Respuesta :
Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
Explanation:
[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]
Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]
The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
[tex]K_a[/tex] for [tex]CH_3COOH[/tex] :
[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]
[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]
[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]
