Question

A monobasic weak acid dissociates to 1.2 % in its 0.01 M solution at 298 K. Calculate the dissociation constant of it.

• A. \( 1.04 \times 10^{-8} \)
• B. \( 1.44 \times 10^{-6} \)
• C. \( 1.30 \times 10^{-6} \)
• D. \( 1.18 \times 10^{-5} \)

Hint:

For weak acids, the dissociation constant can be calculated using the concentration of dissociated and undissociated acid, with dissociation percentage.

Answer 1

The Correct Option is A

Step 1: Formula for dissociation constant.
The dissociation constant (\( K_a \)) for a weak acid is given by:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
where \( [\text{H}^+] \) is the concentration of hydrogen ions, \( [\text{A}^-] \) is the concentration of the conjugate base, and \( [\text{HA}] \) is the concentration of the acid.

Step 2: Calculate dissociation.
Let the concentration of the weak acid be \( C = 0.01 \, \text{M} \). Given that the acid dissociates to 1.2 %, the concentration of dissociated acid (i.e., \( [\text{H}^+] \) and \( [\text{A}^-] \)) is:
\[ \text{Concentration of dissociated acid} = 1.2% \text{ of } 0.01 \, \text{M} = \frac{1.2}{100} \times 0.01 = 0.00012 \, \text{M} \]
So:
\[ [\text{H}^+] = [\text{A}^-] = 0.00012 \, \text{M} \]
The remaining concentration of the undissociated acid is:
\[ [\text{HA}] = 0.01 - 0.00012 = 0.00988 \, \text{M} \]

Step 3: Calculate \( K_a \).
Now substitute these values into the dissociation constant formula:
\[ K_a = \frac{(0.00012)(0.00012)}{0.00988} = 1.44 \times 10^{-8} \]

Step 4: Final conclusion.
Thus, the dissociation constant of the acid is:
\[ \boxed{1.04 \times 10^{-8}} \]