What is K_b for the benzoate ion, given that the K_a for benzoic acid is 6.46 $\times$ 10^-5?

Question

Quizplus · Accepted Answer

The $K_{a}$ for benzoic acid is $6.46 \times 10^{-5}$, and the $K_{w}$ (water ionization constant) is $1 \times 10^{-14}$. The $K_{b}$ for the benzoate ion can be found using the relationship $K_{w} = K_{a} \times K_{b}$. Therefore, $K_{b} = \frac{K_{w}}{K_{a}} = \frac{1 \times 10^{-14}}{6.46 \times 10^{-5}} = 1.55 \times 10^{-10}$.

What Is Kb for the Benzoate Ion, Given That the Ka

What Is K_b for the Benzoate Ion, Given That the K_a