Question

n the laboratory, a general chemistry student measured the pH of a 0.342 M aqueous solution of ethylamine, C2H5NH2 to be 12.067. Use the information she obtained to determine the Kb for this base.

Answer 1

Answer:

The value of $K_b$ of the an ethylamine is $4.121\times 10^{-4}$.

Explanation:

The pH of the solution = 12.067

The pOH of the solution = 14 - pH =14-12.607 =1.933

$pOH=-\log[OH^-]$

$1.933=-\log[OH^-]$

$[OH^-]=0.0117 M$

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

Initially

0.342 M                              0     0

At equilibrium

(0.342-x)                             x      x

The value of x = $[OH^-]=0.0117 M$

The expression of $K_b$is given as:

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$

$K_b=\frac{x^2}{(0.342-x)}$

$K_b=\frac{(0.0117 )^2}{(0.342-0.0117)}=4.121\times 10^{-4}$

The value of $K_b$ of the an ethylamine is $4.121\times 10^{-4}$.