Answer:
The value of [tex]K_b[/tex] of the an ethylamine is [tex]4.121\times 10^{-4}[/tex].
Explanation:
The pH of the solution = 12.067
The pOH of the solution = 14 - pH =14-12.607 =1.933
[tex]pOH=-\log[OH^-][/tex]
[tex]1.933=-\log[OH^-][/tex]
[tex][OH^-]=0.0117 M[/tex]
[tex]C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-[/tex]
Initially
0.342 M 0 0
At equilibrium
(0.342-x) x x
The value of x = [tex][OH^-]=0.0117 M[/tex]
The expression of [tex]K_b[/tex]is given as:
[tex]K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}[/tex]
[tex]K_b=\frac{x^2}{(0.342-x)}[/tex]
[tex]K_b=\frac{(0.0117 )^2}{(0.342-0.0117)}=4.121\times 10^{-4}[/tex]
The value of [tex]K_b[/tex] of the an ethylamine is [tex]4.121\times 10^{-4}[/tex].
