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Question: What is \( Ka \) at \( 25^0C \) for the following equilibrium ? \( CH_3NH^{ 3+ }(aq.) + H_2O(l) \rig...

What is KaKa at 250C25^0C for the following equilibrium ? CH3NH3+(aq.)+H2O(l)→CH3NH2(aq.)+H3O+(aq.)CH_3NH^{ 3+ }(aq.) + H_2O(l) \rightarrow CH_3NH_2(aq.) + H_3O^+(aq.) Kb(CH3NH2)=4.4×10−4Kb(CH_3NH_2) = 4.4 \times 10^{ -4 } at 250C25^0C.
A.4.4×10−44.4 \times 10^{ -4 }
B.2.3×1032.3 \times 10^3
C.4.4×10−104.4 \times 10^{ -10 }
D.4.4×1044.4 \times 10^4
E.2.3×10−112.3 \times 10^{ -11 }

Explanation

Solution

For solving the given problem, we should have knowledge about salt hydrolysis, pHpH , acidic constant KaKa, and calculation of loglog.
Salt hydrolysis is a process where the reactants are salt and water forming an acidic or basic solution. pHpH 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, KaKa 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×Kb=Kw=10−14Ka \times Kb = Kw = 10^{ -14 }
Here, Ka→Ka \rightarrow acid dissociation constant
Kb→Kb \rightarrow Base dissociation constant
Kw→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.
CH3NH3+(aq.)+H2O(l)→CH3NH2(aq.)+H3O+(aq.)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 CH3NH2CH_3NH_2 is the conjugate base here and H3O+H_3O^+ is the conjugate acid. Here, base dissociation constant is 4.4×10−44.4 \times 10^{ -4 } .
Step-3 :
Using the given formula, we have :
Ka×Kb=KwKa \times Kb = Kw
and KaKa at 250C25^0C is .
So, Ka×Kb=10−14Ka \times Kb = 10^{ -14 }
Ka=10−144.4×10−4Ka = \dfrac { 10^{ -14 } }{ 4.4 \times 10^{ -4 } }
=2.27×10−11= 2.27 \times 10^{ -11 }
2.3×10−112.3 \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.