Chemistry (4th Edition)

Original acid molarity : $2.329 \times 10^{-3}M$
1. Find $[H_3O^+]:$ $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.26}$ $[H_3O^+] = 5.495 \times 10^{- 4}M$ 2. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [HCOO^-] = x$ -$[HCOOH] = [HCOOH]_{initial} - x$ 3. Now, use the Ka and x values and equation to find the initial concentration value. $Ka = \frac{[H_3O^+][HCOO^-]}{ [Initial HCOOH] - x}$ $1.7\times 10^{- 4}= \frac{[x^2]}{ [Initial HCOOH] - x}$ $1.7\times 10^{- 4}= \frac{( 5.5\times 10^{- 4})^2}{[Initial HCOOH] - 5.5\times 10^{- 4}}$ $[Initial HCOOH] - 5.5\times 10^{- 4} = \frac{ 3.025\times 10^{- 7}}{ 1.7\times 10^{- 4}}$ $[Initial HCOOH] - 5.5\times 10^{- 4} = 1.779\times 10^{- 3}$ $[Initial HCOOH] = 1.779\times 10^{- 3} + 5.5\times 10^{- 4}$ $[Initial HCOOH] = 2.329\times 10^{- 3}M$