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

Step3: Set up the equilibrium expression for ethylamine

Ethylamine \((C_{2}H_{5}NH_{2})\) reacts with water as \(C_{2}H_{5}NH_{2}+H_{2}O
ightleftharpoons C_{2}H_{5}NH_{3}^{+}+OH^{-}\). Let \(x = [OH^{-}]=[C_{2}H_{5}NH_{3}^{+}]\) and the initial concentration of \(C_{2}H_{5}NH_{2}\) is \(c = 0.413\ M\). At equilibrium, \([C_{2}H_{5}NH_{2}]=c - x\approx c\) (because \(x\ll c\), \(0.413\gg1.38\times 10^{-2}\))

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

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

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

Step4: Calculate \(K_{b}\)

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

Answer:

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