Question

What is the base dissociation constant for a weak base at equilibrium, B + H2O <===> BH+ + OH–?

Answer 1

Answer: The base dissociation constant for the equation is $K_b=\frac{[BH^=][OH^-]}{[B]}$

Explanation:

Base dissociation constant, $K_b$ exists when a weak base is dissolved in water. It is expressed as the ratio of molar concentration of the products and the molar concentration of the reactants raised to power their respective stoichiometric coefficients.

For the dissociation of a weak base, the equation follows:

$B+H_2O\rightarrow BH^++OH^-$

The equilibrium constant for the above equation:

$K_{eq}=\frac{[BH^=][OH^-]}{[B][H_2O]}$

Concentration of water is very large and is taken as constant.

$K_{eq}\times [H_2O]=\frac{[BH^=][OH^-]}{[B]}$

Hence, the equation becomes:

$K_b=\frac{[BH^=][OH^-]}{[B]}$

Answer 2

Depends on what the base is. You would reference the base dissociation chart for that value.