Question

The dissociation constant of weak acid HA is \(1.5\times 10^{-5}\). Find the percent dissociation, containing \(0.2\) moles per \(2\) liters of solution.

• A. \(1.22\%\)
• B. \(1.18\%\)
• C. \(1.14\%\)
• D. \(1.26\%\)

Hint:

Use C = 0.1 M and alpha = sqrt(Ka/C), then multiply by 100.

Answer 1

The Correct Option is A

Step 1: Set up the equilibrium:
Let the acid be at $0.1$ M, because $0.2$ mol sits in $2$ L. If a fraction $\alpha$ ionises, then $[H^+] = [A^-] = 0.1\alpha$ and $[HA] \approx 0.1$.

Step 2: Write the constant:
\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(0.1\alpha)^2}{0.1} = 0.1\alpha^2 \]

Step 3: Solve:
\[ \alpha^2 = \frac{1.5\times 10^{-5}}{0.1} = 1.5\times 10^{-4} \]
\[ \alpha = 0.01225 \]

Step 4: Read off the percentage:
$0.01225 \times 100 = 1.225$, which rounds to $1.22\%$. The other choices (1.18, 1.14, 1.26) do not match this value.

Final Answer:
So the acid is about 1.22 percent ionised. \[ \boxed{1.22\%} \]