What is the hydroxide-ion concentration at equilibrium in a 0.66 M solution of ethylamine (C₂H₅NH₂,K_b = 4.7 × 10^-4)at 25^oC? A)1.7 × 10^-2 M B)2.7 × 10^-2 M C)1.5 × 10^-14 M D)5.8 × 10^-13 M E)6.6 × 10^-1 M

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Quizplus · Accepted Answer

Ethylamine is a weak base, and its ionization in water can be represented as: $$ C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^- $$Given the base dissociation constant ($K_b$) for ethylamine is $4.7 \times 10^{-4}$, we can set up an expression for $K_b$ as follows: $$ K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]_{initial} - [OH^-]} $$Assuming $x$ is the concentration of $OH^-$ at equilibrium, the equation simplifies to: $$ 4.7 \times 10^{-4} = \frac{x^2}{0.66 - x} $$Given that $K_b$ is relatively small, we can assume $x$ is much smaller than $0.66$, simplifying the equation to: $$ 4.7 \times 10^{-4} \approx \frac{x^2}{0.66} $$Solving for $x$, we find: $$ x^2 = 4.7 \times 10^{-4} \times 0.66 $$ $$ x^2 = 3.102 \times 10^{-4} $$ $$ x \approx 1.76 \times 10^{-2} $$Therefore, the concentration of $OH^-$ at equilibrium is approximately $1.7 \times 10^{-2} M$, which corresponds to choice A.

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