Answer
This concentration is $0.072M$
Work Step by Step
1. Write the percent ionization equation, and relate the initial acid concentration to 'x':
%ionization = $\frac{x}{[Initial Acid]} \times 100$
2.5= $\frac{x}{[Initial Acid]} \times 100$
0.025= $\frac{x}{ [Initial Acid]}$
$ [Initial Acid]= \frac{x}{0.025} = \frac{1}{0.025}x = 40x $
2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x$
For approximation, we consider: $[Acid] = [Acid]_{initial}$
3. Use the Ka equation to find "x":
$Ka = \frac{[H_3O^+][[Conj. Base]]}{ [[Acid]]}$
$Ka = 4.5 \times 10^{- 5}= \frac{x * x}{[Acid]}$
$Ka = 4.5 \times 10^{- 5}= \frac{x^2}{ 40x}$
$4.5 \times 10^{-5} = \frac{x}{40}$
$4.5 \times 10^{-5} * 40 = x$
$x = 1.8 \times 10^{-3}$
4. Calculate the initial acid concentration:
$[Initial Acid] = 40 * x = 40 * 1.8 \times 10^{-3} = 0.072M$
