## General Chemistry: Principles and Modern Applications (10th Edition)

$Ka = 1.3 \times 10^{-5}$
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 $[Acid] = 0.44M -x = 0.44M - 0.00239M = 0.438M$ $Ka = \frac{[H_3O^+][A^-]}{[HA]}$ $Ka = \frac{0.00239 * 0.00239}{0.438}$ $Ka = 1.3 \times 10^{-5}$