use the references to access important values if needed for this question.\nin the laboratory, a general…

use the references to access important values if needed for this question.\nin 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.\nuse the information she obtained to determine the ( k_{b} ) for this base.\n( k_{b}(\text{experiment}) = )

use the references to access important values if needed for this question.\nin 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.\nuse the information she obtained to determine the ( k_{b} ) for this base.\n( k_{b}(\text{experiment}) = )

Answer

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\rightleftharpoons 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})