Question

question 4 of 10
what is $k_b$ for $c_5h_5n(aq)+h_2o(l)
ightleftharpoons c_5h_5nh^+(aq)+oh^-(aq)$?
a. $k_b = \frac{c_5h_5nh^+}{c_5h_5n}$
b. $k_b = c_5h_5nh^+oh^-$
c. $k_b = \frac{c_5h_5nh^+oh^-}{c_5h_5n}$
d. $k_b = \frac{c_5h_5n}{c_5h_5nh^+oh^-}$

Explanation:

Step1: Recall base - dissociation constant formula

The base - dissociation constant ($K_b$) for a weak base $B$ reacting with water $B + H_2O
ightleftharpoons BH^++OH^-$ is given by $K_b=\frac{[BH^+][OH^-]}{[B]}$.

Step2: Identify the base and its products in the given reaction

In the reaction $C_5H_5N(aq)+H_2O(l)
ightleftharpoons C_5H_5NH^+(aq) + OH^-(aq)$, $C_5H_5N$ is the base ($B$), $C_5H_5NH^+$ is the conjugate acid ($BH^+$), and $OH^-$ is the hydroxide ion.

Answer:

C. $\frac{[C_5H_5NH^+][OH^-]}{[C_5H_5N]}$