## Chemistry 9th Edition

The initial concentration of formic acid is equal to $0.024M$.
1. Calculate the hydronium ion concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 2.7}$ $[H_3O^+] = 2.0 \times 10^{- 3}M$ 1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $HCOOH(aq) + H_2O(l) \lt -- \gt HCOO^-(aq) + H_3O^+(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $HCOO^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [HCOO^-] = 0 + x = x$ ** Therefore: $x = 2.0 \times 10^{-3}M$ -$[HCOOH] = [HCOOH]_{initial} - x = [HCOOH]_{initial} - 2.0 \times 10^{-3}M$ For approximation, we are going to consider $[HCOOH]_{initial} = [HCOOH]$ 2. Now, use the Ka value and equation to find the initial $[HCOOH]$ value. $Ka = \frac{[H_3O^+][HCOO^-]}{ [HCOOH]}$ $1.8 \times 10^{- 4}= \frac{x * x}{ [HCOOH]_{initial} - x}$ $1.8 \times 10^{- 4}= \frac{(2.0 \times 10^{-3})^2}{ [HCOOH]_{initial} - 2.0 \times 10^{-3}}$ ** Solve for "$[HCOOH]_{initial}$" $[HCOOH]_{initial} - 2.0 \times 10^{-3} = \frac{(2.0 \times 10^{-3})^2}{1.8 \times 10^{-4}}$ $[HCOOH]_{initial} = \frac{(2.0 \times 10^{-3})^2}{1.8 \times 10^{-4}} + 2.0 \times 10^{-3}$ $[HCOOH]_{initial} = 0.024M$