Answer
$[HCOO^-] = 0.77M$
Work Step by Step
1. Calculate $[H_3O^+]$:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.06}$
$[H_3O^+] = 8.7 \times 10^{- 5}M$
2. Write the $K_a$ equation, and find the $HCOO^-$ concentration:
$K_a = \frac{[H_3O^+][HCOO^-]}{[HCOOH]}$
$1.8 \times 10^{-4} = \frac{8.7 \times 10^{-5}*[HCOO^-]}{0.366}$
$\frac{1.8 \times 10^{-4}}{8.7 \times 10^{-5}} = \frac{[HCOO^-]}{0.366}$
$2.1 * 0.366 = [HCOO^-]$
$[HCOO^-] = 0.77M$
