Question

Calculate the \([\text{CH}_{3}\text{COO}^{-}]\) in \(0.1\)M solution of acetic acid if \(K_{a}\) for acetic acid is \(1.8\times 10^{-5}\).

• A. \(1.34\times 10^{-3}\ \text{M}\)
• B. \(1.68\times 10^{-3}\ \text{M}\)
• C. \(2.12\times 10^{-3}\ \text{M}\)
• D. \(2.34\times 10^{-3}\ \text{M}\)

Hint:

For a weak acid the ion concentration is the square root of Ka times the concentration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Acetic acid is a weak acid: \(\text{CH}_{3}\text{COOH}\rightleftharpoons\text{CH}_{3}\text{COO}^{-}+\text{H}^{+}\).
For small dissociation, \(K_a=C\alpha^{2}\).

Step 2: Set up the Expression
\([\text{CH}_{3}\text{COO}^{-}]=C\alpha=\sqrt{K_aC}\)

Step 3: Calculation
\(=\sqrt{1.8\times 10^{-5}\times 0.1}=\sqrt{1.8\times 10^{-6}}=1.34\times 10^{-3}\ \text{M}\)

Step 4: Compare the options
This matches option (A).

Final Answer:
The acetate ion concentration is 1.34 times 10 to the power minus 3 M. \[ \boxed{\text{(A) }1.34\times 10^{-3}\ \text{M}} \]