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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 \(4.545\times 10^{-5}\).

pstnonsonjoku

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

Learn more about Kb: https://brainly.com/question/1301963

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