Question

The acetate ion is the conjugate base of the weak acid acetic acid. The value of Kb for CH3COO-, is 5.56×10-10. Write the equation for the reaction that goes with this equilibrium constant.

Answer 1

Answer: The equation for the reaction that goes with this equilibrium constant is $5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-+H^+$

Here $CH_3COOH$ donates a proton and thus behaves as an acid and forms $CH_3COO^-$ which is called as the conjugate base of $CH_3COOH$

The dissociation constant of acids is given by the term $K_a$ and the dissociation constant of bases is given by the term $K_b$ and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

$K_a$ for  $CH_3COOH$ :

$K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}$

$CH_3COO^-+H^+\rightarrow CH_3COOH$

$K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}$

$5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}$

The equation for the reaction that goes with this equilibrium constant is $K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}$