## Chemistry: Atoms First (2nd Edition)

$[Initial HCOOH] = 0.02411M$
1. Find the $[H_3O^+]$ value: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.7}$ $[H_3O^+] = 1.995 \times 10^{- 3}$ 2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [HCOO^-] = x = 1.995 \times 10^{-3}$ -$[HCOOH] = [HCOOH]_{initial} - x$ 3. Now, use the Ka and x values and equation to find the initial concentration value. $Ka = \frac{[H_3O^+][HCOO^-]}{ [Initial HCOOH] - x}$ $1.8\times 10^{- 4}= \frac{[x^2]}{ [Initial HCOOH] - x}$ $1.8\times 10^{- 4}= \frac{( 1.995\times 10^{- 3})^2}{[Initial HCOOH] - 1.995\times 10^{- 3}}$ $[Initial HCOOH] - 1.995\times 10^{- 3} = \frac{ 3.981\times 10^{- 6}}{ 1.8\times 10^{- 4}}$ $[Initial HCOOH] - 1.995\times 10^{- 3} = 0.02212$ $[Initial HCOOH] = 0.02212 + 1.995\times 10^{- 3}$ $[Initial HCOOH] = 0.02411$