# A solution of formic acid has a pH of 2.70, calculate the initial concentration of formic acid?

## ${K}_{a} = 1.8 \times {10}^{-} 4$ for formic acid My work: Jul 20, 2018

${\underbrace{\left[H C \left(= O\right) O H\right] = 0.0241 \cdot m o l \cdot {L}^{-} 1}}_{\text{INITIALLY}}$

#### Explanation:

$H C \left(= O\right) O H \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s H C {O}_{2}^{-} + {H}_{3} {O}^{+}$

And at equilibrium...${K}_{a} = \frac{\left[H C {O}_{2}^{-}\right] \left[{H}_{3} {O}^{+}\right]}{\left[H C \left(= O\right) O H \left(a q\right)\right]} = 1.80 \times {10}^{-} 4$

But $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = 2.70$...

And so $\left[{H}_{3} {O}^{+}\right] = \left[H C {O}_{2}^{-}\right] = {10}^{- 2.70} \cdot m o l \cdot {L}^{-} 1 =$

And so AT EQUILIBRIUM...$1.80 \times {10}^{-} 4 = \frac{{\left\{{10}^{- 2.70}\right\}}^{2}}{\left[H C \left(= O\right) O H \left(a q\right)\right]}$

${\left[H C \left(= O\right) O H \left(a q\right)\right]}_{\text{at equilibrium}} = \frac{{\left\{{10}^{- 2.70}\right\}}^{2}}{1.80 \times {10}^{-} 4}$

$\equiv 0.0221 \cdot m o l \cdot {L}^{-} 1$...but this is the equilibrium value....and $\left[{H}_{3} {O}^{+}\right] = 1.995 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$ by definition...

And since the hydronium ion is PRESUMED to derive from the formic acid...${\left[H C \left(= O\right) O H\right]}_{\text{initially}} = \left(0.0221 + 1.995 \times {10}^{- 3}\right) \cdot m o l \cdot {L}^{-} 1 = 0.0241 \cdot m o l \cdot {L}^{-} 1$...the degree of dissociation was miniscule....

Pleas check my 'rithmetik...