the acid - dissociation constant, $k_a$, for benzoic acid is $6.5\times10^{-5}$. which will you use to…

the acid - dissociation constant, $k_a$, for benzoic acid is $6.5\times10^{-5}$. which will you use to calculate the base - dissociation constant, $k_b$, for the conjugate base of benzoic acid?\n$\frac{k_a}{k_b}=k_w$\n$k_a\times k_b = k_w$\n$k_a+k_b = k_w$\n$\frac{k_b}{k_a}=k_w$\ndone

the acid - dissociation constant, $k_a$, for benzoic acid is $6.5\times10^{-5}$. which will you use to calculate the base - dissociation constant, $k_b$, for the conjugate base of benzoic acid?\n$\frac{k_a}{k_b}=k_w$\n$k_a\times k_b = k_w$\n$k_a+k_b = k_w$\n$\frac{k_b}{k_a}=k_w$\ndone

Answer

Explanation:

Step1: Recall the relationship formula

The relationship between the acid - dissociation constant ($K_a$), base - dissociation constant ($K_b$) and the ion - product constant of water ($K_w$) is $K_a\times K_b = K_w$. This formula is used to calculate the $K_b$ of a conjugate base when the $K_a$ of the acid is known.

Answer:

$K_a\times K_b = K_w$