For the reaction of deprotonation of formic acid, the concentration of HCOO⁻ at equilibrium is 0.0151 M if the initial concentration of formic acid is 1.35 M.
The reaction of deprotonation of formic acid is the following:
CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺
At the equilibrium we have:
CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺ (1)
1.35 - x x x
The acid equilibrium constant for this reaction is:
[tex] K_{a} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = 1.7 \cdot 10^{-4} [/tex] (2)
Entering the values of [CHOOH] = 1.35-x, [HCOO⁻] = [H₃O⁺] = x, into equation (2) we have:
[tex] 1.7 \cdot 10^{-4} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = \frac{x^{2}}{(1.35 - x)} [/tex]
[tex]1.7 \cdot 10^{-4}(1.35 - x) - x^{2} = 0[/tex]
After solving the above quadratic equation and taking the positive value for x (concentrations cannot be negative), we have:
[tex] x = [HCOO^{-}] = [H_{3}O^{+}] = 0.0151 M [/tex]
Therefore, the concentration of HCOO⁻ at equilibrium is 0.0151 M.
Learn more about the equilibrium constant here:
- https://brainly.com/question/7145687?referrer=searchResults
- https://brainly.com/question/9173805?referrer=searchResults
I hope it helps you!
