Answer
$Ka = 1.3 \times 10^{-5}$
Work Step by Step
1. Find the concentration of the acid:
$C(M) = \frac{n(moles)}{volume(L)} = \frac{0.275}{0.625} = 0.44M$
2. Calculate the Ka:
** Remember that: (Drawing the ICE table:)
$[H_3O^+] = [A^-] = x = 0.00239M$
and
$[Acid] = 0.44M -x = 0.44M - 0.00239M = 0.438M$
$Ka = \frac{[H_3O^+][A^-]}{[HA]}$
$Ka = \frac{0.00239 * 0.00239}{0.438}$
$Ka = 1.3 \times 10^{-5}$
