# Chapter 16 - Acids and Bases - Exercises - Percent Ionization - Page 740: 40

(a) 0.0225 (b) 2.25%

#### Work Step by Step

* Kb for ethylamine = $4.3 \times 10^{-4}$ 1. Drawing the ICE table we get: -$[OH^-] = [B^+] = x$ -$[Base] = [Base]_{initial} - x$ For approximation, we consider: $[Base] = [Base]_{initial}$ 2. Now, use the Kb value and equation to find the 'x' value. $Kb = \frac{[OH^-][B^+]}{ [Base]}$ $Kb = 4.3 \times 10^{- 4}= \frac{x * x}{ 8.5 \times 10^{- 1}}$ $Ka = 4.3 \times 10^{- 4}= \frac{x^2}{ 8.5 \times 10^{- 1}}$ $3.65 \times 10^{- 4} = x^2$ $x = 1.91 \times 10^{- 2}$ - Degree of ionization: $\frac{ 1.91 \times 10^{- 2}}{ 8.5 \times 10^{- 1}} = 0.0225$ - Percent ionization: 0.0225 * 100% = 2.25%

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