# Chapter 13 - Exercises - Page 575d: 76

$Ka \approx 0.03333$

#### Work Step by Step

1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x = 0.30M$ $[Acid]_{initial} = 0.30 + x$ 2. Write the percent dissociation equation, and find 'x': %dissociation = $\frac{x}{0.30 + x} \times 100$ 25= $\frac{x}{ 0.30 + x} \times 100$ 0.25= $\frac{x}{ 0.30 + x}$ $0.075 + 0.25x = x$ $0.075 = x - 0.25x = 0.75x$ $\frac{0.075}{0.75} = x$ $x = 0.1M$ Therefore: $[Conj. Base] and [H_3O^+] = 0.1M$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{0.30}$ $Ka = \frac{( 0.1)^2}{ 0.30}$ $Ka = \frac{ 0.01}{ 0.30}$ $Ka = 0.03333$

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