \, \text{M} $
• D. $ 0.15 \, \text{M} $

Hint:

When mixing solutions, the total moles of solute are conserved. Use the formula for percentage ionization to relate concentration and dissociation constant.

Answer 1

The Correct Option is B

Step 1: Known Information.
Solution "A":
Concentration of acetic acid: \( x \, M \)
Percentage of ionization: \( 4.242\% \)
Volume: \( 1 \, L \)
Solution "B":
Concentration of acetic acid: \( y \, M \)
Percentage of ionization: \( 3\% \)
Volume: \( 1 \, L \)
Acetic acid dissociation constant (\( K_a \)): \( 1.8 \times 10^{-5} \)
Step 2: Relationship Between Ionization Percentage and Concentration.
For a weak acid like acetic acid, the percentage of ionization (\( \alpha \)) is related to the concentration (\( C \)) and the acid dissociation constant (\( K_a \)) by the formula: $$ \alpha = \sqrt{\frac{K_a}{C}} $$ Step 3: Calculate \( x \) (Concentration of Solution "A").
For solution "A":
Percentage of ionization (\( \alpha_A \)) = \( 4.242\% = 0.04242 \)
Using the formula:
$$ \alpha_A = \sqrt{\frac{K_a}{x}} $$ Substitute \( K_a = 1.8 \times 10^{-5} \): $$ 0.04242 = \sqrt{\frac{1.8 \times 10^{-5}}{x}} $$ Square both sides: $$ (0.04242)^2 = \frac{1.8 \times 10^{-5}}{x} $$ $$ 0.001799 = \frac{1.8 \times 10^{-5}}{x} $$ Solve for \( x \): $$ x = \frac{1.8 \times 10^{-5}}{0.001799} \approx 0.01 \, M $$ Step 4: Calculate \( y \) (Concentration of Solution "B").
For solution "B":
Percentage of ionization (\( \alpha_B \)) = \( 3\% = 0.03 \)
Using the formula: $$ \alpha_B = \sqrt{\frac{K_a}{y}} $$ Substitute \( K_a = 1.8 \times 10^{-5} \): $$ 0.03 = \sqrt{\frac{1.8 \times 10^{-5}}{y}} $$ Square both sides: $$ (0.03)^2 = \frac{1.8 \times 10^{-5}}{y} $$ $$ 0.0009 = \frac{1.8 \times 10^{-5}}{y} $$ Solve for \( y \): $$ y = \frac{1.8 \times 10^{-5}}{0.0009} \approx 0.02 \, M $$ Step 5: Calculate the Resultant Concentration.
When solutions "A" and "B" are mixed:
Volume of each solution: \( 1 \, L \)
Total volume of the mixture: \( 1 \, L + 1 \, L = 2 \, L \)
Total moles of acetic acid in the mixture: $$ \text{Moles of acetic acid from A} = x \times 1 = 0.01 \, \text{mol} $$ $$ \text{Moles of acetic acid from B} = y \times 1 = 0.02 \, \text{mol} $$ $$ \text{Total moles of acetic acid} = 0.01 + 0.02 = 0.03 \, \text{mol} $$ Concentration of acetic acid in the resultant solution: $$ \text{Concentration} = \frac{\text{Total moles of acetic acid}}{\text{Total volume}} = \frac{0.03 \, \text{mol}}{2 \, \text{L}} = 0.015 \, M $$ Final Answer: \( \boxed{0.015 \, M} \)