Question

A solution of 0.18mol of the chlorine salt of protinated atropine (AH+), a weak organic base, in 1.0L of solution has pH = 5.20.

Find the Kb of atropine (A).

Answer 1

Answer:The $K_b$ of atropine is $4.545\times 10^{-5}$.

Explanation:

$AH^+\rightleftharpoons A+H^+$

$[AH^+]=\frac{0.18 mol}{1 L}=0.18 M$

$[A]=[H^+]$

The pH of the solution:

$5.20=-\log[H^+]$

$[H^+]=6.3\times 10^{-6} M$

The expression of$K_a$ for atropine is given as:

$K_a=\frac{[A][H^+]}{[AH^+]}$

$K_a=\frac{6.3\times 10^{-6} M\times 6.3\times 10^{-6} M}{0.18 M}=2.20\times 10^{-10}$

$K_a\times K_b=K_w$

$K_b=\frac{K_w}{K_a}=\frac{1\times 10^{-14}}{2.20\times 10^{-10}}=4.545\times 10^{-5}$

The $K_b$ of atropine is $4.545\times 10^{-5}$.

Answer 2

The Kb of the reaction is obtained as 2.2  * 10^-10.

Let the protonated salt be represented with the general formula BH^+. We can set up a reaction as follows;

BH^+(aq) + H2O(l) ----> H3O^+(aq) + B^-(aq)

We are told that the pH of the solution = 5.20

Hence; [ H3O^+]= Antilog (-5.20) = 6.3 * 10^-6 M

Also,  [ H3O^+] = [ B^-]= 6.3 * 10^-6 M

Kb = [ H3O^+] [ B^-]/BH^+

[BH^+] =  0.18mol / 1.0L   = 0.18M

Kb = (6.3 * 10^-6)^2/0.18

Kb = 2.2  * 10^-10

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