Question

use the references to access important values if needed for this question.
in the laboratory, a general chemistry student measured the ph of a 0.413 m aqueous solution of ethylamine, ( c_{2}h_{5}nh_{2} ) to be 12.140.
use the information she obtained to determine the ( k_{b} ) for this base.
( k_{b}(\text{experiment}) = )

Explanation:

Step1: Calculate \(pOH\)

We know that \(pH + pOH=14\). Given \(pH = 12.140\), then \(pOH=14 - pH\)
\(pOH=14 - 12.140=1.86\)

Step2: Calculate \([OH^{-}]\)

Since \(pOH=-\log[OH^{-}]\), then \([OH^{-}]=10^{-pOH}\)
\([OH^{-}]=10^{- 1.86}\approx1.38\times10^{-2}\space M\)

Step3: Set up the equilibrium expression for ethylamine (\(C_{2}H_{5}NH_{2}\))

The reaction of ethylamine with water is \(C_{2}H_{5}NH_{2}+H_{2}O
ightleftharpoons C_{2}H_{5}NH_{3}^{+}+OH^{-}\)
Let \(c = 0.413\space M\) (initial concentration of \(C_{2}H_{5}NH_{2}\)). At equilibrium, \([C_{2}H_{5}NH_{3}^{+}]=[OH^{-}]\) (from the stoichiometry of the reaction) and \([C_{2}H_{5}NH_{2}]=c-[OH^{-}]\)
Since \(c = 0.413\space M\) and \([OH^{-}]=1.38\times 10^{-2}\space M\), and \(c\gg[OH^{-}]\) (because \(0.413\gg1.38\times 10^{-2}\)), we can approximate \([C_{2}H_{5}NH_{2}]\approx c\)

The base - dissociation constant \(K_{b}=\frac{[C_{2}H_{5}NH_{3}^{+}][OH^{-}]}{[C_{2}H_{5}NH_{2}]}\)

Substituting \([C_{2}H_{5}NH_{3}^{+}]=[OH^{-}]=1.38\times 10^{-2}\space M\) and \([C_{2}H_{5}NH_{2}]\approx0.413\space M\) into the \(K_{b}\) formula

\(K_{b}=\frac{(1.38\times 10^{-2})(1.38\times 10^{-2})}{0.413}\)

\(K_{b}=\frac{1.9044\times 10^{-4}}{0.413}\approx4.61\times 10^{-4}\)

Answer:

\(4.61\times 10^{-4}\)