Answer:[tex]1.8\times 10^{-5}[/tex]
Explanation:
Concentration of [tex]CH_3COOH[/tex] = [tex]2.0\times 10^{-1}[/tex]
Concentration of [tex]CH_3COO^-[/tex] = [tex]1.9\times 10^{-3}[/tex]
Concentration of [tex]H_3O^+[/tex] = [tex]1.9\times 10^{-3}[/tex]
The balanced equilibrium reaction will be,
[tex]CH_3COOH+H_2O\rightleftharpoons CH_3COO^-+H_3O^+[/tex]
The expression for equilibrium reaction will be,
[tex]K=\frac{[H_3O^+]\times [CH_3COO^-]}{[CH_3COOH]}[/tex]
Now put all the given values in this expression
[tex]K=\frac{(1.9\times 10^{-3})\times (1.9\times 10^{-3})}{(2.0\times 10^{-1})}[/tex]
[tex]K=1.8\times 10^{-5}[/tex]
Therefore, the value of Keq for this reaction is [tex]1.8\times 10^{-5}[/tex]
