Answer
(a) $[H_3O^+] = 1.585 \times 10^{- 4}M$
(b) This acid is moderately weak.
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x$
2. Calculate the hydronium concentration.
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 3.8}$
$[H_3O^+] = 1.585 \times 10^{- 4}M$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[InitialAcid] - x}$
$Ka = \frac{( 1.585\times 10^{- 4})^2}{ 0.0025- 1.585\times 10^{- 4}}$
$Ka = \frac{ 2.512\times 10^{- 8}}{ 2.342\times 10^{- 3}}$
$Ka = 1.073\times 10^{- 5}$
- Therefore, it is a moderately weak acid.
