Answer
The initial concentration of formic acid is equal to $0.024M$.
Work Step by Step
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 "$[HCOOH]_{initial}$"
$[HCOOH]_{initial} - 2.0 \times 10^{-3} = \frac{(2.0 \times 10^{-3})^2}{1.8 \times 10^{-4}}$
$[HCOOH]_{initial} = \frac{(2.0 \times 10^{-3})^2}{1.8 \times 10^{-4}} + 2.0 \times 10^{-3}$
$[HCOOH]_{initial} = 0.024M$
