Answer:
10.64
Explanation:
Let's consider the basic reaction of cyclohexamine, C₆H₁₁NH₂.
C₆H₁₁NH₂(aq) + H₂O(l) ⇄ C₆H₁₁NH₃⁺(aq) + OH⁻ pKb = 3.36
C₆H₁₁NH₃⁺ is its conjugate acid, since it donates H⁺ to form C₆H₁₁NH₂. C₆H₁₁NH₃⁺ acid reaction is as follows:
C₆H₁₁NH₃⁺(aq) + H₂O(l) ⇄ C₆H₁₁NH₂(aq) + H₃O⁺(aq) pKa
We can find the pKa of C₆H₁₁NH₃⁺ using the following expression.
pKa + pKb = 14.00
pKa = 14.00 - pKb = 14.00 - 3.36 = 10.64
The pKa of conjugate acid of cyclohexamine has been 10.64.
The pKa has been acid dissociation constant, and has been describing the hydrogen ion concentration in the sample.
The pKb has been the base dissociation constant and has been describing the hydroxide ion concentration in the sample.
The sum of pKa and pKb has been found to be 14.
[tex]\rm pKa \;+\;pKb=14[/tex]
The pKb of cyclohexamine has been 3.36. The pKa of conjugate acid in the sample has been given by substituting the values as:
[tex]\rm pKa\;+\;3.36=14\\pKa=14\;-\;3.36\\pKa=10.64[/tex]
The pKa of conjugate acid of cyclohexamine has been 10.64.
For more information about pKa and pKb, refer to the link:
https://brainly.com/question/12639696
