#### Answer

$Ka= 2.603 \times 10^{-3}$

#### Work Step by Step

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:
$Ka= \frac{[H3O+][A−]}{[HA]}$
$Ka= \frac{0.0275∗0.0275}{0.2905}$
$Ka= 2.603 \times 10^{-3}$