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

$Ka= 2.603 \times 10^{-3}$
1. Calculate $[H^+]$ $[H^+] = 10^{-pH}$ $[H^+] = 10^{- 1.56}$ $[H^+] = 0.0275$ 2. Drawing the ICE table, we get: $[H3O+]=[A−]=x= 0.0275M$ and $[Acid]=0.318M − x= 0.318M − 0.0275M = 0.2905M$ 3. Calculate the Ka: $Ka= \frac{[H3O+][A−]}{[HA]}$ $Ka= \frac{0.0275∗0.0275}{0.2905}$ $Ka= 2.603 \times 10^{-3}$