Question

What is $ Ka $ at $ 25^0C $ for the following equilibrium ? $ CH_3NH^{ 3+ }(aq.) + H_2O(l) \rightarrow CH_3NH_2(aq.) + H_3O^+(aq.) $ $ Kb(CH_3NH_2) = 4.4 \times 10^{ -4 } $ at $ 25^0C $.
A.$ 4.4 \times 10^{ -4 } $
B.$ 2.3 \times 10^3 $
C.$ 4.4 \times 10^{ -10 } $
D.$ 4.4 \times 10^4 $
E.$ 2.3 \times 10^{ -11 } $

Answer 1

Hint:
For solving the given problem, we should have knowledge about salt hydrolysis, $ pH $ , acidic constant $ Ka $, and calculation of $ log $.
Salt hydrolysis is a process where the reactants are salt and water forming an acidic or basic solution. $ pH $ is described as power of hydrogen or potential of hydrogen. It is used to describe the concentration of hydrogen and is inversely proportional to it.
Acidic constant or Acid Dissociation constant, $ Ka $ is used to quantitatively measure the acidic strength in a solution.

Formula Used :
To calculate the acid dissociation constant in the following reaction, the formula used are :
$ Ka \times Kb = Kw = 10^{ -14 } $
Here, $ Ka \rightarrow $ acid dissociation constant
$ Kb \rightarrow $ Base dissociation constant
$ Kw \rightarrow $ Autolysis water dissociation constant

Complete step by step answer:
Step-1 :
First, we have to look forward to the reaction and find out about the conjugate acid and base here.
$ CH_3NH^{ 3+ }(aq.) + H_2O(l) \rightarrow CH_3NH_2(aq.) + H_3O^+(aq.) $
When base in the reactant side gains proton in the product side, it is known as conjugate acid and vice versa.
Step-2 :
We here know that $ CH_3NH_2 $ is the conjugate base here and $ H_3O^+ $ is the conjugate acid. Here, base dissociation constant is $ 4.4 \times 10^{ -4 } $ .
Step-3 :
Using the given formula, we have :
$ Ka \times Kb = Kw $
and $ Ka $ at $ 25^0C $ is .
So, $ Ka \times Kb = 10^{ -14 } $
$ Ka = \dfrac { 10^{ -14 } }{ 4.4 \times 10^{ -4 } } $
$ = 2.27 \times 10^{ -11 } $
$ 2.3 \times 10^{ -11 } $

So, option (D) is correct.

Note:
While solving the above question, remember the bronsted and lowry concept of acid-base and find the conjugate acid and base correctly. Those compounds that lose protons are acids or which gain protons are bases. An acid after losing a proton from a conjugate base and a base after gaining a proton from conjugate acid.